How fast must a 3000 kg elephant move to have the same kinetic energy as a 65 spinter running at 10ms?

Answers

Answer 1

Q. : How fast must a 3000 kg elephant move to have the same kinetic energy as a 65.0 kg sprinter running at 10m/s?

Answer:

The elephant must be 1.47 m/s fast to have the same kinetic energy with the sprinter

Explanation:

Kinetic Energy: This is the energy of a body in motion. It can be expressed mathematically as

Ek = 1/2mv²

Since the kinetic energy the elephant = the kinetic energy of the sprinter.

1/2m₁v₁² = 1/2m₂v₂².......................... Equation 1

Making v₁ the subject of the equation,

v₁ = √(m₂v₂²/m₁)................................. Equation 2

Where v₁ = speed of the elephant, m₁ = mass of the elephant, v₂ = speed of the sprinter, m₂ = mass of the sprinter.

Given: m₁ = 3000 kg, m₂ 65 kg, v₂ = 10 m/s

Substituting these values into equation 2,

v₁ =√(65×10²/3000)

v₁ = [tex]\sqrt{(6500)/3000[/tex]

v₁ =[tex]\sqrt{2.167}[/tex]

v = 1.47 m/s

Therefore The elephant must be 1.47 m/s fast to have the same kinetic energy with the sprinter


Related Questions

When you throw a ball up in the air, it travels up and then stops instantaneously before falling back down. At the point where it stops and changes directions to fall back down is

Answers

Answer:

The ball stops instantaneously at the topmost point of the motion.

Explanation:

Assume we have thrown a ball up in the air. For that we have given a force on the ball and it acquires an initial velocity in the upward direction.

The forces that resist the motion of the ball in the upward direction are the force of gravity and air resistance. The ball will instantaneously come to rest when the velocity of the ball reduces to zero.

The two forces acting in the downward direction reduces its speed continuously and it becomes zero at the topmost point.

Final answer:

The highest point of a ball's upward trajectory is when its velocity is zero, and it momentarily stops before falling back to Earth under the influence of gravity. This is the point where the ball has its maximum potential energy and is the peak of the trajectory.

Explanation:

When you throw a ball up in the air, it travels upwards and slows down under the influence of gravity. At the highest point of its trajectory, the ball's velocity momentarily becomes zero before it reverses direction and falls back to the ground. This moment is known as the peak or the apex of the ball's trajectory. The ball's vertical velocity increases in the downward direction as it descends due to the acceleration caused by gravity. At the peak, not only does the velocity become zero, but this is also the point where the ball has its maximum potential energy.

When calculating the time it takes for the ball to reach its highest point or how high it goes, you can use the initial velocity and the acceleration due to gravity. For example, if a ball is thrown upwards with a velocity of 10 m/s, you can calculate the maximum height using the formula for the displacement under constant acceleration, considering that the final velocity at the highest point is zero and the acceleration due to gravity is -9.8 m/s² (negative as it is in the opposite direction to the ball's initial motion).

A bullet with mass 4.53 g is fired horizontally into a 1.953-kg block attached to a horizontal spring. The spring has a constant 5.73 102 N/m and reaches a maximum compression of 6.20 cm. Find the initial speed of the bullet-block system.

Answers

Answer:

The initial speed of the bullet-block system is 113 m/s

Explanation:

Hi there!

Let´s apply the theorem of conservation of energy. If we neglect friction, all the kinetic energy of the bullet-block system is used to compress the spring. In other words, the kinetic energy is converted into elastic potential energy. Then, the initial kinetic (KE) energy of the bullet-block system will be equal to the final elastic potential energy (EPE) of the spring:

KE = EPE

1/2 · m · v² = 1/2 · k · x²

Where:

m = mass of the bullet-block system.

v = initial speed of the bullet-block system.

k = spring constant.

x = compression of the spring

Then, solving for v:

v² = k · x² / m

v² = 5.73 × 10² N/m · (0.620 m)² / 1.957 kg

v = 113 m/s

As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff springs arranged side by side so that they are parallel to each other. When you push the platform, you compress the springs. You do an amount of work of 85.0 J when you compress the springs a distance of 0.250 m from their uncompressed length. What magnitude of force must you apply to hold the platform in this position? How much additional work must you do to move the platform a distance 0.220 m farther? What maximum force must you apply to move the platform a distance 0.220 m farther?

Answers

Answer:

a) F = 680 N, b)  W = 215 .4 J , c)  F = 1278.4 N

Explanation:

a) Hooke's law is

              F = k x

To find the displacement (x) let's use the elastic energy equation

            [tex]K_{e}[/tex] = ½ k x²

             k = 2 [tex]K_{e}[/tex]  / x²

             k = 2 85.0 / 0.250²

             k = 2720 N / m

We replace and look for elastic force

            F = 2720  0.250

            F = 680 N

b) The definition of work is

          W = ΔEm

          W = [tex]K_{ef}[/tex] - [tex]K_{eo}[/tex]

          W = ½ k ( [tex]x_{f}[/tex]² - x₀²)

The final distance

         [tex]x_{f}[/tex] = 0.250 +0.220

        [tex]x_{f}[/tex] = 0.4750 m

We calculate the work

          W = ½ 2720 (0.47² - 0.25²)

          W = 215 .4 J

We calculate the strength

          F = k [tex]x_{f}[/tex]

          F = 2720 0.470

          F = 1278.4 N

Which of the following statements is true?
1. The 2s orbital in the hydrogen atom is larger than the 3s orbital also in the hydrogen atom.
2. The hydrogen atom has quantized energy levels.
3. The Bohr model of the hydrogen atom has been found to be incorrect.
4. An orbital is the same as a Bohr orbit.
5. The third energy level has three sublevels - the s, p, and d sublevels.

