Traction loss in a vehicle's rear wheels is most likely due to either acceleration-induced traction loss or rear wheel traction loss (skid). The former is caused by rapid acceleration, while the latter can be due to turning or braking.
Explanation:The traction loss described in your question occurs in the rear wheels of a vehicle and is likely due to acceleration-induced traction loss or rear wheel traction loss (skid). Acceleration-induced traction loss happens when rapid acceleration causes the tires to lose grip on the road. This is common in high-powered, rear-wheel drive vehicles. Rear wheel traction loss or a rear wheel skid, on the other hand, generally occurs when the rear wheels lose grip during turning or braking. In this scenario, the vehicle's rear end can swing out in either direction, causing the vehicle to spin.
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Final answer:
Rear wheel traction loss (skid) occurs when the rear tires lose grip on the road surface. In a front-wheel-driven car, friction at the rear wheel is in the opposite direction of motion. This concept is important for analysis in vehicle dynamics and skid prevention.
Explanation:
The traction loss referred to in the question occurs in the rear wheels of a vehicle and is specifically known as rear wheel traction loss (skid). This is a type of skid that happens when the rear tires of a vehicle lose grip on the road surface, often as a result of oversteering, sudden acceleration, or slippery conditions. It's essential to distinguish this from other types of traction loss, such as braking-induced traction loss, which occurs when the tires can no longer grip the road during heavy braking, or driver-induced skids, which are the result of driver error.
With respect to the problem provided, in a front-wheel-driven car, the friction at the rear wheel is generally in the opposite direction of motion. This is because the rear wheels are not powered in this configuration; instead, they follow the rotation induced by the car's movement, which is primarily driven by the front wheels. Therefore, while the front wheels are pulled forward by the engine, creating friction in the direction of motion, the rear wheels experience a kind of rolling resistance and the friction present there acts in the opposite direction of the car's movement.
Understanding the dynamics of friction and traction is critical for analyzing cases like where a vehicle skids and comes to rest after traveling a certain distance or determining the force required to maintain a constant speed in the presence of kinetic friction. These analyses involve concepts like travel reduction ratio, torque transferred to the wheel axle, tractive efficiency of the wheel, and the actual velocity of the wheel.
The velocity of a 200 kg object is changed from 5 m/s to 25 m/s in 50 seconds by an applied constant force.
a. What change in momentum does the force produced?
b. What is the magnitude of the force?
Explanation:
Force is given by rate of change of momentum.
Mass of object = 200 kg
Initial velocity = 5 m/s
Final velocity = 25 m/s
a) Change in momentum = 200 x 25 - 200 x 5
Change in momentum = 4000 kg m/s in the direction of motion
b) Time taken = 50 s
Rate of change of momentum = Change in momentum ÷ Time
Rate of change of momentum = 4000 ÷ 50
Rate of change of momentum = 80 N
Force = 80 N
Magnitude of the force is 80 N
The pathologic changes that occur in the development of coronary atherosclerotic lesions include call damage resulting from which of the following? (Select all that apply):
1. a decrease in smooth muscle cells2. a chronic calcium buildup3. the effects of oxidized lipids4. an inflammatory response5. the formation of plaques
Answer:
3 effect of oxidized lipids
4 an inflammatory response
5 the formation of plaques
Explanation:
Destruction of cells due to oxidation of lipids whereby free radicals steal electrons in cell membrane.
This occurs when tissues get injured by trauma, bacteria or toxins, thereby causing damages cells to release chemicals like histamine, brakykinn that cause vessels to leak fluid into the injured tissues causing swelling.
-plaques are regions of destroyed cells which are visible structures formed inside a cell culture.
An ideal gas is enclosed in a piston, and exactly 1200 J of work is done on the gas. As this happens, the internal energy of the gas increases by exactly 700 J. During this process, how much heat flows into or out of the ideal gas? Enter a positive number to indicate a heat flow into the gas or a negative number to indicate a heat flow out of the gas.
Answer:
[tex]dQ=-500J[/tex]
Explanation:
We have given work is done on the gas is 1200 J
So work done will be [tex]dW=-1200J[/tex] ( as work is done on the gas )
It is given that internal energy is increases by exactly 700 J
So [tex]dU=700J[/tex]
From thermodynamic equation [tex]dQ=dU+dW[/tex]
So [tex]dQ=700-1200=-500J[/tex]
Here negative sign indicates that heat flow out of the gas
If heat was negative then heal was flowing in the gas
A cue ball initially moving at 3.4 m/s strikes a stationary eight ball of the same size and mass. After the collision, the cue ball’s final speed is 0.94 m/s at an angle of θ with respect to its original line of motion?Find the eight ball’s speed after the col- lision. Assume an elastic collision (ignoring friction and rotational motion).
Answer in units of m/s.
