Explanation:
1. The universe consisted of hydrogen and helium initially. This statement is true.
2. It is important to understand radioactive decay in order to understand this question, here's a good analogy:
A snake will shed it's skin, just as an atom will shoot off different parts of itself. It would be very difficult to force that snake to reenter it's skin once it sheds, just as it takes a lot of energy to force fusion of atoms and the parts mentioned. In normal circumstances, nuclear decay is one-way.
3. The earth is a giant floating rock in space. It took many many years to gather a bunch of asteroids and dust to make this planet.
4. I shouldn't have to explain this one, it doesn't make much sense.
Key elements of which Earth and life are made, including carbon, oxygen, and iron, did not exist when the universe was born and were created later in stars through nucleosynthesis.
The statement that is true regarding the history of the universe is that key elements of which Earth and life are made, including carbon, oxygen, and iron, did not exist when the universe was born and were created later in stars.
Initially, after the Big Bang, all matter consisted mainly of hydrogen and helium.
As stars formed and evolved, they produced other elements through nuclear reactions, including the elements necessary for Earth and life as we know it.
This process, known as nucleosynthesis, enriched the universe with heavier elements over time.
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A Roller Derby Exhibition recently came to town. They packed the gym for twoconsecutive weekend nights at South's field house. On Saturday evening, the68-kg Anna Mosity was moving at 17 m/s when she collided with 76-kg SandraDay O'Klobber who was moving forward at 12 m/s and directly in Anna's path.Anna jumped onto Sandra's back and the two continued moving together atthe same speed. Determine their speed immediately after the collision.
Answer:
14.4 m/s
Explanation:
mass of Anna (Ma) = 68 kg
speed of Anna (Va) = 17 m/s
mass of SandraDay (Ms) = 76 kg
speed of SandraDay (Vs) = 12 m/s
We can find their speed (V) immediately after collision from the conservation of momentum where
(Ma x Va) + (Ms + Vs) = (Ma + Ms) x V
where V = speed immediately after collision
(68 x 17) + (76 + 12) = (68 + 76) x V
2068 = 144 V
V = 2068 / 144 = 14.4 m/s
What would occur if there was a gain-of-function mutation in the promoter for the cyclin E gene such that cyclin E protein was always made at high levels even under conditions in which cyclin E would not normally be made?
Answer:
b) Cells will pass through the G1/S checkpoint even if conditions are not ideal for cell division.
Explanation:
In the given problem, if there exists a gain-of-function mutation for the given cell, there would not be the formation of cyclin E when there is the possibility of cells movement via the checkpoint of the G1/S, even when there are non-deal conditions for the division of cell. Thus, the correct option in the lists of options is the option b.
_______ describes the relationship between a stimulus and its resulting sensation by proposing that the JND is a constant fraction of the stimulus intensity.
Answer:
Weber's Law
Explanation:
Weber’s law is an important law in psychology.
This law quantifies the perception of change within a provided stimulus.
This law is also known as Weber-Fechner law as it relates two hypotheses in the field of psychophysics which are Weber law and the Fechner law.
Weber’s law describes that an observed change in a stimulus is a constant ratio of the original stimulus.
Final answer:
Weber's Law describes the relationship between a stimulus and the resulting sensation, stating that the Just Noticeable Difference (JND) is a constant fraction of stimulus intensity. It is a key principle in psychophysics, useful for experiments such as determining the JND for different weights of rice in bags.
Explanation:
The concept you're inquiring about is Weber's Law, which is a fundamental principle in the field of psychophysics, a branch of psychology. Weber's Law describes the relationship between a stimulus and its resulting sensation by proposing that the Just Noticeable Difference (JND) is a constant fraction of the stimulus intensity.
To put this into practice, for example, if one were testing the JND of various weights of rice in bags, they could use incremental percentage increases of the weight to determine the smallest difference that can be perceived. By choosing increments like 10 percent or 20 percent between certain weight thresholds, a researcher can apply Weber's Law to determine the JND in a standardized way.
This law helps explain why when you have a cup of coffee with little sugar, adding a teaspoon can make a noticeable difference, but in a cup with much more sugar, that same teaspoon becomes less detectable, requiring a proportionally larger amount of sugar to notice a change in sweetness.
In 1977, Kitty O’Neil drove a hydrogen peroxide–powered rocket dragster for a record time interval (3.22 s) and final speed (663 km/h) on a 402-m-long Mojave Desert track. Determine her average acceleration during the race and the acceleration while stopping (it took about 20 s to stop). What assumptions did you make?