Answers

Answer:

2. The hydrogen atom has quantized energy levels.

Explanation:

The Bohr model of the atom states that the structure of the atom is quantized, that is, that electrons can only orbit the nucleus in specific orbits with a fixed radius. Therefore, the electron cannot be in energy levels that do not correspond to these quantized levels.

A centripetal force of [tex]F_c[/tex] acts on a car going around a curve. If another car goes around the same curve but with twice the speed, the centripetal force acting on that car is ___________.

Answers

Answer:

The centripetal on the car will become 4 times when the velocity gets twice.

Explanation:

As we know that centripetal force on the car of mass m and moving with constant speed v given as

[tex]F_c=\dfrac{mv^2}{r}[/tex]

m=mass

v=velocity

r=radius of the circular arc

We are assuming that the mass of the both the car is same.

If the velocity of the car gets twice 2 v

The new centripetal force on the car

[tex]F_c'=\dfrac{m(2v)^2}{r}[/tex]

[tex]F_c'=4\dfrac{mv^2}{r}[/tex]

[tex]F_c'=4\ F_c[/tex]

Therefore we can say that centripetal on the car will become 4 times when the velocity gets twice.

What is the distance from the moon to the point between Earth and the Moon where the gravitational pulls of Earth and Moon are equal? The mass of Earth is 5.97 x 10^24 kg, the mass of the Moon is 7.35 x 10^22 kg, the distance between Earth and the Moon is 3.84 x 10^8 m, and >G= 6.67x10^-11N x m^2/kg^2
A)3.83 x 10^6 mB)3.83 x 10^7 mC)4.69 x 10^6 mD)4.69 x 10^7 mE)3.45 x 10^8 m

Answers

Answer:

the point between Earth and the Moon where the gravitational pulls of Earth and Moon are equal is E)3.45 × 10⁸ m

Explanation:

The force that the Earth exerts on a mass m is

F_e = (G M_e m) / R_e²

where

G is the universal gravitational constantM_e is the mass of EarthR_e is the radius of Earth

The force that the Moon exerts on a mass m is

F_m = (G M_m m) / R_m²

where

G is the universal gravitational constantM_m is the mass of the MoonR_m is the radius of the Moon

Therefore, the point where the gravitational pulls of Earth and Moon are equal is:

F_e = F_m

R_e + R_m = R = 3.84×10⁸ m

Thus,

(G M_e m) / R_e² = (G M_m m) / R_m²

M_e / R_e² = M_m / (R - R_e²)

(R - R_e²) / R_e² = M_m / M_e

(R - R_e) / R_e = (M_m / M_e)^1/2

R_e(R/R_e -1) / R_e = (M_m / M_e)^1/2

R/ R_e = (M_m / M_e)^1/2 + 1

R_e = R / [(M_m / M_e)^1/2 + 1]

R_e = (3.84×10⁸ m) / [(7.35 x 10²² kg / 5.97 x 10²⁴ kg )^1/2 + 1]

R_e = 3.45 × 10⁸ m

Therefore, the  point between Earth and the Moon where the gravitational pulls of Earth and Moon are equal is 3.45 × 10⁸ m.

A green object will absorb ____________________ light and reflect ____________________ light.

Answers

Answer:Magenta , Green

Explanation:

A green object will absorb its complementary color i.e. Magenta (mixture of red and blue) and thus reflect the remaining green color.

This can be understood by the fact that a particular light filter absorb its complementary color and reflects its color for example

Blue filter will absorb yellow light (mixture of red and green)and reflects blue color.

Final answer:

A green object absorbs light in the red portion of the visible spectrum and reflects green light, as the observed color is complementary to the color that is absorbed.

Explanation:

The question is asking which colors of light a green object will absorb and which it will reflect. A green object will absorb light in the red portion of the visible spectrum. This is due to the fact that the color observed from an object comes from the light that is transmitted or reflected, not the light that is absorbed. The reflected or transmitted light is always complementary in color to the light that is absorbed. Thus, for a green object, the absorbed light is predominantly red while it reflects green light.

An example that helps explain this concept is how different complexes absorb light depending on their ligand field strength and the structure of the complex, which influences the observable color. For instance, the complex [Cr(NH3)6]3+ has strong-field ligands which cause it to absorb high-energy photons corresponding to blue-violet light, making the complex appear yellow, which is the complementary color to blue-violet.

A mass m is oscillating with amplitude A at the end of a vertical spring of spring constant k. The mass is increased by a factor of four.1. What is true about the period? (period does what factor)2. The mass is increased by a factor of four, keeping A and k constant. What is true about the total mechanical energy?3. The mass is increased by a factor of four, keeping A and k constant. What is true about the maximum speed?4. The amplitude is doubled, keeping mass and k constant. What is true about the period?5. The amplitude is doubled, keeping mass and k constant. What is true about the total mechanical energy? Choose the best answer.6. The amplitude is doubled, keeping mass and k constant. What is true about the maximum speed?