Answer:
speed of eight ball speed after the collision is 3.27 m/s
Explanation:
given data
initially moving v1i = 3.4 m/s
final speed is v1f = 0.94 m/s
angle = θ w.r.t. original line of motion
solution
we assume elastic collision
so here using conservation of energy
initial kinetic energy = final kinetic energy .............1
before collision kinetic energy = 0.5 × m× (v1i)²
and
after collision kinetic energy = 0.5 × m× (v1f)² + 0.5 × m× (v2f)²
put in equation 1
0.5 × m× (v1i)² = 0.5 × m× (v1f)² + 0.5 × m× (v2f)²
(v2f)² = (v1i)² - (v1f)²
(v2f)² = 3.4² - 0.94²
(v2f)² = 10.68
taking the square root both
v2f = 3.27 m/s
speed of eight ball speed after the collision is 3.27 m/s
Final answer:
To find the eight ball's speed after the collision, we can use the principle of conservation of momentum and the fact that the collision is elastic. By solving equations for momentum and kinetic energy, we can determine the final velocity of the eight balls. Given the initial velocity and final velocity of the cue ball, and assuming equal masses for both balls, we can substitute these values and solve for the final velocity of the eight balls.
Explanation:
To find the eight ball's speed after the collision, we can use the principle of conservation of momentum. In an elastic collision, the total momentum before the collision is equal to the total momentum after the collision. Since the cue ball initially strikes the stationary eight ball, the total initial momentum is the momentum of the cue ball, and the total final momentum is the momentum of both the cue ball and the eight ball after the collision.
Let's denote the mass of both balls as 'm'. The initial momentum of the cue ball is given by: p_initial = m * v_cue (where v_cue is the initial velocity of the cue ball).
The final momentum of both balls is given by: p_final = m * v_cue_final + m * v_eight_final (where v_cue_final is the final velocity of the cue ball and v_eight_final is the final velocity of the eighth ball).
Since the collision is elastic, there is no loss of kinetic energy, so the total kinetic energy before the collision is equal to the total kinetic energy after the collision. The initial kinetic energy of the cue ball is given by: KE_initial = (1/2) * m * v_cue^2.
The final kinetic energy of both balls is given by: KE_final = (1/2) * m * v_cue_final^2 + (1/2) * m * v_eight_final^2.
We can solve these equations to find the final velocity of the eight balls. Given that the cue ball's initial velocity (v_cue) is 3.4 m/s, its final velocity (v_cue_final) is 0.94 m/s at an angle θ (concerning its original line of motion), and the masses of both balls are the same, we can substitute these values into the equations and solve for the final velocity of the eight ball.
In 1932, James Chadwick bombarded beryllium-9 with alpha particles. One product was a neutron. This led to the direct experimental verification of the existence of neutrons. The other product of Chadwick's nuclear reaction was carbon-12. boron-12. helium-4. nitrogen-14.
In 1932, James Chadwick proved the existence of neutrons by bombarding Beryllium-9 with alpha particles. The reaction resulted in the production of neutrons and likely also Carbon-12.
Explanation:
In the experiment conducted by James Chadwick in 1932, he bombarded a sample of Beryllium-9 with alpha particles. The reaction resulted in two products: a neutron, which was significant as it directly demonstrated the existence of neutrons, and another product. This missing product has not been mentioned in the question, but based on the options provided, one can infer that the most likely answer is Carbon-12. This was identified because the carbon-12 isotope is a stable one usually formed in nuclear reactions like Chadwick's.
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When comparing saturated and naturally occurring unsaturated fats, the unsaturated fats have __________ and are __________ at room temperature.
Answer: have "cis C=C double bonds" and "liquid" at room temperature.
Explanation:
The unsaturated fatty acids have one or more C=C double bonds in the cis formation. Thus, this results in the molecules not been as stable as the saturated fats. They have weaker intermolecular bonds thus resulting in lower melting point . The consequently results in it being liquid at room temperature.
Unsaturated fats have one or more double bonds in their carbon chain and are usually liquid at room temperature. In contrast, saturated fats, which have no double bonds, are typically solid at room temperature.
Explanation:When comparing saturated and naturally occurring unsaturated fats, the unsaturated fats have one or more double bonds in their carbon chain and are typically liquid at room temperature. Saturated fats, on the other hand, have no double bonds and are usually solid at room temperature. This is due to the straight, closely packed chains of saturated fat molecules, whereas the double bonds in unsaturated fats cause bends, preventing them from packing closely together and remaining fluid at room temperature.
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Tech A says quartz crystals under mechanical stress will produce a voltage. Tech B says the piezoelectric principle is used in coolant temp sensors. Who is correct?
Answer:
Tech A is right.
Explanation:
Its the ability of quartz crystal that if we apply a mechanical stress on them by an applied force then they can produce the electricity or voltage. This phenomenon of producing electricity by squeezing a crystal is termed as piezoelectricity or a piezoelectric effect.
Tech A is correct. Quartz crystals under mechanical stress can produce a voltage. The piezoelectric principle is used in coolant temperature sensors.
Explanation:Tech A is correct. Quartz crystals, when subjected to mechanical stress, exhibit piezoelectric behavior, which means they can generate a voltage. This phenomenon is based on the piezoelectric effect, where the crystal structure of quartz generates an electric charge when deformed. This property is extensively used in various applications, including sensors, resonators, and oscillators.
Tech B is also correct. The piezoelectric principle is indeed used in coolant temperature sensors. These sensors utilize a piezoelectric material, such as quartz, to convert temperature changes into an electric signal. When the temperature of the coolant changes, it causes the piezoelectric material to deform, resulting in a change in voltage or current that can be measured and interpreted to determine the coolant temperature.
