A train travels 80 kilometers in 2 hours, and then 79 kilometers in 4 hours. What is its average speed?

What is the guage pressure of an object with a mass of 30 kg and 5ft height?

wariber [46]

Answer: D=pg

Explanation:

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8 0

3 years ago

A certain planet has an escape speed V. If another planet has twice the radius and twice the mass of the first planet, its escap

Ber [7]

Answer:

option A

Explanation:

Escape velocity of the planet

$v = \sqrt{\dfrac{GM}{R}}$

now, it is given that

Second Planet

R₂ = 2R and M₂ = 2M

now escape velocity of the second planet

now,

$v'= \sqrt{\dfrac{G(2M)}{(2R)}}$

on solving

$v'= \sqrt{\dfrac{GM}{R}}$

$v'= v$

escape velocity of the second planet is equal to first.

The correct answer is option A

4 0

3 years ago

A 75 gram ball is fired into a 200 gram pendulum and the center of gravity of the pendulum-ball rises to a height of 2.5 cm. The

pantera1 [17]

Answer:

The initial velocity of the ball is 2.567 m/s

Solution:

Mass of the ball, m = 75 gm = 0.075 kg

Mass of the pendulum, m' = 200 gm = 0.2 kg

Height of rise of the ball, h = 2.5 cm = 0.025 m

Length, l = 30 cm = 0.3 m

Now,

Total mass, M = m + m' = 0.075 + 0.2 = 0.275 kg

Now,

Suppose the initial velocity of the ball be $v_{i}$

The total mass is raised to a height of 0.025 m after the ball has been fired, thus the total potential energy of the system is given by:

$PE = Mgh = 0.275\times 9.8\times 0.025 = 0.0674\ J$

Now, the initial kinetic energy of the system:

$KE = \frac{1}{2}Mv^{2}$

Now, using law of conservation of energy:

PE = KE

$\frac{1}{2}Mv^{2} = 0.0674$

$v = \sqrt{\frac{0.13475}{0.275}} = 0.7\ m/s$

Now by using the principle of momentum conservation:

$mv_{i} = Mv$

$v_{i} = \frac{Mv}{m}$

$v_{i} = \frac{0.275\times 0.7}{0.075} = 2.567\ m/s$

5 0

3 years ago

A coin rests 11.0 cm from the center of a turntable. The coefficient of static friction between the coin and turntable surface i

9966 [12]

Answer:

5.5 rad/s

Explanation:

The friction between the coin and the turntable provides the centripetal force that keeps the coin in circular motion. Therefore, we can write:

$\mu mg = m\omega^2 r$

where

$\mu = 0.340$ is the coefficient of friction

m is the mass of the coin

$g=9.8 m/s^2$ is the acceleration of gravity

$\omega$ is the angular speed

r is the distance of the coin from the centre of rotation

In this problem,

r = 11.0 cm = 0.11 m

The coin starts to slip when the centripetal force becomes larger than the maximum frictional force:

$m\omega^2 r > \mu m g$

Solving for $\omega$ , we find the angular speed at which this happens:

$\omega = \sqrt{\frac{\mu g}{r}}=\sqrt{\frac{(0.340)(9.8)}{0.11}}=5.5 rad/s$

5 0

3 years ago

If someone is traveling the speed of light in a car and if and if a car is coming towards you and puts there head lights on can

Mama L [17]

The speed of light is c = 3 * 10E8 m/s in all reference frames.

Therefore, no matter how fast the car is approaching you

you will receive light photons traveling at speed c and the

approaching car will see photons traveling away from it with speed c.

8 0

3 years ago