Answers

Answer:

1. The period doubles

2. The mechanical energy is unchanged

3. The speed is halved

4. The period is unchanged

5. The energy is quadrupled

6. The maximum is speed doubled

Explanation:

1.  From Hooke's law we have [tex]T\propto\sqrt{\frac{m}{k} }[/tex]

where T is period, m is mass and k is the spring constant

So If the mass is increased by a factor of 4 then period doubles while k constant

2 According to law of conservation of energy, the energy remains unchanged so therefore the total mechanical energy remain unchanged.

3. From the  [tex]w = \sqrt{\frac{k}{m} }[/tex]

where m is  mass and w is speed

so we see that mass and speed is inversely proportional therefore if we increase mass speed decreases.

4. The period is independent of amplitude

5. [tex]E=A^{2} [/tex]

where E is energy and A is amplitude

So if the amplitude is doubled the energy is quadrupled.

6.  we have the relation

[tex]v_{max}=A [/tex]

where [tex]v_{max}[/tex] is maximum speed and A is amplitude  

From the formular we can see they are directly proportional, so if we double amplitude then [tex]v_{max}[/tex] doubles also.

The period would double when the mass of the spring is increased by a factor of four (4).

How to calculate the period.

Since all springs obey Hooke’s law, the period is given by this formula:

[tex]T\alpha \sqrt{\frac{m}{k}}\\\\T = 2\pi \sqrt{\frac{m}{k} }[/tex]

Where:

k is the spring constant.m is the mass.

We can deduce that, the period is directly proportional to mass of the spring. Thus, the period would double when the mass of the spring is increased by a factor of four (4).

In accordance with the law of conservation of energy, the total mechanical energy of the spring would remain the same because energy can neither be created nor destroyed.

How to calculate the maximum speed.

The speed of a spring in simple harmonic motion is an angular speed and it is given by this formula:

[tex]\omega = \sqrt{\frac{k}{m} }[/tex]

We can deduce that, the angular speed is inversely proportional to mass of the spring. Thus, the angular speed would decrease when the mass of the spring is increased by a factor of four (4).

In simple harmonic motion, the period of a spring is independent of its amplitude.

How to calculate the total mechanical energy.

Mathematically, the the total mechanical energy of a spring is given by this formula:

[tex]E=A^2[/tex]

We can deduce that, the total mechanical energy is directly proportional to the square of amplitude of the spring. Thus, the total mechanical energy would quadruple when the amplitude of the spring is doubled.

Also, the maximum speed is directly proportional to the amplitude of the spring. Thus, the maximum speed would also double when the amplitude of the spring is doubled.

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dante is leading a parade across the main street in front of city hall. Starting at city hall, he marches the parade 4 blocks east, then 3 blocks south. From there, the parade marches 1 block west and 9 blocks north and finally stops. What is the vector displacement and direction of the the parade, starting from the city hall and the stopping point? (1 point)

Displacement: 6.71 m, Direction: 63.4 degrees north of east

Displacement: 8.01 m, Direction: 21.9 degrees north of east

Displacement: 2.56 m, Direction: 39.7 degrees north of east

Displacement: 4.31 m, Direction: 88.1 degrees north of east

Answers

Answer:

The correct option is A)

Displacement: 6.71 m, Direction: 63.4 degrees north of east

Explanation:

Given that Dante is leading a parade across the main street in front of city hall.

Let, Initial location of parade is 0i+0j

One block of city is one units on the XY- graph

Statement 1: Parade marches the parade 4 blocks east, then 3 blocks south

New location of parade is 4i-3j

Statement 2: The parade marches 1 block west and 9 blocks north and finally stops.

Final location of parade is (4i-3j)+(-1i+9j)=3i+6j

Displacement is given by

Displacement = (Final destination)-(Initial destination)

Displacement = (3i+6j)-(0i+0j)=3i+6j

Thus,

Magnitude of displacement = [tex]\sqrt{3^{2}+6^{2}}[/tex]

                                              = 6.71 m

Direction of displacement =  [tex]tan^{-1}(\frac{Y}{X} )[/tex]

                                           =  [tex]tan^{-1}(\frac{6}{3} )[/tex]

                                           = 63.43 NE

Therefore, the correct option is A) Displacement: 6.71 m, Direction: 63.4 degrees north of east

One long wire lies along an x axis and carries a current of 43 A in the positive x direction. A second long wire is perpendicular to the xy plane, passes through the point (0, 5.9 m, 0), and carries a current of 41 A in the positive z direction. What is the magnitude of the resulting magnetic field at the point (0, 1.7 m, 0)?