Answer:
57.19461 m/s²
-9.20833 m/s²
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
Equation of motion
[tex]v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{\dfrac{663}{3.6}-0}{3.22}\\\Rightarrow a=57.19461\ m/s^2[/tex]
The acceleration during the race is 57.19461 m/s²
[tex]v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{0-\dfrac{663}{3.6}}{20}\\\Rightarrow a=-9.20833\ m/s^2[/tex]
The acceleration while stopping is -9.20833 m/s²
The assumptions made here are:
The initial velocity of the dragster at the start of the race is zero
The acceleration is constant in both cases of acceleration.
The final velocity when it stops is zero
Motion is in a straight line path
The voltage between the two plates will induce an electric field between them. Suppose you wanted the object to remain at rest halfway between the two plates. In what direction should the electric field be (upward, downward, to the right, or to the left? Explain.
Answer:Upward
Explanation:
To hang the object in the halfway we need to direct the Electric field toward upward direction
Suppose object has mass m and charge q
If electric field is Pointing towards upward direction then the force experienced by charged particle is in the upward direction which will be balanced by weight of particle(pointing downward) such that
qE=mg
where q=charge of particle
E=Electric field
m=mass of particle
g=acceleration due to gravity
DETERMINE THE LAUNCH ELEVATION AND MAXIMUM RANGE: The test vehicle should obtain a burnout velocity of 7200 m/s at 180 km altitude (RBurnout = 6558 km). What would the flight path angle (fBurnout) be?
Answer:
Flight path angle= 15.12°, maximum range= 5.29× 10*6 km
Explanation:
u= 7200m/s, H= 180km= 180000m
Recall that
Maximum height, H= (u*2sin*2∆)/2g
180000= (7200×7200sin*2∆)/2×9.8
(18000×2×98)/7200×7200= sin*2∆
Sin∆= 0.2609
∆= 15.12°
Maximum range, R= u*2/g
(7200×7200)/9.8
= 5289795.92km
= 5.29× 10*6 km
It is relatively easy to strip the outer electrons from a heavy atom like that of uranium (which then becomes a uranium ion), but it is very difficult to remove the inner electrons. Discuss why this is so.
Answer: it is very difficult to remove the inner electrons from an atom because of the strong electrostatic force of attraction by the nucleus. In contrast, it is very easy to remove the outer electrons from an atom because the electrostatic force of attraction from the nucleus is not strong enough to hold the outer electrons hence it is removed.
When bridges are built, special joints must be used because the material of the bridge shrinks, and without these joints, the material would break. Which of the following properties is described here? A. thermal decay B. thermal expansion C. thermal stasis D. thermal contraction
Answer:
Option D.
Explanation:
The correct answer is Option D.
The shrinkage of the bridge material is because of thermal contraction.
Thermal contraction and thermal expansion are the phenomena of the bridge material which takes place due to the change in temperature of the atmosphere.
When the temperature of the surrounding increases expansion of the bridge material takes place and when temperature decreases the contraction of the material takes place.
This phenomenon sometimes damages the structure because due to continuous expansion and contraction of materials strength of the bridge decreases.
Thermal expansion joints in bridges prevent material breakage due to temperature changes.
Thermal expansion joints are used in bridges to allow for the changing length of the bridge due to temperature fluctuations. Without these joints, the bridge material would break due to thermal stress caused by expansion and contraction.
A revolutionary war cannon, with a mass of 2240 kg, fires a 15.5 kg ball horizontally. The cannonball has a speed of 131 m/s after it has left the barrel. The cannon carriage is on a flat platform and is free to roll horizontally. What is the speed of the cannon immediately after it was fired?Answer in units of m/s.
Answer:
v' = -0.0906 m/s
Explanation:
given,
mass of cannon, M = 2240 Kg
mass of the ball, m = 15.5 Kg
speed of cannon ball, v = 131 m/s
speed of he cannon = ?
initial speed of cannon and the cannon ball is equal to 0 m/s
using conservation of energy
(M+m)V = M v' + m v
(M+m) x 0= 22400 v' + 15.5 x 131
22400 v' = -2030.5
v' = -0.0906 m/s
negative sign represent the canon will move in opposite direction of the ball.
hence, speed of cannon is equal to 0.0906 m/s
Using the law of conservation of momentum and the given values, the recoil velocity of the cannon is found to be approximately 0.9058 m/s after firing a cannonball.
According to the law of conservation of momentum, the total momentum before an event must equal the total momentum after the event, if no external forces act on the system. In this case, the cannon and the cannonball system is isolated and free to recoil, so the momentum before the cannon fires must be equal to the momentum after the cannon fires.
Momentum is the product of mass and velocity. Before the cannon fires, both the cannon and the cannonball are at rest and have a momentum of zero. After the cannon fires, the momentum of the cannonball is mcannonball × vcannonball, which must be equal and opposite to the momentum of the cannon. Therefore, mcannonball × vcannonball = mcannon × vcannon.