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What four processes make up the simple ideal Rankine cycle? Constant heat addition in a boiler, Isothermal compression in a pump, constant heat rejection in a condenser and Isentropic expansion in a turbine Constant pressure heat addition in a boiler, Isentropic compression in a pump, constant pressure heat rejection in a condenser and Isentropic expansion in a turbine Constant heat addition in a boiler, Isentropic compression in a pump, constant heat rejection in a condenser and Isobaric expansion in a turbine Constant temperature heat addition addition in a boiler, Isentropic compression in a pump, constant heat rejection in a condenser and Isentropic expansion in a turbine
Answer:
Constant pressure heat addition in a boiler,
Isentropic compression in a pump,
constant pressure heat rejection in a condenser and Isentropic expansion in a turbine
B
Explanation:
The Rankine cycle is a model used to predict the performance of steam turbine systems. It involves the use of these four components
1. Pump
2. Boiler
3. Turbine
4. Condenser
It includes the following processes;
1. Isentropic compression in pump
2. Constant pressure heat addition in boiler
3. Isentropic expansion in turbine
4. Constant pressure heat rejection in condenser
All this processes make up a Rankine cycle.
A ball of mass 0.165 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.940 m. What impulse was given to the ball by the floor?
Answer:
I = 1.525 kg.m/s
Explanation:
given,
mass of the ball = 0.165 Kg
height of drop, h = 1.25 m
ball rebound and reach to height, h' = 0.940 m
impulse = ?
using conservation of energy
Potential energy is converted into kinetic energy
[tex]mgh = \dfrac{1}{2}mv^2[/tex]
[tex]v=\sqrt{2gh}[/tex]
[tex]v=\sqrt{2\times 9.8 \times 1.25}[/tex]
v = 4.95 m/s
velocity of the ball after rebound
again using conservation of energy
[tex]mgh = \dfrac{1}{2}mv'^2[/tex]
[tex]v'=\sqrt{2gh}[/tex]
[tex]v'=\sqrt{2\times 9.8 \times 0.94}[/tex]
v' = 4.29 m/s
impulse is equal to change in momentum
I = m ( v' - v )
I = 0.165 x ( 4.29 - (-4.95))
I = 1.525 kg.m/s
The astronomer who, at the turn of the century, measured the spectra of hundreds of thousands of stars (leaving a catalog that astronomers used for the rest of the century) was:______________
Answer: The astronomer who, at the turn of the century, measured the spectra of hundreds of thousands of stars (leaving a catalog that astronomers used for the rest of the century) was: Annie Jump Cannon.
Annie Jump CannonIntroduction:Annie Jump Cannon was born on 11th December 1863 in Dover, Delaware. She was an american astronomer whose work contributes in the expansion of current stellar classification. She made the first accurate attempt to organize and classify stars on the basis of their temperature and spectral form. She suffered from hearing loss during her childhood. She was almost deaf during her career. She was member of National women's party and by her nature, she was a extreme suffragist.
Early life :Annie Jump was the eldest of the three daughters of Wilson Cannon (shipbuilder and state senator) and Mary Jump (second wife of Wilson Cannon). It was Annie's mother who encouraged her to peruse her education and career in her interests. She suggested her to study in Wellesley college, perusing in mathematics, biology and chemistry. Cannon's mother teaching like household economics, helped her to organize her research later on.
Education:She took her early education and was brilliant student of mathematics, then went to Wellesley College of Massachusetts in 1880 and studied physics and astronomy, later she went to work at Harvard Observatory.
Career:Astronomer Annie Jump was the pioneer, who develop a simple spectral classification system. She categorized nearly 400,000 stars in total. She for the very first time found a double star, three hundred variable stars and 5 novas. She worked as astronomer for more then forty years till 1940, until she got retired. During her career she also work as suffragist, and helped many women to attain respect and reputation in scientific society. She was a hard working woman with a calm nature who build the ways for upcoming women in the field of scientific research and astronomy.
Post Life and Death:Annie Jump Canon died on 13th April, 1941 in Cambridge, Massachusetts. she died at the age of 72. She was hospitalized prior to her death due extreme illness. After her death American Astronomical Society announce an award named after her (The Annie Jump Cannon Award), yearly to those female astronomers, who worked remarkably in astronomy.
The astronomer who, at the start of the 20th century, measured and catalogued the spectra of hundreds of thousands of stars was Annie Jump Cannon. She worked under Edward C. Pickering at the Harvard Observatory and her detailed catalogue served as a major reference for astronomers for the rest of the century.
Explanation:The astronomer you're referring to is Annie Jump Cannon who worked at the Harvard Observatory under Edward C. Pickering. In the late 1800s, Pickering started an ambitious project of classifying stellar spectra, the foundation of this work was a remarkable collection of close to one million photographic spectra of stars. Cannon was hired by Pickering to assist in the classification of spectra and her efficiency was such, that she could visually identify and determine the spectral types of several hundred stars per hour.
This relentless work resulted in Cannon creating a comprehensive catalog containing the spectral types for hundreds of thousands of stars. This seminal piece of work became the blueprint for much of twentieth-century astronomy, and served to influence and inform a whole generation of astronomers. Therefore, Cannon's detailed and systematic study was instrumental in developing a deeper understanding of the universe as we know it today.
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At an outdoor market, a bunch of bananas attached to the bottom of a vertical spring of force constant 16.0 N/m is set into oscillatory motion with an amplitude of 20.0 cm. It is observed that the maximum speed of the bunch of bananas is 44.0 cm/s. What is the weight of the bananas in newtons?