Answers

Answer:

Bnet=1.006*10^-6T

Explanation:

One long wire lies along an x axis and carries a current of 43 A in the positive x direction. A second long wire is perpendicular to the xy plane, passes through the point (0, 5.9 m, 0), and carries a current of 41 A in the positive z direction. What is the magnitude of the resulting magnetic field at the point (0, 1.7 m, 0)?

the magnetic field Bnet=[tex]\sqrt{b1^2+b2^2}[/tex]

the magnetic field due this long wire is given by

B1=∨I1/[tex](2\pi *R1)[/tex]..............................1

B2=∨I2/[tex](2\pi *R2)[/tex]............................2

Bnet=[tex]\sqrt{(vI1/2*pi*R1)^2+(vI2/2*pi*R2)^2}[/tex].......................3

Bnet=v/2*pi[tex]\sqrt{(I1/R1)^2+(i2/R2)^2}[/tex]

Bnet=4*pi*10^-7/(2[tex]\pi[/tex])[tex]\sqrt{(43/1.7)^2+(41/29.5)^2}[/tex]

Bnet=0.0000002*(641.72)^.5

Bnet=1.006*10^-6T

A 10 kilogram object suspended from the end of a vertically hanging spring stretches the spring 9.8 centimeters. At time t=0, the resulting mass-spring system is disturbed from its rest state by the force F(t)=140cos(8t). The force F(t) is expressed in Newtons and is positive in the downward direction, and time is measured in seconds.

Answers

Answer:

K= 1000 N-m

Explanation:

It is assumed that we asked to find the spring constant k of the spring

We know that under equilibrium condition

weight of the object = force applied by the spring

given m =10 Kg

x= extension in the spring = 9.8 cm

mg=kx

[tex]10\times9.8=k\times9.8\times10^(-2)[/tex]

K= 1000 N-m

The spring constant, k of the vertically suspended spring measured in Newton meter is 1000 N/m

Given the Parameters :

Acceleration due to gravity, g = 9.8 m/s²

Mass of object, m = 10 kg

Extension, e = 9.8 centimeters

Using the Relation :

F = ke ; k = spring constant

Force, F = mass × acceleration due to gravity

The expression can be written thus :

mg = ke

Converting, extension to meters = (9.8/100) = 0.098 m

(10 × 9.8) = 0.098k

98 = 0.098k

k = 98/0.098

k = 1000 N/m

Therefore, the spring constant is 1000 N/m

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A person is straining to lift a large crate, without success because it is too heavy. We denote the forces on the crate as follows: P is the upward force the person exerts on the crate, C is the vertical contact force exerted on the crate by the floor, and W is the weight of the crate. How are the magnitudes of these forces related while the person is trying unsuccessfully to lift the crate?

Answers

Answer:

P + C = W if in equilibrium?

P + C - W = 0; P + C = W

Explanation:

Action and reaction are equal and opposite according to newton's third law of motion. a force is that which tends to change a body's state of rest or uniform motion in a straight line, recall

P is the upward force the person exerts on the crate,

C is the vertical contact force exerted on the crate by the floor,

W is the weight of the crate

let the sum of upward forces be on the left hand side of the equation and the sum of downward forces on the right hand side.

P+C=W

P+C-W=0 if they are in equilibrium

the forces are related by the above.

Final answer:

When a person is unsuccessfully attempting to lift a crate, the forces acting on the crate are balanced. The sum of the upward force (P) applied by the person and the contact force (C) from the floor equals the weight (W) of the crate, represented by the equation P + C = W.

Explanation:

In the situation described, the person is trying to lift a crate but is unable to do so. This means that the forces acting on the crate are in equilibrium, or balanced, as there is no resulting upward or downward motion of the crate. In this case, the force exerted by the person (P) and the vertical contact force from the floor (C) is opposing the weight of the crate (W).

In terms of magnitudes, the sum of P and C equals the weight W. This is based on the principle of equilibrium which states that an object at rest has equal and opposite forces acting on it. Therefore, P + C = W. This means the upward forces are equal to the downward force, hence the person can't lift the crate successfully.

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Which of the following types of light cannot be studied with telescopes on the ground?a. Visible light b. X-rays c. Radio waves d. All of the above

Answers

Answer:

Which of the following types of light cannot be studied with telescopes on the ground?

The Answer is X-rays

Explanation:

Earth’s atmosphere blocks most of the radiation from space preventing some electromagnetic spectrum from reaching the Earth because they are absorbed or reflected by the Earth's atmosphere. Visible light and radio waves get through to telescopes on the ground, while X-rays are absorbed by most molecules in the Earth’s atmosphere making it visible only from above the atmosphere.

Some Terms Explained:

Telescope:  is an optical instrument that makes distant objects appear magnified. The purposes of a telescope are to gather light using either a lens or a mirror and resolve detail. There are two basic types of telescopes, refractors and reflectors.

Visible light: covers the range of wavelengths from 400–700 nanometers.

X-rays: range in wavelength from 0.001–10 nanometers. They are shorter in wavelength than Ultraviolet rays and longer than gamma rays. They will pass through most substances, and this makes them useful to see inside things.

Radio waves: the longest waves, longer than 1 meter they have the lowest energy. It is used for long distance communication.

Final answer:

X-rays cannot be studied with telescopes on the ground due to Earth's atmosphere blocking them. Space telescopes are needed for such observations.

Explanation:

The type of light that cannot be studied with telescopes on the ground is X-rays. Ground-based telescopes can effectively observe visible light and radio waves, but X-rays are mostly blocked by Earth's atmosphere. To study X-ray emissions, astronomers use space telescopes. Other ranges of the electromagnetic spectrum, such as ultraviolet and gamma rays, are similarly studied using telescopes positioned in space to bypass atmospheric interference.

Here are the names of some substances: Sulphur, copper, oxygen, iron, water, magnesium, mercury Which substance: Is at gas room temperature? Is a liquid metal? Is a solid non-metal? is a chemical compound?