Using the given values for mass and velocity of the cannonball (15.5 kg and 131 m/s), and the mass of the cannon (2240 kg), we can calculate the cannon's recoil velocity using the equation:
mcannonball × vcannonball = mcannon × vcannon
(15.5 kg × 131 m/s) / 2240 kg = vcannon
The calculated recoil velocity of the cannon is approximately 0.9058 m/s.
A battery-operated car utilizes a 12.0 V system. Find the charge the batteries must be able to move in order to accelerate the 750 kg car from rest to 25.0 m/s, make it climb a 2.00×102 m high hill, and then cause it to travel at a constant 25.0 m/s by exerting a 5.00×102 N force for an hour.
Answer:
3894531 coulombs
Explanation:
1 hour = 3600 seconds
Let g = 10m/s2
The distance that the car travel at 25 m/s over an hour is
s = 25 * 3600 = 90000 m
The total mechanical energy of the car is the sum of its kinetic energy to reach 25 m/s, its potential energy to climb up 200m high hill and it work to travel a distance of s = 90000m with F = 500 N force:
[tex] \sum E = E_k + E_p + E_W[/tex]
[tex]\sum E = mv^2/2 + mgh + Fs[/tex]
[tex]\sum E = 750*25^2/2 + 750*10*200 + 500*90000 = 46734375 J[/tex]
This energy is drawn from the battery over an hour (3600 seconds), so its power must be
[tex]P = E / t = 46734375/3600 = 12982 W[/tex]
The system is 12V so its current is
[tex]I = P/U = 12982 / 12 = 1081.8 A[/tex] or 1081.8 Coulombs/s
The the total charge it needs for 1 hour (3600 s) is
C = 1081.8 * 3600 = 3894531 coulombs
The quantity of charge the batteries must be able to move is equal to 3.9 μC.
Given the following data:
Voltage = 12.0 VoltsMass = 750 kgSpeed = 25.0 m/sHeight = 200 meters.Force = 500 Newton.Time = 1 hour = 3600 seconds.Scientific data:
Acceleration due to gravity = 9.8 [tex]m/s^2[/tex]To find the quantity of charge the batteries must be able to move:
In this scenario, we would calculate the distance traveled and the total energy that is possessed by this battery-operated car.
For the distance.Mathematically, the distance covered by an object is given by this formula:
[tex]Distance = speed \times time\\ \\ Distance = 25 \times 3600[/tex]
Distance = 90,000 meters.
For the total energy:[tex]E = mgh + \frac{1}{2} mv^2 + Fd\\ \\ E=[750\times 9.8 \times 200] + \frac{1}{2} \times 750 \times 25^2 + [500 \times 90000]\\ \\ E=1470000+234375+45000000[/tex]
Total energy = 46,704,375 Joules.
Next, we would calculate the power consumed by this battery-operated car:
[tex]Power = \frac{Energy}{time}\\ \\ Power = \frac{46,704,375}{3600} [/tex]
Power = 12,973.44 Watts.
Also, we would calculate the current:
[tex]Current = \frac{power}{voltage} \\ \\ Current = \frac{12,973.44}{12}[/tex]
Current = 1,081.12 Amperes.
Now, we can calculate the quantity of charge the batteries must be able to move:
[tex]Q = current \times time\\ \\ Q = 1081.12 \times 3600[/tex]
Q = 3,892,032 Coulombs.
Note: 1 μC = [tex]1 \times 10^6 \;C[/tex]
Q = 3.9 μC
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Suppose you first walk 12.0 m in a direction 20 owest of north and then 20.0 m in a direction 40.0osouth of west. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements A and B , then this problem asks you to find their sum R = A + B . The two displacements A and B add to give a total displacement R having magnitude R and direction in degrees.
Answer:
R=19.5m
[tex]\theta[/tex] = 4.65° S of W
Explanation:
Refer the attached fig.
displacement of the x and y components
x-component displacement is ([tex]R_{x}[/tex]) = [tex]A_{x}+B_{x}[/tex]
= A [tex]\sin[/tex](20°) + B [tex]\cos[/tex](40°)
= -12.0[tex]\sin[/tex](20°) + 20.0[tex]\cos[/tex](40°)
= -19.425m
x-component displacement is ([tex]R_{y}[/tex]) = [tex]A_{y}+ B_{y}[/tex]
= A [tex]\cos[/tex](20°) - B [tex]\sin[/tex](40°)
= 12.0[tex]\cos[/tex](20°) - 20.0[tex]\sin[/tex](40°)
= -1.579
resultant displacement
∴
R = [tex]\sqrt{R_{x}^{2} +R_{y}^{2} } }[/tex]
=[tex]\sqrt{(-19.425)^{2}+(-1.579)^{2} }[/tex]
=19.5m
[tex]\theta[/tex] = [tex]\tan^{-1}\left | \frac{R_{x}}{R_{y}} \right |[/tex]
[tex]\theta[/tex] = [tex]\tan^{-1}\left | \frac{1.579}{19.425} \right |[/tex]
[tex]\theta[/tex] = 4.65° S of W
This problem involves finding the result of two vector displacements using vector addition. Using the Pythagorean theorem, the magnitude of the overall displacement can be determined. The direction of the displacement can be calculated using the tangent formula.
