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Crazy boy [7]
3 years ago
14

At some airports there are speed ramps to help passengers get from one place to another. A speed ramp is a moving conveyor belt

on which you can either stand or walk. Suppose a speed ramp has a length of 120 m and is moving at a speed of 2.2 m/s relative to the ground. In addition, suppose you can cover this distance in 72 s when walking on the ground. If you walk at the same rate with respect to the speed ramp that you walk on the ground, how long does it take for you to travel the 120 m using the speed ramp?
Physics
1 answer:
Harrizon [31]3 years ago
8 0

Answer:

It will take you 30.8 s to travel the 120 m of the ramp.

Explanation:

Hi there!

The equation for the position of an object moving in a straight line is:

x = x0 + v * t

Where:

x = position at time t

x0 = initial position

v = velocity

t = time

In this case, we will consider the start of the ramp as the origin of our reference system so that x0 = 0.

Now, let´s calculate the speed of the person walking on the ground:

x = v * t

120 m = v * 72 s

v = 120 m / 72 s

v = 1.7 m/s

If you walk on the ramp with that speed, your total speed will be your walking speed plus the speed of the ramp because both are in the same direction. Then, using the equation for the position:

x = v * t

In this case, v = speed of the ramp + walking speed

v = 2.2 m/s + 1.7 m/s = 3.9 m/s

120 m = 3.9 m/s * t

t = 120 m / 3.9 m/s = 30.8 s

It will take you 30.8 s to travel the 120 m

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The magnitude of the upthrust depends on the density of the liquid.

When the liquid is denser the boat will experience a great upthrust as compared to when in a less dense liquid.

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A firework is ignited, and explodes with a flash and a loud bang as it is blown apart. The system consists of: the firework, the
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The complete question is:

Study the scenario. A firework is ignited, and explodes with a flash and a loud bang as it is blown apart. The system consists of: the firework, the earth, and the air. which choice best describes how energy is transformed in the system?

A) When the firework is ignited, a chemical reaction absorbs energy from the surrounding environment. the energy is in several forms including sound and light, and mechanical energy of the fragments of the firecracker that are launched through the air. Eventually all the energy released is transformed into thermal energy.

B) When the firework is ignited, a chemical reaction releases energy in several forms, including sound, light, and the mechanical energy of the fragments being launched through the air. Eventually all the energy released is transformed into mechanical energy.

C) When the firework is ignited, a chemical reaction releases energy in several forms, including sound, light, and the mechanical energy of the fragments being launched through the air. Eventually all the energy released is transformed into thermal or mechanical energy.

D) When the firework is ignited, a chemical reaction absorbs energy from the surrounding environment. The energy is taken in in several forms including sound and light, and mechanical energy of the fragments being launched through the air. Eventually all the energy is transformed into thermal energy.

Answer:

C) When the firework is ignited, a chemical reaction releases energy in several forms, including sound, light, and the mechanical energy of the fragments being launched through the air. Eventually all the energy released is transformed into thermal or mechanical energy.

Explanation:

Energy is released from the system, not absorbed.

All the sound, light and movement of the debris is as a result of energy transformation from the chemical energy.

Eventually, most of the energy are finally wasted away as heat energy.

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A particle, whose acceleration is constant, is moving in the negative x direction at a speed of 4.91 m/s, and 12.9 s later the p
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Answer:

The particle’s velocity is -16.9 m/s.

Explanation:

Given that,

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Before 12.4 sec,

Velocity of particle in negative x direction= 5.32 m/s

We need to calculate the acceleration

Using equation of motion

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a=\dfrac{v-u}{t}

Where, v = final velocity

u = initial velocity

t = time

Put the value into the equation

a=\dfrac{7.12-(-4.91)}{12.9}

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We need to calculate the initial speed of the particle

Using equation of motion again

v=u+at

u=v-at

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u=-16.9\ m/s

Hence, The particle’s velocity is -16.9 m/s.

4 0
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