Answers

Answer:

Gas at room temperature: Oxygen

Liquid metal: Mercury

Solid non-metal: Sulphur, Copper

Chemical compound: Water

Explanation:

Oxygen(O): atomic no.-8

                    It is highly reactive non-metal and third most abundant element   in the universe after hydrogen and helium. Diatomic oxygen(O₂) constitutes 20.8% of earth atmosphere that used in respiration in living organisms.

Mercury(Hg): atomic no.-80

                    It is shiny, silvery and only metal which is found in liquid state at normal room temperature. Most common use of mercury is in thermometer, barometer, sphygmomanometer etc.

Sulphur(S): atomic no.-16

                    It is bright yellow crystalline solid non-metal at room temperature in elemental form. Combined sulphur exists in sulphates and sulphides i.e. CaSO₄.2H₂o, MgSO₄.7H₂O, PbS, ZnS etc. Abundance in earth crust is only 0.03-0.1%.

Copper(Co): atomic no.-29

                   It is pinkish-orange, malleable and ductile metal found in solid state at room temperature. It is commonly used in making wire, alloy etc.

Water(H₂O):

                   Water covers 71% of earth surface. A molecule contains one oxygen and two hydrogen atoms connected by covalent bonds. It is crucial for all forms of life.

A coin is placed 32 cm from the center of a horizontal turntable, initially at rest. The turntable then begins to rotate. When the speed of the coin is 110 cm/s (rotating at a constant rate), the coin just begins to slip. The acceleration of gravity is 980 cm/s^2 . What is the coefficient of static friction between the coin and the turntable?

Answers

Answer:

0.39

Explanation:

distance from the center (r) = 32 cm = 0.32 m

speed of the coin (v) = 110 cm/s = 1.1 m/s

acceleration due to gravity (g) = 980 m/s^{2} = 9.8 m/s^{2}

find the coefficient of static friction (k) between the coin and the turn table

frictional force = kmg

before the table begins to move, the frictional force balances the centripetal force ([tex]\frac{mv^{2} }{r}[/tex])

therefore

frictional force = centripetal force

kmg = [tex]\frac{mv^{2} }{r}[/tex]

kg = [tex]\frac{v^{2} }{r}[/tex]

k = [tex]\frac{v^{2} }{r}[/tex] ÷ g

k = [tex]\frac{1.1^{2} }{0.32}[/tex] ÷ 9.8 = 0.39

Anti-lock brakes (ABS) __________
A. prevent skidding and allows drivers to steer during an emergency braking situation.
B. sense a skid and automatically apply the brakes.
C. stop the vehicle faster than normal brakes.
D. all of the above

Answers

Answer:

A. prevent skidding and allows drivers to steer during an emergency braking situation.

Explanation:

In an urgent stopping case, anti-lock brakes avoid skidding and allow drivers to steer. ABS can also help improve automobile balance (evitating spinouts), steering (that is directing the vehicle where the driver needs it to go) and braking (distance needed to stop the vehicle). Hence, the correct answer is A.

Final answer:

Anti-lock brakes (ABS) primarily prevent skidding and allow drivers to steer during an emergency braking situation. They do not necessarily stop the vehicle faster. The main benefit of ABS is that it enables steering while braking.

Explanation:

The correct answer to your question is A. Anti-lock brakes (ABS) primarily prevent skidding and allow drivers to maintain steering control during an emergency braking situation. When the system senses a wheel is about to lock up and skid, it automatically modulates brake pressure to that wheel. This process repeats rapidly, often several times per second.

Note that ABS does not necessarily stop the vehicle faster. In fact, in certain conditions like gravel or snow, stopping distances can be longer. However, the crucial advantage is that ABS allows you to steer while braking, potentially avoiding a collision.

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An ice-skater is moving at a constant velocity across an icy pond, The skater throws a snowball directly ahead.
Which of the following correctly describes the velocity of the center of mass of the skater-snowball system immediately after the snowball is thrown?
Assume friction and air resistance are negligible.
(A) It is equal to the velocity of the snowball.
(B) It is equal to the new velocity of the skater.
(C) It is equal to half the original velocity of the skater.
(D) It is equal to the original velocity of the skater.

Answers

Answer:

(D) It is equal to the original velocity of the skater.

Explanation:

The velocity of the center of mass of a system is

[tex]\vec{v}_{cm} = \frac{m1\vec{v}_1 + m_2\vec{v}_2}{m_1 + m_2}[/tex]

The velocity of the center of mass is constant if there is no external force, because the total momentum of the whole system is conserved.

So, before the snowball is thrown, the velocity of the center of mass is equal to that of the skater. This velocity will always be equal to the velocity of the center of mass of the system.

Immediately after the snowball is thrown, the velocity of the center of mass is equal to that of the skater. Option D is correct.

Velocity of the center of mass:

It is defined as the ratio of the sum of the momentum of the masses and total mass.

The velocity of the center of mass of a system given as

[tex]\bold {V_c_m = \dfrac {p_1 + p_2 } {m_1+m_2}}[/tex]

Where,

p1 - momentum of the first mass

p2 - momentum of the second mass

m1 + m1 - total mass of the system

 

If there is no external force, the velocity of the center of mass is constant because the total momentum of the whole system is conserved.

Therefore, immediately after the snowball is thrown, the velocity of the center of mass is equal to that of the skater.