Explanation:This is a vector addition problem in physics. We can solve it using Pythagorean theorem for finding the magnitude and tangent formula for finding the direction.
For the first leg of the journey(A), you are going 20° W of N, or 70° clockwise from the x-axis. For the magnitude of this vector: Ax = 12.0m cos(70°) and Ay = 12.0m sin(70°).
For the second leg of the journey(B), you are going 40° S of W, or 130° clockwise from the x-axis. For the magnitude of this vector: Bx = 20.0m cos(130°) and By = 20.0m sin(130°).
Summing the x and y components gives: Rx = Ax + Bx and Ry = Ay + By. The magnitude R can then be found with Pythagorean theorem formula: R = sqrt(Rx² + Ry²).
The direction can be found using tangent formula: θ = atan(Ry / Rx). As Rx is negative and Ry is positive, θ will fall in the second quadrant.
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A ball is thrown from the top of a building with an initial velocity of 20.0 m/s straight upward, at an initial height of 50.0 m above the ground. The ball just misses the edge of the roof on its way down.
Determine
(a) the time needed for the ball to reach its maximum height,
(b) the maximum height,
(c) the time needed for the ball to return to the height from which it was thrown and the velocity of the ball at the instant,
(d) the time needed for the ball to reach the ground, and
(e) the velocity and position of the ball at t=5.00 s. Neglect air drag.
Answer:
2.03873 s
70.38735 m
4.07747 seconds
5.82688 seconds
27.37482 m from the ground
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
g = Acceleration due to gravity = 9.81 m/s² = a
[tex]v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{0-20}{-9.81}\\\Rightarrow t=2.03873\ s[/tex]
time needed for the ball to reach its maximum height is 2.03873 s
The time taken to go up and the time taken to reach the point from where it was thrown is the same.
So, time needed for the ball to return to the height from which it was thrown is 2.03873+2.03873 = 4.07747 seconds
[tex]v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-20^2}{2\times -9.81}\\\Rightarrow s=20.38735\ m[/tex]
The maximum height the ball will reach is 50+20.38735 = 70.38735 m
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 70.38735=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{70.38735\times 2}{9.81}}\\\Rightarrow t=3.78815\ s[/tex]
Time needed to reach the ground is 2.03873+3.78815 = 5.82688 seconds
The time from the maximum height that is required is 5-2.03873 = 2.96127 seconds
[tex]s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 2.96127^2\\\Rightarrow s=43.01253\ m[/tex]
The ball will be 70.38735-43.01253 = 27.37482 m from the ground
The problem is solved using kinematic equations. The ball reaches its maximum height in approximately 2.04 s, with the maximum height reaching about 70.4 m. It returns to the original height in around 4.08 s with a velocity at -39.79 m/s, reaches the ground in about 5.82 s, and at t = 5 s, its position is approximately 49.4 m with velocity of around -29 m/s.
Explanation:The calculations are based on the physics of kinematics, specifically related to the motion under constant acceleration which in this context is gravity (designated 'g'), typically -9.8 m/s².
(a) You can use the equation v = u + gt to solve for time (t). Since the velocity (v) at the maximum height is 0, initial velocity (u) is 20 m/s and g is -9.8m/s², the equation becomes 0 = 20 - 9.8t. Solving for t, we get t = 20/9.8 ≈ 2.04 s.
(b) The maximum height can be found using the equation s = ut + 0.5gt². Substituting the known values we get the maximum height to be about 70.4m
(c) The time taken for the ball to return to its original height will be double the time to reach maximum height, so it's 2*2.04s = 4.08s. The velocity is simply gt, yields -39.79m/s (negative because it's moving downwards).
(d)We can utilize the equation s = ut + 0.5gt² to find out the time to hit the ground. The total distance covered is 50m + 20.4m = 70.4m. Solving the equation we get t = 3.78s (above the throwing point) and to find the total time on air, add t = 2.04s and value calculated you get t = 5.82 s.
(e)Using equations of motion, we can find out that at t=5s, the position of the ball is 49.4 m above the ground and its velocity is -29m/s.
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Force = mass x acceleration : __________.a. for any force, there is an equal and opposite reaction force. b. the orbit of each planet about the Sun is an ellipse with the Sun at one focus. c. a planet moves faster in the part of its orbit nearer the Sun and slower when farther from the Sun, sweeping out equal areas in equal times. d. an object moves at constant velocity if there is no net force acting upon it. e. more distant planets orbit the Sun at slower average speeds, obeying the precise mathematical relationship p2 =a3.
