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3 years ago

A cup falls off of a table from a height of 0.75 m. What is the impact speed of the cup?

Physics

2 answers:

Artemon [7]3 years ago

3 0

3.83 meters per second.

andrew-mc [135]3 years ago

3 0

Answer:

The impact speed of the cup is 3.83 m/s.

Explanation:

Given that,

Height = 0.75 m

We need to calculate the impact speed of the cup

Using kinetic and potential energy

$K.E=P.E$

$\dfrac{1}{2}mv^2=mgh$

Where,

M = mass of the cup

v = velocity of cup

g = acceleration due to gravity

h = height

Put the value into the formula

$\dfrac{1}{2}v^2=gh$

$v=\sqrt{2gh}$

$v=\sqrt{2\times9.8\times0.75}$

$v=3.83\ m/s$

Hence, The impact speed of the cup is 3.83 m/s.

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A truck traveling at a velocity of 33m/s comes to a halt by decelerating at 11m/s^2. How far does the truck travel in the proces

snow_lady [41]

Answer:

The truck was moving 16.5 m/s during the time it took to stop, which was 3 seconds.

Initial velocity = 33 m/s

Final velocity = 0 m/s

Average velocity = (33 + 0) / 2 m/s = 16.5 m/s

Explanation:

First, how long does it take the truck to come to a complete stop?

( 33 m/s ) / ( 11 m / s^2 ) = 3 seconds

Then we can look at the average velocity between when the truck started decelerating and when it came to a complete stop. Because the deceleration is constant (always 11m/s^2) we can use this trick.

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Lelu [443]

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3 years ago

Find the momentum of a particl with a mass of one gram moving with half the speed of light.

joja [24]

Answer:

129900

Explanation:

Given that

Mass of the particle, m = 1 g = 1*10^-3 kg

Speed of the particle, u = ½c

Speed of light, c = 3*10^8

To solve this, we will use the formula

p = ymu, where

y = √[1 - (u²/c²)]

Let's solve for y, first. We have

y = √[1 - (1.5*10^8²/3*10^8²)]

y = √(1 - ½²)

y = √(1 - ¼)

y = √0.75

y = 0.8660, using our newly gotten y, we use it to solve the final equation

p = ymu

p = 0.866 * 1*10^-3 * 1.5*10^8

p = 129900 kgm/s

thus, we have found that the momentum of the particle is 129900 kgm/s

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3 years ago

A starship blasts past the earth at 2.0*10^8 m/s .Just after passing the earth, the starship fires a laser beam out its back of

Vilka [71]

Answer:

at the speed of light ( $c=3.0\cdot 10^8 m/s$ )

Explanation:

The second postulate of the theory of the special relativity from Einstein states that:

"The speed of light in free space has the same value c in all inertial frames of reference, where $c=3.0\cdot 10^8 m/s$ "

This means that it doesn't matter if the observer is moving or not relative to the source of ligth: he will always observe light moving at the same speed, c.

In this problem, we have a starship emitting a laser beam (which is an electromagnetic wave, so it travels at the speed of light). The startship is moving relative to the Earth with a speed of 2.0*10^8 m/s: however, this is irrelevant for the exercise, because according to the postulate we mentioned above, an observer on Earth will observe the laser beam approaching Earth with a speed of $c=3.0\cdot 10^8 m/s$ .

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3 years ago

What is the final velocity of the ball that is dropped from a height of 200m?

Hoochie [10]

The final velocity of the ball that is dropped from a height of 200m is v = 44.73 m/s .

<h3>What is velocity with example?</h3>

The rate at which an object is travelling in one direction is referred to as its velocity. an automobile traveling north on a highway, or a rocket taking off. Its velocity vector's absolute value always is equal to the motion's speed because it is a scalar.

<h3>Briefing:</h3>

Given the initial velocity of the ball (u) = 0

Distance travelled by the ball (s) = 200m

Acceleration (a) = 10 m/s²

As we know:

v² = u² + 2as

Putting values:

v² = 0+2 × (10 m/s²) × (200 m)

v = 44.73 m/s.

To know more about Velocity visit:

brainly.com/question/18084516

#SPJ9

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