Answer:
32.4289 N
Explanation:
A = Amplitude = 20 cm
[tex]v_m[/tex] = Maximum velocity = 44 cm/s
k = Spring constant = 16 N/m
g = Acceleration due to gravity = 9.81 m/s²
m = Mass of object
Maximum velocity is given by
[tex]v_m=A\omega[/tex]
Angular velocity is given by
[tex]\omega=\sqrt{\dfrac{k}{m}}[/tex]
[tex]v_m=A\sqrt{\dfrac{k}{m}}\\\Rightarrow m=\dfrac{A^2k}{v_m^2}\\\Rightarrow m=\dfrac{0.2^2\times 16}{0.44^2}\\\Rightarrow m=3.3057\ kg[/tex]
Weight is given by
[tex]W=mg\\\Rightarrow W=3.3057\times 9.81\\\Rightarrow W=32.4289\ N[/tex]
The weight of the bananas is 32.4289 N
A flying saucer lifts the Physical Science building 10,000 ft into the air before discovering it is useless and discards the remains. If the building weighs 1,000,000 pounds and the ascent takes 20 seconds, what is one saucer power?
Answer:
500000000 lbft/s
Explanation:
F = Force or weight = 1000000 lbf
s = Displacement = 10000 ft
t = Time taken = 20 seconds
Work done is given by
[tex]W=Fs\\\Rightarrow W=1000000\times 10000\\\Rightarrow W=10000000000\ lb-ft[/tex]
Power is given by
[tex]P=\dfrac{W}{t}\\\Rightarrow P=\dfrac{10000000000}{20}\\\Rightarrow P=500000000\ lbft/s[/tex]
One Saucer power is 500000000 lbft/s
The time spent on design will vary, dependent on the complexity of the problem. Why is it important to give the design process adequate time?
Answer:To ensure significant dialogue which will guarantee that the needed Products meets specifications
It will help to develop a better Operational model for the users.
Explanation: The design phase is one of the most critical phase in product development,it is in this phase that the product structure, contents and specifications are put into action. It is necessary that the end users have a significant time to dialogue with the project or product development team to guarantee that the end product meets specifications.
It is also necessary to spent adequate time in order to make the product user friendly (easy to operate and maintain).
Suppose you kick a soccer ball straight up to a height of 10 meters. Which of the following is true about the gravitational potential energy of the ball during its flight?
A) The ball's gravitational potential energy is greatest at the instant when the ball is at its highest point.
B) The ball's gravitational potential energy is always the same.
C) The ball's gravitational potential energy is greatest at the instant the ball leaves your foot.
D) The ball's gravitational potential energy is greatest at the instant it returns to hit the ground.
Answer:
A) The ball's gravitational potential energy is greatest at the instant when the ball is at its highest point.
Explanation:
Suppose you kick a soccer ball straight up to a height of 10 meters. Which of the following is true about the gravitational potential energy of the ball during its flight?
A) The ball's gravitational potential energy is greatest at the instant when the ball is at its highest point. (true)
B) The ball's gravitational potential energy is always the same. (false)
because the gravitational potential energy is changed as the height changed.
C) The ball's gravitational potential energy is greatest at the instant the ball leaves your foot. (false)
Because The ball's gravitational potential energy is greatest at the instant when the ball is at its highest point.
D) The ball's gravitational potential energy is greatest at the instant it returns to hit the ground. (false)
The ball's gravitational potential energy is greatest at the instant when the ball is at its highest point. not when return to the ground
Final answer:
The gravitational potential energy of a soccer ball kicked straight up is A) The ball's gravitational potential energy is greatest at the instant when the ball is at its highest point.
Explanation:
If you kick a soccer ball straight up to a height of 10 meters, the gravitational potential energy of the ball is greatest at the instant when the ball is at its highest point. At the highest point, all the energy in the system is gravitational potential energy.
This is because gravitational potential energy is calculated by the formula PE = mgh, where m is the mass of the ball, g is the acceleration due to gravity, and h is the height above the reference point, which in this case is the ground. At the ball's highest point, the h value is at its maximum of 10 meters. Therefore, the correct answer to the question is A) The ball's gravitational potential energy is greatest at the instant when the ball is at its highest point.
You are on the roof of the physics building, 46.0 above the ground . Your physics professor, who is 1.80 tall, is walking alongside the building at a constant speed of 1.20m/s. If you wish to drop an egg on your professor's head, how far from the building should the professor be when you release the egg? Assume that the egg is in free fall.
There is one mistake in the question as unit of height of building is not given.So I assume it as meter.The complete question is here
You are on the roof of the physics building, 46.0 m above the ground. Your physics professor, who is 1.80 m tall, is walking alongside the building at a constant speed of 1.20 m/s. If you wish to drop an egg on your professor’s head, where should the professor be when you release the egg? Assume that the egg is in free fall.
Answer:
d=3.67 m
Explanation:
Height of building=46.0 m
First we need to find time taken by egg to reach 1.80 m above the surface
So to find time use below equation
[tex]S=vt+\frac{1}{2} gt^{2}\\ (46.0-1.80)m=(om/s)t+\frac{1}{2}(9.8m/s^{2} )t^{2}\\t=\sqrt{\frac{(46.0-1.80)m}{4.9} }\\ t=3.06s[/tex]
As velocity 1.20m/s is given and we have find time.So we can easily find the distance
So
[tex]distance=velocity*time\\d=v*t\\d=(1.20m/s)*(3.06s)\\d=3.67m[/tex]
A pipe 0.770 m long and closed at one end is filled with an unknown gas. The third lowest harmonic frequency for the pipe is 750 Hz. (a) What is the speed of sound in the unknown gas?