Answer:
a. for any force, there is an equal and opposite reaction force.
Explanation:
Newton third law of motion states that " for every action, there is an equal and opposite reaction.
Action force= reaction force =
M₁ * a₁ = M₂ * a₂
where M₁ = mass of object 1
a₁ = acceleration due to object 1
M₂ = mass of object 2
a₂ = corresponding acceleration due to object 2
this happen in everyday life of an individual. an example is seen in a groups of three man exerting a push force on a static car, the push force on the car make the car to exhibit some motion equivalent to the force applied by the three man.
A rocket is continuously firing its engines as it accelerates away from Earth. For the first kilometer of its ascent, the mass of fuel ejected is small compared to the mass of the rocket. For this distance, which of the following indicates the changes, if any, in the kinetic energy of the rocket, the gravitational potential energy of the Earth-rocket system, and the mechanical energy of the Earth-rocket system? System Gravitational System Potential Energy Rocket Kinetic Energy Mechanical Energy
(A) Increasing IncreasingIncreasing
(B) Increasing Increasing Constant IncreasingDecreasing Decreasing Decreasing Increasing Constant
Answer:
A) Increasing, Increasing, Increasing
Explanation:
The greater the rate of fuel ejection higher will be the kinetic energy of the rocket. As rocket is fire upward its fuel rejection rate increases and hence it's kinetic energy increases.
Gravitational potential energy,
as the rocket moves further it's r from the Earth increases. Hence the Gravitational potential energy increases as its height from the Earth increases.
Therefore, mechanical energy of the rocket must also increase as it is sum of kinetic and potential energy.
Hence Option A is correct.
For the first kilometer of the rocket's ascent, the kinetic energy, gravitational potential energy, and mechanical energy of the Earth-rocket system increase.
Explanation:The changes in the kinetic energy, gravitational potential energy, and mechanical energy of the Earth-rocket system for the first kilometer of the rocket's ascent can be determined.
Since the mass of fuel ejected is small compared to the mass of the rocket, the rocket's kinetic energy will increase as it accelerates away from Earth. The gravitational potential energy of the Earth-rocket system will also increase as the rocket moves higher against the pull of gravity. Therefore, the changes in the three quantities are as follows:
Kinetic Energy: IncreasingGravitational Potential Energy: IncreasingMechanical Energy: IncreasingThe electromagnetic interaction _______.A. applies only to charges at rest B. applies only to charges in motion C. is responsible for sliding friction and contact forces D. all of the above E. none of the above
Answer:
C. is responsible for sliding friction and contact forces
What amount of heat is required to raise the temperature of 25 grams of copper to cause a 15ºC change? The specific heat of copper is 0.39 J/gºC. A. 115 J B. 150 J C. 250 J D. 300 J
The amount of heat required is B) 150 J
Explanation:
The amount of heat energy required to increase the temperature of a substance is given by the equation:
[tex]Q=mC\Delta T[/tex]
where:
m is the mass of the substance
C is the specific heat capacity of the substance
[tex]\Delta T[/tex] is the change in temperature of the substance
For the sample of copper in this problem, we have:
m = 25 g (mass)
C = 0.39 J/gºC (specific heat capacity of copper)
[tex]\Delta T = 15^{\circ}C[/tex] (change in temperature)
Substituting, we find:
[tex]Q=(25)(0.39)(15)=146 J[/tex]
So, the closest answer is B) 150 J.
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The distance between a carbon atom (m = 12 u) and an oxygen atom (m = 16 u) in the CO molecule is 1.13 × 10–10 m. How far from the carbon atom is the center of mass of the molecule?
The distance between the center of mass of the molecule and carbon atom is 0.65 × 10⁻¹⁰ m
To answwer the question, we need to know the center of mass of CO molecule
What is the center of mass of CO molecule?The center of mass of the CO molecule is given by
x = m₁x₁ + m₂x₂/(m₁ + m₂) where
m₁ = mass of carbon atom = 12 u, x₁ = position of carbon atom = 0 m (assuming its at the origin), m₂ = mass of oxygen atom = 16 u and x₂ = position of oxygen atom = distance between molecules = 1.13 × 10⁻¹⁰ mSubstituting the values of the variables into the equation, we have
x = (m₁x₁ + m₂x₂)/(m₁ + m₂)
x = (12 u × 0 m + 16 u × 1.13 × 10⁻¹⁰ m)/(12 u + 16 u)
x = (0 + 16 u × 1.13 × 10⁻¹⁰ m)/28 u
x = 18.08 × 10⁻¹⁰ um/28 u
x = 0.646 × 10⁻¹⁰ m
x ≅ 0.65 × 10⁻¹⁰ m
The distance between the center of mass of the molecule and carbon atom
Since the carbon atom is at x₁ = 0 m and the center of mass is at x = 0.65 × 10⁻¹⁰ m, the distance between the carbon atom and the center of mass of the molecule is d = x - x₁
= 0.65 × 10⁻¹⁰ m - 0 m
= 0.65 × 10⁻¹⁰ m
So, the distance between the center of mass of the molecule and carbon atom is 0.65 × 10⁻¹⁰ m
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A sheet of custom-size copy paper measures 3.5 in. by 14 in. If a ream (500 sheets) of this paper has a volume of 98 in. cubed, how thick is the ream?