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Water has a density of 1.0 g/cm3. What is the mass of 10.0 cm3 of water?

Answers

Answer:

The mass of 10 cm³ of water, is 10 grams

Explanation:

Density of water = 1 g/cm³

Density = Mass / volume

Volume of water = 10 cm³

1 g/cm³ = Mass of water / 10 cm³

10 g = mass of water

Final answer:

The mass of 10.0 cm3 of water, given the density is 1.0 g/cm3, is found by multiplying the volume by the density, which gives us 10.0 g.

Explanation:

The question is asking for the mass of a certain volume of water. Given that the density of water is 1.0 g/cm3, and you have 10.0 cm3 of water, the mass can be found by multiplying the volume by the density. So, it's simply 1.0 g/cm3 * 10.0 cm3 = 10.0 g. Therefore, the mass of 10.0 cm3 of water is 10.0 g.

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A satellite with an orbital period of exactly 24.0 h is always positioned over the same spot on Earth.This is known as a geosynchronous orbit. Television, communication, and weather satellites use geosynchronous orbits. At what distance would a satellite have to orbit Earth in order to have a geosynchronous orbit?

Answers

Answer:

35870474.30504 m

Explanation:

r = Distance from the surface

T = Time period = 24 h

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

m = Mass of the Earth =  5.98 × 10²⁴ kg

Radius of Earth = [tex]6.38\times 10^6\ m[/tex]

From Kepler's law we have relation

[tex]T^2=\dfrac{4\pi^2r^3}{GM}\\\Rightarrow r^3=\dfrac{T^2GM}{4\pi^2}\\\Rightarrow r=\left(\dfrac{(24\times 3600)^2\times 6.67\times 10^{-11}\times 5.98\times 10^{24}}{4\pi^2}\right)^{\dfrac{1}{3}}\\\Rightarrow r=42250474.30504\ m[/tex]

Distance from the center of the Earth would be

[tex]42250474.30504-6.38\times 10^6=\mathbf{35870474.30504\ m}[/tex]

35870474.30504 m

A satellite in geosynchronous orbit remains over the same point on Earth, requiring an altitude of approximately 35,793 kilometers. This is calculated using the orbital mechanics involving the gravitational constant and Earth's mass and radius.

A geosynchronous orbit means that the satellite has an orbital period of [tex]24[/tex] hours, remaining over the same point on Earth as it rotates. To find the distance at which a satellite must orbit to achieve this, we use the universal law of gravitation and centripetal force.

We know:

Orbital period [tex]T = 86400 \text{ seconds}[/tex]

Gravitational constant [tex]G = 6.67430 \times 10^{-11} \, \text{N} \left( \text{m/kg} \right)^2[/tex]

Mass of Earth [tex]M = 5.972 \times 10^{24} \, \text{kg}[/tex]

Radius of Earth [tex]R = 6.371 \times 10^{6} \, \text{meters}[/tex]

The formula for the geostationary orbit (orbital radius r) is given by:

[tex]r^3 = \frac{GMT^2}{4\pi^2}[/tex]

Substituting the values, we get:

[tex]GM T^2 = 6.67430 \times 10^{-11} \times 5.972 \times 10^{24} \times (86400)^2\\\\GM T^2 = 6.67430 \times 10^{-11} \times 5.972 \times 10^{24} \times 7464960000\\\\GM T^2 = 2.963 \times 10^{14}\\\\r^3 \approx \frac{2.963 \times 10^{14}}{4 \times 9.8696}\\\\r^3 \approx 7.507 \times 10^{12} \, \text{m}^3\\\\r \approx \sqrt[3]{7.507 \times 10^{12}}\\\\r \approx 42164 \, \text{km}\\\\[/tex]

To find the altitude (h) above Earth’s surface:

[tex]h = r - \text{Earth's radius} = 42164 \, \text{km} - 6371 \, \text{km} \approx 35793 \, \text{km}[/tex]

Thus, a satellite needs to orbit at an altitude of approximately [tex]35,793 km[/tex] to maintain a geosynchronous orbit.

In one case, a sports car, its engine running, is driven up a hill at a constant speed. In another case, a truck approaches a hill, and its driver turns off the engine at the bottom of the hill. The truck then coasts up the hill. Which vehicle is obeying the principle of conservation of mechanical energy? Ignore friction and air resistance.

Answers

Final answer:

The truck coasting up the hill with its engine turned off is obeying the principle of conservation of mechanical energy because the kinetic energy is converted into gravitational potential energy without external work, while the sports car with the running engine is not conserving mechanical energy since it is adding energy to the system.

Explanation:

The principle of conservation of mechanical energy states that the total mechanical energy in a system remains constant if only conservative forces are doing work. When friction and air resistance are ignored, as requested, the truck that coasts up the hill with the engine turned off obeys this principle because its initial kinetic energy is converted into gravitational potential energy without any external work being done on or by the system. On the other hand, the sports car with its engine running is continually adding energy to the system to maintain a constant speed up the hill, which means it is not conserving mechanical energy as in the first case.

Marcie wants to work as a freelance editor. She purchases a desk, computer, widescreen monitor, and ergonomic keyboard. What kind of resources are these?

Answers

Answer:Capital Resources

Explanation:

Desk, computer, widescreen monitor, and ergonomic keyboard are an example of capital resources.