Answer:
speed of sound in unknown gas will be 360m/sec
Explanation:
We have given length of the pipe l = 0.770 m
It is given that pipe is open at one end and closed at one end
Frequency of third harmonic f = 750
Third harmonic frequency of pipe which one end is open and one end is closed is given by
[tex]f=\frac{5v}{4L}[/tex], here v is the speed of the sound, L is the length of the pipe
So [tex]750=\frac{5v}{4\times 0.770}[/tex]
v = 360 m /sec
So speed of sound in unknown gas will be 360m/sec
A hypothetical square shrinks at a rate of 49 squared meters per minute. At what rate are the sides of the square changing when the sides are 13m each?
Answer:
-3.769 m/min
Explanation:
[tex]\dfrac{dA}{dt}[/tex] = Rate of change of area = -49 m²/min (negative due to shrinking)
s = Side length = 13 m
[tex]\dfrac{ds}{dt}[/tex] = Rate of change of side
Area of a square is given by
[tex]A=s^2[/tex]
Differentiating with respect to time
[tex]\dfrac{dA}{dt}=\dfrac{ds^2}{dt}\\\Rightarrow \dfrac{dA}{dt}=2s\dfrac{ds}{dt}\\\Rightarrow \dfrac{ds}{dt}=\dfrac{dA}{dt}\times \dfrac{1}{2s}\\\Rightarrow \dfrac{ds}{dt}=-49\times \dfrac{1}{13}\\\Rightarrow \dfrac{ds}{dt}=-3.769\ m/min[/tex]
The rate of change of the sides of the square is -3.769 m/min
In terms of conservation of momentum explain why it is not a good idea to jump from a rowboat to a dock that seems just within jumping distance, if the boat is not tied to the dock.
Answer:
Explanation:
It is dangerous to jump from rowboat to a dock because you may end up hurting yourself.
As momentum is conserved, the momentum of the boat after you jump off from boat is equal and opposite to your momentum. So you end up landing far behind than you expected and hurt yourself.
Explains why jumping from a rowboat to a dock violates the conservation of momentum principle, leading to potential boat instability.
In terms of conservation of momentum, when you jump from a rowboat to a dock, your momentum changes suddenly as you leave the boat, but the boat still retains its momentum. This violates the principle of conservation of momentum, which states that in the absence of external forces, the total momentum of a system remains constant.
For instance, if the rowboat (system) and you initially have zero momentum while at rest, the moment you jump, your momentum changes, but the boat will move in the opposite direction to compensate, due to Newton's third law of motion, causing instability and potentially capsizing the boat.
At an accident scene on a level road, investigators measure a car's skid mark to be 84 m long. It was a rainy day and the coefficient of friction was estimated to be 0.36.
The given data is incomplete. The complete question is as follows.
At an accident scene on a level road, investigators measure a car's skid mark to be 84 m long. It was a rainy day and the coefficient of friction was estimated to be 0.36. Use these data to determine the speed of the car when the driver slammed on (and locked) the brakes. (why does the car's mass not matter?)
Explanation:
Let us assume that v is the final velocity and u is the initial velocity of the car. Let s be the skid marks and [tex]\mu[/tex] be the friction coefficient and m be the mass of car.
Hence, the given data is as follows.
v = 0, s = 84 m, [tex]\mu[/tex] = 0.36
According to Newton's law of second motion the expression for acceleration is as follows.
F = ma
[tex]-\mu N[/tex] = ma
[tex]-\mu mg[/tex] = ma
a = [tex]-\mu g[/tex]
Also,
[tex]v^{2} = u^{2} + 2as[/tex]
[tex](0)^{2} = u^{2} + 2(-\mu g)s[/tex]
[tex]u^{2} = 2(\mu g)s[/tex]
= [tex]\sqrt{2(0.36)(9.81 m/s^{2})(84 m)}[/tex]
= 24.36 m/s
Thus, we can conclude that the speed of the car when the driver slammed on (and locked) the brakes is 24.36 m/s.
Young David who slew Goliath experimented with slings beforetackling the giant. He found that he could revolve a sling oflength 0.600 m at the rate of9.00 rev/s. If he increased the lengthto 0.900 m, he could revolve the slingonly 7.00 times per second.(a) What is the speed of the stone for eachrate of rotation?1 m/s at 9.00 rev/s2 m/s at 7.00 rev/s(b) What is the centripetal acceleration of the stone at9.00 rev/s?3 m/s2(c) What is the centripetal acceleration at 7.00 rev/s?4 m/s2
Answer:
(a) velocity at 9 rev/s is 34.3m/s and the velocity at 7rev/s is 40.4 m/s. These were values can be gotten by using the formula v = 2(pi)R/T. Where T is the period. T = 1/w(angular velocity).
(b) acceleration at 9.0rev/s is 1961 m/s²
(c) acceleration at 7.0rev/s is 1814 m/s²
These values can be gotten by using the formula a = v²/R
Explanation:
The full solution can be found in the attachment below.
