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3 years ago

The kinetic energy of a book on a shelf is equal to the work done to lift the book to the shelf. t/f

2 answers:

7 0

The kinetic energy of a book on a shelf is equal to the work done to lift the book to the shelf is false. The kinetic energy on the shelf is zero because it is not in action.

algol [13]3 years ago

6 0

Answer: The given statement is false.

Explanation:

Kinetic energy is defined as the energy obtained by an object because of its motion.

For example, moving a pulley on the road.

Whereas energy of an object obtained due to its position is known as potential energy.

For example, lifting a book from table to the shelf shows position of the book has changed.

Therefore, we can conclude that the statement kinetic energy of a book on a shelf is equal to the work done to lift the book to the shelf, is false.

You might be interested in

Two ice skaters, paula and ricardo, push off from each other. ricardo weighs more than paula. part a which skater, if either, ha

ahrayia [7]

Apply the law of conservation of momentum for this situation. The law states that the momentum of a system is constant (in absence of external forces acting on it).

The 'system' in this case are the two skaters. There is no external force on the skaters. Suppose the skaters are initially standing still. The momentum in the system is 0. This value will need to remain constant, even after the mutual push (which is a set of forces from <em>inside</em> the system). So we know that

(total momentum before) = (total momentum after)

Indexing the masses and velocities by the first letter of the skaters' names:

$0 = m_P\vec v_P+m_R\vec v_R\\m_P\vec v_P = m_R(-\vec v_R)$

From the last row, you can see that the skaters will have momentum of same magnitude but opposite direction, after the push off. That answers the first question: neither will have a greater momentum (both will have one of same magnitude).

Since Ricardo is heavier, from the above equality it follows that

$m_R>m_P\implies|\vec v_R|$

In words, Paula has the greater speed, after the push-off.

7 0

3 years ago

You travel at a speed of 24 m/s. How much time will it take you to travel 127m (round to two decimal places if needed)

shtirl [24]

Answer: 5.29 seconds

Explanation:

if the speed is 24 meters per second than it would take you 5.29 seconds or 127meters/24m/s to get 5.29s

4 0

3 years ago

In 2000, NASA placed a satellite in orbit around an asteroid. Consider a spherical asteroid with a mass of 1.40×1016 kg and a ra

Arturiano [62]

A) 8.11 m/s

For a satellite orbiting around an asteroid, the centripetal force is provided by the gravitational attraction between the satellite and the asteroid:

$m\frac{v^2}{(R+h)}=\frac{GMm}{(R+h)^2}$

where

m is the satellite's mass

v is the speed

R is the radius of the asteroide

h is the altitude of the satellite

G is the gravitational constant

M is the mass of the asteroid

Solving the equation for v, we find

$v=\sqrt{\frac{GM}{R+h}}$

where:

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$

$M=1.40\cdot 10^{16}kg$

$R=8.20 km=8200 m$

$h=6.00 km = 6000 m$

Substituting into the formula,

$v=\sqrt{\frac{(6.67\cdot 10^{-11})(1.40\cdot 10^{16}kg)}{8200 m+6000 m}}=8.11 m/s$

B) 11.47 m/s

The escape speed of an object from the surface of a planet/asteroid is given by

$v=\sqrt{\frac{2GM}{R+h}}$

where:

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$

$M=1.40\cdot 10^{16}kg$

$R=8.20 km=8200 m$

$h=6.00 km = 6000 m$

Substituting into the formula, we find:

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(1.40\cdot 10^{16}kg)}{8200 m+6000 m}}=11.47 m/s$

5 0

3 years ago

A wave passes along the surface of the water in a ripple tank. Describe the motion of a molecule on the surface as the water pas

laila [671]

Answer:

Explanation:

Water waves are generally a transverse wave which do not cause permanent displacement of molecules of the medium. Transverse waves are waves in which the direction of propagation of the wave is perpendicular to the direction of vibration of the particles of the medium.

As the wave propagates from one point to another on the surface of water transferring energy, a molecule of water on its surface vibrates upwards and downwards. Its motion is perpendicular to the direction of propagation of the wave. After the vibration, it comes back to its initial position.

8 0

3 years ago

This is physics 11th grade and a homework question I don’t understand how to do this or what the question is asking me

Alexxx [7]

a) Frequency is the number of complete oscillations per second. Looking at the graph, there are 9 complete oscillations in 5 seconds. Thus,

Frequency = 9/5 = 1.8 oscillations per second

Frequency = 1.8 Hz

Period = 1/frequency = 1/1.8

Period = 0.056 s

b) When we differenctiate displacement with respect to time, the result is velocity.

Recall, period = 1/f = 5/9 cycles

1/4 cycle behind = 1/4 x 5/9 = 5/36

It is delayed with 5/36 sec with respect to displacement.

5/36 sec = 0.139 sec

Acceleration = first derivative of velocity = second derivative of displacement = 1/4 cycle behind velocity = 1/2 cycle behind displacement =

5/36 = 0.139 sec delayed with respect to velocity

= 5/18 = 0.2777 secs delayed with respect to displacement

Thus, the number of seconds out of phase with the displacements is 0.278 seconds

c) The formula for calculating the period of an ideal pendulum anywhere is

T = 2π√length/local gravity). We would calculate the local gravity.

From the information given,

length = 0.2

T = P = 5/9

Thus,

5/9 = 2π√0.2/local gravity)

(5/9)/2π = √0.2/local gravity

Square both sides. It becomes

[(5/9)/2π]^2 = 0.2/local gravity

local gravity = 0.2/[(5/9)/2π]^2

local gravity = 25.56 m/s^2

Thus,

acceleration due to gravity = 25.56 m/s^2

Recall, earth's gravity = 9.8 m/s^2

number of g forces = 25.56/9.8

number of g forces = 2.61

6 0

1 year ago