Capital resources are goods produced and used for the production of other goods and services. Basic items in capital goods are tools, machinery, and building. However,  any good being utilized by a business to produce other good and services are under the category of capital goods.

Here Desk, computer and other items help Marcie to design her  artwork  as freelancer.                              

The Bernoulli effect can have important consequences for the design of buildings. For example, wind can blow around a skyscraper at remarkably high speed, creating low pres- sure. The higher atmospheric pressure in the still air inside the buildings can cause windows to pop out. As originally constructed, the John Hancock Building in Boston popped windowpanes that fell many stories to the sidewalk below. (a) Suppose a horizontal wind blows with a speed of 11.2 m/s outside a large pane of plate glass with dimensions 4.00 m 3 1.50 m. Assume the density of the air to be constant at 1.20 kg/m3. The air inside the building is at atmospheric pressure. What is the total force exerted by air on the win- dowpane?

Answers

Answer:

9483.26399 N

Explanation:

[tex]\rho[/tex] = Density of air = 1.2 kg/m³

v = Velocity of wind = 11.2 m/s

A = Area = [tex]4\times 31.5\ m^2[/tex]

The force on the pane is

[tex]F=\dfrac{1}{2}\rho v^2A\\\Rightarrow F=\dfrac{1}{2}\times 1.2\times 11.2^2\times 4\times 31.5\\\Rightarrow F=9483.26399\ N[/tex]

The force on the pane of glass is 9483.26399 N

A mass is attached to an ideal spring. At time t = 0 the spring is at its natural length and the mass is given an initial velocity; the period of the ensuing (one-dimensional) simple harmonic motion is T . At what time is the power delivered to the mass by the spring first a maximum?

Answers

SHM in a mass-spring system involves a restoring force proportional to displacement, with power being maximal at extreme positions.

Simple Harmonic Motion (SHM) is a type of periodic motion defined by a restoring force proportional to displacement. In the context of a mass attached to a spring, the period of SHM can be calculated using the mass and force constant of the spring.

The power delivered to the mass by the spring is first a maximum when the restoring force is at its maximum, which occurs when the mass achieves its greatest displacement. This typically happens when the mass is at the equilibrium point after being displaced.

The time at which the power delivered by the spring is first a maximum aligns with the moments when the mass reaches its extreme positions during the oscillatory motion, typically when it's at the maximum displacement from equilibrium.

When a mass is attached to an ideal spring and set into simple harmonic motion (SHM) with an initial velocity from its natural length, the power delivered by the spring to the mass is first maximized when the mass passes through the equilibrium position at maximum speed. Since the period T is the time it takes for one complete cycle of the motion, the mass reaches its maximum velocity at equilibrium position at one quarter of the period. Therefore, the power delivered to the mass by the spring is first a maximum at t = T/4, where T is the period of the simple harmonic motion

. A 2.00 kg ball is attached to a ceiling by a string. The distance from the ceiling to the center of the ball is 1.00 m, and the height of the room is 3.00 m. What is the gravitational potential energy associated with the ball relative to each of the following? (a) the ceiling (b) the floor (c) a point at the same elevation as the ball

Answers

a) Potential energy of ball relative to ceiling is 19.62 Jb) Potential energy of ball relative to floor is 39.24 Jc) Potential energy of ball relative to same elevation is 0 J

Explanation:

Mass of ball, m = 2 kg

Acceleration due to gravity, g = 9.81 m/s²

Height of ball from ground = 3 - 1 = 2 m

Potential energy, PE = mgh

Potential energy of ball, PE = 2 x 9.81 x 2 = 39.24 J

a) Potential energy of ball at ceiling = 2 x 9.81 x 3 = 58.86 J

Potential energy of ball relative to ceiling = 58.86 - 39.24 = 19.62 J

b) Potential energy of ball at floor = 2 x 9.81 x 0 = 0 J

Potential energy of ball relative to floor = 39.24 - 0 = 39.24 J

c) Potential energy of ball at same elevation = 2 x 9.81 x 2 = 39.24 J

Potential energy of ball relative to same elevation = 39.24 - 39.24 = 0 J

Final answer:

The gravitational potential energy of the ball relative to the ceiling and a point at the same elevation as the ball is 0 Joules because they are at the same height. The gravitational potential energy of the ball relative to the floor is 39.24 Joules.

Explanation:

The gravitational potential energy of an object, in this case, the 2.00 kg ball, can be calculated using the formula: PE = mgh, where m is the mass of the object in kg, g is acceleration due to gravity (on Earth, it's about 9.81 m/s²), and h is the height in meters.

(a) the ceiling: The distance from the ceiling to the center of the ball is 1.00m, thus the potential energy relative to the ceiling is PE = 2.00 kg * 9.81 m/s² * 0 m = 0 J, since the ball is at the same level as the ceiling.

(b) the floor: The height of the ball from the floor can be calculated as the height of the room minus the distance from the ceiling to the center of the ball, which is 3.00 m - 1.00 m = 2.00 m. Thus, the potential energy relative to the floor is PE = 2.00 kg * 9.81 m/s² * 2.00 m = 39.24 J.

(c) a point at the same elevation as the ball: Since the point is at the same elevation as the ball, therefore, the potential energy would be PE = 2.00 kg * 9.81 m/s² * 0 m = 0 J.