Thank you for reading.
Final answer:
To find the speed of the stone for each rate of rotation, use the formula Speed = (2 * pi * radius * rate of rotation). The centripetal acceleration at 9.00 rev/s is approximately 3.1 m/s^2 and at 7.00 rev/s is approximately 3.9 m/s^2.
Explanation:
To find the speed of the stone for each rate of rotation, we can use the formula:
Speed = (2 * pi * radius * rate of rotation)
Using a length of 0.600 m and a rate of rotation of 9.00 rev/s, the speed of the stone is approximately 3.4 m/s. Using a length of 0.900 m and a rate of rotation of 7.00 rev/s, the speed of the stone is approximately 4.9 m/s.
The centripetal acceleration of the stone at 9.00 rev/s can be found using the formula:
Centripetal acceleration = (speed)^2 / radius
Substituting the values, we get a centripetal acceleration of approximately 3.1 m/s^2.
The centripetal acceleration at 7.00 rev/s can be found using the same formula. Substituting the values, we get a centripetal acceleration of approximately 3.9 m/s^2.
Which sentence below describes a contact force? A) A baseball bat hits a ball B) A book falls to the floor C) A leaf floats in the air and falls to the ground D) A magnetic force pulls a paper clip toward a magnet Please help!!!
A baseball bat hits a ball is an example of contact force.
Option A.
Explanation:Contact force is defined as those forces which are acting between two bodies by actually touch between two bodies.
In case of the book falling to ground, the force acting is the gravitational force. Gravitational force acts even when two bodies aren't in contact. So its a non contact force.
Similar is the condition of the leaf and ground where the gravitational force acts.
The magnetic force between the paper clip and the magnet is also non contact force, which acts even when two bodies aren't in contact.
But the bat hitting the ball is having direct contact between the two bodies, and their contact is what makes the ball fly off. So its a contact force.
Contact force is the type of force that occurs when the external force act on the system act physically. A baseball bat hitting a ball is the right example of contact forces.
What is contact force?Any force that requires touch is referred to as a contact force. The majority of apparent interactions between macroscopic groupings of matter are caused by contact forces.
Contact forces are used in everyday circumstances. For example, a baseball bat hits a ball.
Hence a baseball bat hitting a ball is the right example of contact forces.
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A cargo helicopter, descending steadily at a speed of 2.3 m/s, releases a small package. Let upward be the positive direction for this problem.
(a) If the package is 71 m above the ground when it is dropped, how long does it take for the package to reach the ground?
1 s
(b) What is its velocity just before it lands? (Indicate the direction with the sign of your answer.)
2 m/s
The package takes approximately 0.117 seconds to reach the ground. The velocity just before it lands is 25.364 m/s (upwards).
Explanation:To find the time it takes for the package to reach the ground, we can use the equation for uniform acceleration: acceleration (a) = g - (-g) = 2g, where g is the acceleration due to gravity (9.8 m/s^2) and the negative sign represents the direction of motion. Using v = u + at, where v is the final velocity (0 m/s), u is the initial velocity (2.3 m/s), and a is the acceleration, we can find the time (t) it takes for the package to reach the ground. Rearranging the equation, we have t = (v - u)/a. Plugging in the values, we have t = (0 - 2.3)/(2g) = -2.3/19.6 = 0.117 seconds. The package takes approximately 0.117 seconds to reach the ground.
To find the velocity just before it lands, we can use the equation v = u + at, where u is the initial velocity, a is the acceleration, and t is the time. Rearranging the equation, we have v = u + at. Plugging in the values, we have v = 2.3 + (2g)(0.117) = 2.3 + 23.064 = 25.364 m/s (upwards).
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A car moving at 10.0 m/s crashes into a barrier and stops in 0.050 s. There is a 20.0-kg child in the car. Assume that the child’s velocity is changed by the same amount as that of the car, and in the same time period. What is the impulse needed to stop the child?
Answer:
The impulse needed to stop the child = -200 Ns
Explanation:
Impulse: This can be defined as the product of force and time of a body. The S.I unit of impulse is kgm/s or Ns
From newton's second law,
I = mΔv ................... Equation 1.
Where I = Impulse needed to stop the child, m = mass of the child, Δ = change in velocity of the child.
Given: m = 20.0 kg,
Δv = v-u where v =0 m/s, u = 10.0 m/s
Δv = 0-10 = -10 m/s.
Substituting into equation 1
I = 20(-10)
I = -200 Ns.
Thus the impulse needed to stop the child = -200 Ns
Note: The negative sign shows that the impulse act against the motion of the child
The correct impulse needed to stop the child is 200 N·s.
The impulse needed to stop the child is 16 N·s.To find the impulse needed to stop the child, we can use the formula for impulse, which is the change in momentum of an object. The momentum of an object is given by the product of its mass and velocity. Impulse is the product of the average force applied to the object and the time interval during which the force is applied. The impulse is equal to the change in momentum.
The initial momentum of the child (p_initial) is given by the mass of the child (m) multiplied by the initial velocity of the car (v_initial), since the child's velocity is assumed to be the same as that of the car. The final momentum of the child (p_final) is zero because the child comes to a stop.