Answer:
2 in
Explanation:
Area of a sheet = 3.5 x 14 in²
Volume of ream, V = 98 in³
Let t be the thickness of ream
So, Volume of ream Area of a sheet x thickness of ream
98 = 3.5 x 14 x t
t = 2 in
Thus, the thickness of ream is 2 in.
Answer:
t = 2 in
Explanation:
given,
width of sheet, W = 3.5 in
Length of the sheet, L = 14 in
thickness of the ream = ?
volume of paper = 98 in³
we now,
Volume = Length x width x thickness
V = l w t
98 = 14 x 3.5 x t
[tex]t = \dfrac{98}{14\times 3.5}[/tex]
t = 2 in
thickness of the ream is equal to 2 in.
The time in hours it takes a satellite to complete an orbit around the earth varies directly as the radius of the orbit (from the center of the earth) and inversely as the orbital velocity. If a satellite completes an orbit 660 miles above the earth in 13 hours at a velocity of 32,000 mph, how long would it take a satellite to complete an orbit if it is at 1500 miles above the earth at a velocity of 38,000 mph?
To find the new orbital period of a satellite at 1500 miles above Earth with a velocity of 38,000 mph, we use a proportional relationship comparing it to a known orbit. We set up the ratio using the direct variation with the orbit's radius and inverse variation with the orbital velocity, then solve for the unknown orbital period.
Explanation:The duration of a satellite's orbit around Earth can be described as varying directly with the orbit's radius and inversely with the orbital velocity. To calculate the orbital period for a new satellite trajectory, we need to compare it with a known situation using a proportional relationship. Given that a satellite orbits 660 miles (or 1062 kilometers) above Earth in 13 hours at 32,000 mph, we can set up a ratio to determine the orbital period at 1500 miles (or 2414 kilometers) above Earth with a velocity of 38,000 mph.
The ratio for the original orbit is T1 / (r1 / v1) = T2 / (r2 / v2), where T is time, r is radius (distance from Earth's center to the satellite), and v is velocity. Using the provided values, we get:
13 / ((3963 + 660) / 32000) = T2 / ((3963 + 1500) / 38000)
Solving for T2 gives us the new orbital period. Considering that we should convert the altitude to the same units and use Earth's radius in miles (3963 miles), the final calculation will provide us with the new orbital period in hours.
If you are caught outdoors in a thunderstorm, why shouldn’t you stand under a tree? Can you think of a rea- son why you should not stand with your legs far apart? Or why lying down can be dangerous? (Hint: Consider the electric potential difference.)
Answer:
Because its dangerous.
Explanation:
During lighting strikes, there is discharge or energy transfer(electrons) from the clouds to the earth, these electrons flow through the path with least resistance between the cloud and the earth. Also the electric field around the tip of the leaves are strong, which makes trees a great target.
we as humans have lower resistance than trees, that is to say, the lighting may leave the tree and flow trough the body to the earth.
The tree and the ground around it are then raised to a high potential relative to the ground some distance away.
If you stand with your legs far apart, one leg on a higher-potential part of the ground than the other, or if you lie down with a potential difference between your head and your feet, you may find yourself a conducting path.
If it is also raining, the electricity may transfer down the wet tree to the wet ground and shock anyone standing near the tree.
During a thunderstorm, it is advised not to stand under a tree to avoid being struck by lightning, not to stand with legs far apart to reduce the electric potential difference, and not to lie down due to increased risk. Staying inside a car provides safety as it acts as a Faraday cage.
Explanation:Standing under a tree during an electrical storm is dangerous since lightning tends to strike the tallest object in an area, which could be the tree you're under, potentially causing serious injury or death. As for not standing with legs apart, this is because, in the event of a ground strike, electricity can travel through the ground. If your legs are far apart, there could be a significant electric potential difference between them, which can result in a stronger current passing through your body, leading to severe harm. Lying down increases your contact with the ground, and consequently, the risk of current flowing through your body from a ground strike is greater.
During thunderstorms, your car acts as a Faraday cage, which shields you from electric fields if a lightning strike occurs nearby. It's safest to remain inside the car with windows closed. However, it's critical to refrain from touching metal parts inside the car as lightning can transfer its charge through the car's metal frame.