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PLEASE HELP DUE IN 5 MINUTES AND I NEED TO PASS!!!!
What is the significance of Le Châtelier's principle when used in industrial fields?

1) Scientists know that reactions at chemical equilibrium will have consistent reaction rates if the intensity of external factors is steadily increased, so scientists can use external factors to manipulate these chemical reactions into producing specific results.

2) Scientists know that reactions at chemical equilibrium will shift to regain equilibrium if left in the same external conditions long enough, so scientists can stabilize these chemical reactions' environments in order to produce specific results.

3) Scientists know that reactions at chemical equilibrium will have consistent reaction rates despite any change in external conditions, so scientists can depend on these chemical reactions to always produce the same specific results.

4) Scientists know that reactions at chemical equilibrium will shift to regain equilibrium when affected by external factors, so scientists can use external factors to manipulate these chemical reactions into producing specific results.

Answers

scientists can prove the answer of the chemical reaction and in every case

Answer:

4) Scientists know that reactions at chemical equilibrium will shift to regain equilibrium when affected by external factors, so scientists can use external factors to manipulate these chemical reactions into producing specific results.

Explanation:

Le Châtelier's principle states that changes in the temperature, pressure, volume, or concentration of a system will result in predictable and opposing changes in the system in order to achieve a new equilibrium state. Le Châtelier's principle doesn't say anything about reaction rates.

The idea of an industrial process is to obtain an specific product. So, you can apply changes into a system in equilibrium to produce more desirable product. A simple example is to decrease the concentration of products by removing them or increasing the concentration of reactants by adding them, so that, more product is produced and more reactant is used, and the equilibrium is reached again.

Professionals in careers in the exercise and sport sciences need to provide quality programs in specialized settings to prove that a higher degree of competence is needed to fulfill the responsibilities of these careers than may currently be the case.

Answers

Answer:

True

Explanation:

It is true that professionals in exercises and sport science need to provide quality specialized settings to prove that a higher degree of competence is needed to fulfill the responsibility of the careers.

An 880 kg cannon at rest fires a 12.4 kg cannonball forward at 540 m/s. What is the recoil velocity of the cannon

Answers

Answer:

The recoil velocity of the cannon is 7.61 m/s in the opposite direction of the cannonball

Explanation:

Linear Momentum

The principle of conservation of the linear momentum establishes that the sum of the linear momentums of every object in an isolated system (no external forces) is constant, regardless of the interactions between them.

Let's think we have two objects with masses [tex]m_1[/tex] and [tex]m_2[/tex], moving at speeds [tex]v_1[/tex] and [tex]v_2[/tex]. If they collide and change their speeds to [tex]v_1'[/tex] and [tex]v_2'[/tex], then

[tex]m_1v_1+m_2v_2=m_1v_1'+m_2v_2'[/tex]

In our problem, the 880 kg cannon is initially at rest and has the cannonball of 12.4 Kg inside of it. As the initial speed of both joined objects is zero, the initial total momentum is zero. After the ball is fired, the ball moves at v_2=540 m/s. We need to find the recoil velocity of the cannon [tex]v_1'[/tex]

[tex]m_1v_1'+m_2v_2'=0[/tex]

[tex]\displaystyle m_1v_1'=-m_2v_2'[/tex]

[tex]\displaystyle v_1'=-\frac{m_2v_2'}{m_1}[/tex]

[tex]\displaystyle v_1'=-\frac{12.4(540)}{880}=-7.61\ m/s[/tex]

The recoil velocity of the cannon is 7.61 m/s in the opposite direction of the cannonball

Rocks on either side of a strike-slip fault move past each other without much upward or downward movement is called___________.

Answers

Answer:

strike-slip fault

Explanation:

A fault is a crack on the crust of the Earth where forces on the rocks causes the displacement of the rocks.

The movement of the crust determines the type of fault.

When rocks slip pass horizontally with respect to each other then it is called a strike slip fault. The crust moves towards each other causing horizontal compression. The displacement of the rock takes place parallel to the horizontal force.

Hence, the statement here is referring to the strike-slip fault.

A certain chemical reaction experiences a 7.50 x 10-7 kilogram loss in mass. If this mass were to be totally converted to energy, what would be the magnitude of that energy.

Answers

Answer:

E=6.75×10¹⁰J

Explanation:

Given data

Mass (m)=7.50×10⁻⁷kg

To find

Energy (E) = ?

Solution

From Albert Einstein’s theory of special relativity

E = mc²........eq(1)  

Where

E is energy

m is mass

c is speed of light which 3.0×10⁸m/s

put values in eq(1) we get

E=(7.50×10⁻⁷kg)×( 3.0×10⁸m/s)²

E=6.75×10¹⁰J

A mxiture of n2 and H2 has mole fraction of 0.4 and 0.6 respectively. Determine the density of the mixture at one bar and 0 c.

Answers

Answer:

The density of the mixture is 0.55kg/m^3

Explanation:

P = 1bar = 100kN/m^2, T = 0°C = 273K, n = 0.4+0.6 = 1mole

PV = nRT

V = nRT/P = 1×8.314×273/100 = 22.70m^3

Mass of N2 = 0.4×28 = 11.2kg

Mass of H2 = 0.6×2 = 1.2kg

Mass of mixture = 11.2 + 1.2 = 12.4kg

Density of mixture = mass/volume = 12.4/22.7 = 0.55kg/m^3

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