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2 years ago

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PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE H

ELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP
I NEED TO GIVE THIS IN TODAY!!!

1 answer:

Masteriza [31]2 years ago

5 0

Answer:

a) greater

smaller

Explanation:

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Determine the kinetic energy of a 10 Kg roller coaster car that is moving with a speed of 2m/s.

skelet666 [1.2K]

In this question all required information's are already provided. Based on these details the answer to the question can be easily determined. Let us now write down all the information's that are already given.

Mass of the roller coaster = 1000 kg

Velocity of the roller coaster = 20.0 m/s

We know the formula for finding the kinetic energy is

Kinetic energy = 0.5 * mass * (velocity) ^2

= 0.5 * 1000 * (20)^2

= 0.5 * 1000 * 400

= 200000 Joules

So the Kinetic energy of the roller coaster is 200000 joules.

i hope this helps you friend good luck on your quiz or lesson

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How many ohms of resistance must be present in a circuit that has 240 volts and a current of 15 amps?

Mkey [24]

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On another planet gravity has a value of 5.5 m/s . If an object is dropped how long will it take to fall 53 m?

Nikolay [14]

Answer:

4.4 seconds

Explanation:

Given:

a = -5.5 m/s²

v₀ = 0 m/s

y₀ = 53 m

y = 0 m

Find: t

y = y₀ + v₀ t + ½ at²

0 = 53 + 0 + ½ (-5.5) t²

0 = 53 − 2.75 t²

t = 4.39

Rounded to two significant figures, it takes 4.4 seconds for the object to land.

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A tablet is made by crushing several chemicals, combining the chemicals as powders, and pressing the powders together to make a

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A small rubber ball is launched by a compressed-air cannon from ground level with an initial speed of 11.8 m/s directly upward.

Fantom [35]

Answer:

7.09683 m

1.20285 s

2.4057 s

11.8 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² (negative up, positive down)

From equation of motion we have

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-11.8^2}{2\times -9.81}\\\Rightarrow s=7.09683\ m$

The maximum height above the ground that the ball reaches is 7.09683 m

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-11.8}{-9.81}\\\Rightarrow t=1.20285\ s$

Time taken to go up is 1.20285 s it will take the same time to come down so total time taken to reach the ground after it is shot is 1.20285+1.20285 = 2.4057 s

$v=u+at\\\Rightarrow v=0+9.81\times 1.20285\\\Rightarrow v=11.8\ m/s$

The velocity just before it hits the ground is 11.8 m/s

6 0

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