You have a set of calipers that can measure thicknesses of a few inches with an uncertainty of ± 0.005 inches. You measure the thickness of a deck of 52 cards and get 0.590 in: a. If you now calculate the thickness of 1 card, what is your answer, including its uncertainty?
Answer:
The thickness of 1 card is between 0.006in and 0.016in
Explanation:
Thickness of a deck of 52 cards = 0.590in
Thickness of 1 card = 0.590in/52 = 0.011in
Uncertainty = + or - 0.005in
Lower limit of thickness of 1 card = 0.011in - 0.005in = 0.006in
Upper limit of thickness of 1 card = 0.011in + 0.005in = 0.016in
Therefore, thickness of 1 card is between 0.006in and 0.016in
A +7.5 nC point charge and a -2.0 nC point charge are 3.0 cm apart. What is the electric field strength at the midpoint between the two charges?
Answer:
Ep= 3.8 10⁵ N/C
Explanation:
Conceptual analysis
The electric field at a point P due to a point charge is calculated as follows:
E = k*q/d²
E: Electric field in N/C
q: charge in Newtons (N)
k: electric constant in N*m²/C²
d: distance from charge q to point P in meters (m)
Equivalence
1nC= 10⁻⁹C
1cm= 10⁻²m
Data
k= 9*10⁹ N*m²/C²
q₁ =+7.5 nC = +7.5*10⁻⁹C
q₂ = -2.0 nC = -2.0*10⁻⁹C
d₁ =d₂ = 1.5cm = 1.5 *10⁻²m = 0.015 m
Calculation of the electric fieldsat the midpoint (P) between the two charges
Look at the attached graphic:
E₁: Electric Field at point ;Due to charge q₁. As the charge q₁ is positive negative (q₁+), the field leaves the charge .
E₂: Electric Field at point : Due to charge q₂. As the charge q₂ is negative (q₂-) ,the field enters the charge
E₁ = k*q₁/d₁² = 9*10⁹ *7.5 *10⁻⁹/ ( 0.015 )² = 3*10⁵ N/C
E₂ = k*q₂/d₂²= 9*10⁹ *2*10⁻⁹/( 0.015 )² = 0.8*10⁵ N/C
The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.
Ep= E₁ + E₂
Ep= 3*10⁵ N/C + 0.8*10⁵ N/C
Ep= 3.8 10⁵ N/C
The electric field strength at the midpoint between a +7.5 nC point charge and a -2.0 nC point charge that are 3.0 cm apart is 1.56 *10^6 N/C (Newtons per Coulomb) away from the positive charge.
Explanation:The question is asking for the electric field at the midpoint between two point charges. The formula for the electric field created by a point charge is given by E=kQ/r^2, where E is the electric field, k is Coulomb's constant (approximately 9.0 * 10^9 N*m^2/C^2), Q is the charge, and r is the distance from the charge. In this scenario we have two charges, so we calculate the electric field created by each charge and sum the results.
Applying this to both point charges and summing the electric fields we obtain: E_total = |E1| + |E2| = k*|Q1|/d^2 + k*|Q2|/d^2 = ((9.0*10^9 N*m^2/C^2) * 7.5 *10^-9 C/0.015m^2) + ((9.0*10^9 N*m^2/C^2) * 2 *10^-9 C/ 0.015m^2) = 1.8 *10^6 N/C and -0.24 * 10^6 N/C respectively. The total electric field strength at the midpoint is the sum which equals 1.56 *10^6 N/C away from the positive charge.
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You break a piece of Styro foam packing material, and it releases lots of little spheres whose electric charge makes them stick annoyingly to you. If two of the spheres carry equal charges and repel with a force of 22 { mN} when they're 16 { mm} apart, what's the magnitude of the charge on each?
Answer:
The magnitude of charge on each sphere is [tex]q=2.50\times 10^{-8}\ C[/tex].
Explanation:
Given that,
Force of repulsion between the charges, F = 22 mN
The distance between spheres, r = 16 mm = 0.016 m
It is mentioned that both the spheres carry equal charges. The force between charges is given by :
[tex]F=\dfrac{kq^2}{r^2}[/tex]
[tex]q=\sqrt{\dfrac{Fr^2}{k}}[/tex]
[tex]q=\sqrt{\dfrac{22\times 10^{-3}\times (0.016)^2}{9\times 10^9}}[/tex]
[tex]q=2.50\times 10^{-8}\ C[/tex]
So, the magnitude of charge on each sphere is [tex]q=2.50\times 10^{-8}\ C[/tex]. Hence, this is the required solution.