Given:
- Mass of the child, m = 20.0 kg
- Initial velocity of the car (and the child), v_initial = 10.0 m/s
- Time taken to stop, Aat = 0.050 s
- The change in momentum (Aap) is:
Aap = [tex]p_final - p_initial[/tex]
Aap = [tex]0 - (m * v_initial)[/tex]
Aap = - (20.0 kg * 10.0 m/s)
Aap= - 200 kgA·m/s
The negative sign indicates that the momentum is decreasing (the child is stopping).
The impulse (J) is equal to the change in momentum:
J = Aap
J = - 200 kgA·m/s
Since impulse is a vector quantity and we are interested in the magnitude of the impulse needed to stop the child, we take the absolute value:
J = | - 200 kgA·m/s |
J = 200 kgA·m/s
Now, we can express the impulse in Newtons-seconds (NA·s) by noting that 1 kgA·m/s is equivalent to 1 NA·s:
J = 200 NA·s
However, the answer provided initially (16 NA·s) seems to be incorrect. Let's re-evaluate the calculation to ensure accuracy.
The impulse J is the change in momentum Aap over the time interval Aat. The average force F_avg applied to the child is then the impulse divided by the time interval:
F_avg = J / Aat
Rearranging for J gives:
J = F_avg / Aat
Since we know the change in momentum Aap is equal to the impulse J, we can write:
J = Aap = m * Aav
The change in velocity Aav of the car (and thus the child) is the final velocity [tex]v_final[/tex]minus the initial velocity [tex]v_initial[/tex]. Since the car stops, [tex]v_final[/tex]= 0, and Aav is simply [tex]-v_initial[/tex].
Aav = v[tex]_final - v_initial[/tex]
Aav = 0 - 10.0 m/s
Aav= - 10.0 m/s
Now we can calculate the impulse needed to stop the child:
J = m * Aav
J = 20.0 kg * (- 10.0 m/s)
J = - 200 kgA·m/s
Taking the magnitude:
J = | - 200 kgA·m/s |
J = 200 kgA·m/s
Since 1 kgA·m/s is equivalent to 1 NA·s, the impulse in Newtons-seconds is:
J = 200 NA·s
This confirms that the initial answer provided (16 N·s) is indeed incorrect, and the correct impulse needed to stop the child is 200 N·s.
As a spherical ammonia vapor bubble rises in liquid ammonia, its diameter changes from 1 cm to 3 cm. Calculate the amount of work produced by this bubble, in kJ, if the surface tension of ammonia is 0.04 N/m.
Answer:
So amount of work produced will be [tex]10{-4}J[/tex]
Explanation:
We have given diameter of ammonia bubble is changes from 1 cm to 3 cm
So radius changes from 0.5 cm to 1.5 cm
Surface area of bubble[tex]=4\pi r^2[/tex]
So change in area of bubble [tex]=4\pi (0.015^2-0.005^2)=8\times 3.14\times (0.015^2-0.005^2)=0.00251m^2[/tex]
Surface tension of ammonia = 0.04 N/m
So work done will be [tex]Work\ done=surface\ tension\times change\ in\ area=0.04\times 0.00251= 10^{-4}J[/tex]
Based on observations, the speed of a jogger can be approximated by the relation v 5 7.5(1 2 0.04x) 0.3, where v and x are expressed in mi/h and miles, respectively. Knowing that x 5 0 at t 5 0, determine (a) the distance the jogger has run when t 5 1 h, (b) the jogger’s acceleration in ft/s2 at t 5 0, (c) the time required for the jogger to run 6 mi.
a) The jogger has run approximately 7.978 miles after 1 hour. b) The jogger's acceleration at t = 0 is approximately 0.41 ft/s^2. c) The equation does not provide a direct relationship between distance and time, so we cannot determine the time required to run 6 miles.
(a) To find the distance the jogger has run when t = 1 h, we need to substitute t = 1 into the given equation. Plugging in t = 1, we have: v = 7.5(1 + 2(0.04(1)))^0.3. Simplifying this expression, we get v ≈ 7.5(1.08)^0.3. Evaluating this expression, we find v ≈ 7.978 mi/h. To find the distance, we multiply the speed by the time: d = v × t = 7.978 × 1 = 7.978 mi.
(b) The acceleration can be found by taking the derivative of the velocity equation with respect to time. Differentiating the given equation, we have dv/dt = 7.5(1 + 2(0.04x))^(-0.7) × 2(0.04). Substituting t = 0 into this expression, we get: a = 7.5(1 + 2(0.04(0)))^(-0.7) × 2(0.04). Simplifying this expression, we find a ≈ 0.28 mi/h^2. To convert this to ft/s^2, we multiply by 5280 ft/mi and divide by 3600 s/h: a = 0.28 × 5280 / 3600 ≈ 0.41 ft/s^2.
(c) To find the time required for the jogger to run 6 mi, we need to solve the equation v = 7.5(1 + 2(0.04x))^0.3 for t. Plugging in d = 6 and v = 7.5(1 + 2(0.04x))^0.3, we can solve for t. However, the given equation does not provide a direct relationship between distance and time, so we cannot solve it.
A cue ball initially moving at 2.5 m/s strikes a stationary eight ball of the same size and mass. After the collision, the cue ball’s final speed is 1.2 m/s at an angle of ? With respect to its original line of motion.Find the eight ball’s speed after the collision. Assume an elastic collision (ignoring friction and rotational motion).Answer in units of m/s
Final answer:
To find the speed of the eight ball after an elastic collision with the cue ball in motion, apply conservation laws of momentum and kinetic energy.
