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3 years ago

What question could you ask about kinetic energy which will include the variables that affect it?

Physics

2 answers:

Gemiola [76]3 years ago

4 0

Answer:

<u>Kinetic Energy (K.E):</u>

The K.E is acquired or possessed by the body due to its virtue of motion. As the K.E is based upon the mass of the body, m which remains the same for each object, while the K.E also depends upon the square of velocity,v of the same body(while it is in motion), which is a variable.

K.E=1/2 m×v².

Explanation:

<u>K.E and velocity,v:</u>

The K.E acquired by the body depends upon the constant value of mass,m and the variable component of velocity,v. As the velocity of the body changes along time change,Δt.

Juli2301 [7.4K]3 years ago

3 0

KE= (1/2) mv²

So, the variables we need to include in our question would be a varable for a mass(m) of an object at some velocity(v).

My Answer:

(This is just an example question, yours can be different)
What is the Kinetic Energy experienced by an bouncy ball rolling at 7m/s (that's your velocity) across a frictionless surface that has a mass(m) of 10 grams?

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2 years ago

A cyclist is coasting at 15 m/s when she starts down a 450 m long slope that is 30 m high. The cyclist and her bicycle have a co

Flura [38]

Answer:

Her speed at the bottom of the slope is 25.665 m/s

Explanation:

Here we have

Initial velocity, v₁= 15 m/s

Final velocity = v₂

The energy balance present in the system can be represented as

$\frac{1}{2}mv_2^2 -\frac{1}{2}mv_1^2 - mgh = W$

Where:

m = Mass of the cyclist = 70 kg

W = work done by the drag force = $-F_Dd$

Where:

d = Distance traveled = 450 m

Therefore,

$\frac{1}{2}mv_2^2 -\frac{1}{2}mv_1^2 - mgh = -F_Dd$ and

$v_2^2 =\frac{ \frac{1}{2}mv_1^2 + mgh -F_Dd}{ \frac{1}{2}m} = v_1^2 + 2gh -\frac{ 2F_Dd}{ m} = 15^2 + 2\times 9.8\times 30 - \frac{2\times 12\times 450}{70}$

= 658.714 m²/s²

v₂ = 25.665 m/s

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3 years ago