50 J of work was performed in 20 seconds. How much power was used to perform this task? A. 0.4 W B. 2.5 W C. 4 W D. 24.5 W
Answer:
B. 2.5 W
Explanation:
Power: This can be defined as the rate at which energy used or it i the rate at which work is done. The S.I unit of power is Watt (W).
Mathematically,
Power = Energy or work/Time
P = W/t ........................Equation 1
Where P = power used to perform the task, W = work, t = time.
Given: W = 50 J, t = 20 s.
Substitute into equation 1
P = 50/20
P = 2.5 W.
Hence the power is 2.5 W.
The right option is B. 2.5 W
The power used to perform the work of 50J in 20 seconds is 2.5 watts. This much power should be generated by it. The correct answer is B.
Given that 50 J of work was performed in 20 seconds,
The power used to perform a task, you can use the formula:
Power = Work / Time
Power = 50 J / 20 s
Power = 2.5 W
Therefore, the power used to perform the task is 2.5 W.
Therefore, the correct answer is B) 2.5 W.
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Most active galactic nuclei are at large distances from us; relatively few nearby galaxies harbor active galactic nuclei. What does this imply?
Answer:
The milky way does not have an active galactic nuclei and active galactic nuclei reduce their activity as time pass.
Explanation:
An active galactic nucleus (AGN) is a region that is compact and at the galaxy's center. It has a extra high luminosity in which some observations indicate that the luminosity is not as a result of the presence of stars.
Suppose a 58-N sled is resting on packed snow. The coefficient of kinetic friction is 0.11. If a person weighing 655 N sits on the sled, what force is needed to pull the sled across the snow at constant speed?
Answer:
You have to apply force of 78.43 N to pull the sled or person across the snow at constant speed
Explanation:
Given data
Sled force F₁=58 N
Person weight F₂=655 N
Coefficient of kinetic friction u=0.11
To find
Friction Force
Solution
To pull the sled across the snow means the force you apply must be equal to and opposite the force of frictional.
First we need to find the Total normal force
So
[tex]F_{Normal-Force}=F_{1}+F_{2}\\F_{Normal-Force}=58N+655N\\F_{Normal-Force}=713N[/tex]
Now the force of friction
[tex]F_{Friction-Force}=u_{coefficient}*F_{Normal-Force}\\F_{Friction-Force}=0.11*713N\\F_{Friction-Force}=78.43N[/tex]
So you have to apply force of 78.43 to pull the sled or person across the snow at constant speed
7. Two children of mass 20 kg and 30 kg sit balanced on a seesaw with the pivot point located at the center of the seesaw. If the children are separated by a distance of 3 m, at what distance from the pivot point is the small child sitting in order to maintain the balance?
Answer:
Explanation:
Given
mass of children [tex]m_1=20\ kg[/tex]
[tex]m_2=30\ kg[/tex]
distance between two children [tex]L=3\ m[/tex]
suppose small child is at a distance of x m from pivot point
so torque of small child and heavier child must be equal
[tex]20\times (x)=30\times (3-x)[/tex]
[tex]2x=9-3x[/tex]
[tex]5x=9[/tex]
[tex]x=1.8\ m[/tex]
A solar eclipse that occurs when the new moon is too far from Earth to completely cover the Sun can be either a partial solar eclipse or a(n)_____________.
Answer:
ANULAR ECLIPSE
Explanation:
ANULAR ECLIPSE. Because the moon is very far, only a portion of the sun would be obscured, and then only the moon's outer ring will be viewable; this is called the anular eclipse.The ring of fire marks the maximum stage of an annular solar eclipse.
Describe the weather in orlando today and how is it different than describing florida’s climate.
Answer:it is cold
Explanation:because of a winter storm it is very cold which doesn’t
Match Florida’s tropical warm climate
Weather refers to short-term atmospheric conditions while climate refers to long-term predictable conditions. Weather in Orlando today depicts the present conditions, whereas Florida's climate represents the long-term averages and patterns across the state.
Explanation:The weather in Orlando today simply refers to the current atmospheric conditions in Orlando, such as temperature, humidity, precipitation, wind, etc. It's a mere snapshot of what's happening right now in the atmosphere around Orlando. Now, when we're talking about Florida's climate, we're referring to the long-term atmospheric conditions in the state of Florida. This includes consistent seasonal temperature and rainfall patterns over many years. Therefore, the primary difference between describing the weather in Orlando today and Florida's climate lies in the timeframe – weather explains temporary conditions, while climate refers to long-term, predictable conditions.
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heather and matthew walk with an average velocity of 98 m/s eastward. If it takes them 34 min ro walk to the store, what is their displacement?
Answer:
The answer is 199920 meters.
Explanation:
First we need to find how many seconds they walked to the store:
[tex]34*60=2040[/tex]
If they walked 2040 seconds to eastward with velocity of 89 m/s, the displacement will be:
[tex]2040*98=199920[/tex]