Explanation:
A cue ball initially moving at 2.5 m/s strikes a stationary eight ball of the same size and mass. After the collision, the cue ball’s final speed is 1.2 m/s at an angle of ? With respect to its original line of motion. To find the eight ball’s speed after the collision, we can apply the principles of conservation of momentum and conservation of kinetic energy.
Let's denote the velocities: V1 (initial cue ball velocity), V2 (initial eight ball velocity), V1' (final cue ball velocity), and V2' (final eight ball velocity). Considering an elastic collision, where momentum and kinetic energy are conserved, we can set up equations to solve for V2'.
By using the conservation laws, we can determine that the eight ball's speed after the collision is approximately 2.3 m/s.
A film of soapy water (n = 1.33) on top of a sheet of Plexiglas (n = 1.51) has a thickness of 266 nm. What wavelength is most strongly reflected if it is illuminated perpendicular to its surface? (This is the apparent color of the soapy film.) nm
Wavelength of the most strongly reflected is approximately 707 nm, which corresponds to red light in the visible spectrum.
To determine the wavelength most strongly reflected by a soapy water film, we can use the principle of thin film interference. Here,
the film has a refractive index of 1.33 thickness of 266 nm, it is on top of Plexiglas with a refractive index of 1.51.For constructive interference to occur in a thin film illuminated perpendicularly, the condition we use is:
2nt = mλ
where n is the refractive index of the film,
t is the thickness of the film, λ is the wavelength of light in the vacuum, m is an integer representing the order of the interference.Since we need the most strongly reflected wavelength, we'll consider
m = 1 (the first order of constructive interference).
Substituting the values into the equation:
2 x 1.33 x 266 nm = 1 x λ
This simplifies to:
λ = 2 x 1.33 x 266 nm
λ ≈ 707 nm
Therefore, the most strongly reflected wavelength is approximately 707 nm, which corresponds to red light in the visible spectrum.
A student measures the mass of a 1.0 kg standard bar. He obtains measurements of 0.77 kg, 0.78 kg, and 0.79 kg. Which describes his measurements
Complete question
A student measures the mass of a 1.0 kg standard bar. He obtains measurements of 0.77 kg, 0.78 kg, and 0.79 kg. Which describes his measurements
a)precise but not accurate
b)accurate but not precise
c)neither precise nor accurate
d)both precise and accurate
Answer:
The measurement is precise but not accurate
Explanation:
A measurement can either be precise or accurate.
A Precise measurement describes how close the measured values are to each other.An accurate measurement describes how close a measured value is to the true value.In this question, the measured values (0.77 kg, 0.78 kg and 0.79 kg) are far from the true value (1.0 kg), therefore the measurement is not accurate.
However, the measured values (0.77 kg, 0.78 kg and 0.79 kg) are close to each other, therefore the measurement is precise.
Therefore the correct option is 'a' the measurement is precise but not accurate
Answer:
precise but not accurate
Explanation:
precise but not accurate
A satellite is placed in orbit 8.52 x 105 m above the surface of Jupiter. Jupiter has a mass of 1.90 x 1027 kg and a radius of 7.14 x 107 m. Find the orbital speed of the satellite.
The orbital speed of the satellite can be found using the equations for gravitational force and centripetal acceleration. By equating these two equations, we can solve for the orbital speed. Plugging in the given values, we find that the orbital speed of the satellite is approximately 6.43 x 10^3 m/s.
Explanation:To find the orbital speed of the satellite, we can use the equation for gravitational force: F = ma, where F is the gravitational force, m is the mass of the satellite, and a is the centripetal acceleration. We can also use the equation for centripetal acceleration: a = v^2 / r, where v is the orbital speed and r is the distance between the satellite and the center of Jupiter. By equating the two equations, we can solve for v.
The gravitational force acting on the satellite is given by the formula F = G * (m1 * m2) / r^2, where G is the gravitational constant, m1 is the mass of Jupiter, and r is the distance between the satellite and the center of Jupiter.
Plugging in the values given in the question, we can solve for v:
v^2 = (G * m1) / r
v = sqrt((G * m1) / r)
Using the given values of m1 = 1.90 x 10^27 kg, r = 8.52 x 10^5 m, and the gravitational constant G = 6.67 x 10^-11 Nm^2/kg^2, we can calculate the orbital speed of the satellite:
v = sqrt((6.67 x 10^-11 Nm^2/kg^2 * 1.90 x 10^27 kg) / (8.52 x 10^5 m))
v ≈ 6.43 x 10^3 m/s
Rutherford’s α-scattering experiment showed that most α particles directed toward a thin metallic foil passed through with only slight deviations. From this evidence we can conclude that
1. the foil was too thin.
2. an α particle is a type of light which should not be deflected.
3. the mass of the atom is concentrated in a very small area.
4. α particles are uncharged.
5. α particles are too small to hit anything.
Answer:
3. the mass of the atom is concentrated in a very small area.
Explanation:
As the Rutherford experiment has shown that majority of particles has passed the foil with slightest deviations. It shows that the most of the part of atom is empty which doesn't effect the deviation of alpha particles. Few particles are deflected at significant angels shows that there must be a mass occupying the smaller area of atom which is later on called as nucleus.