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polet [3.4K]
3 years ago
14

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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Sam, whose mass is 78 kg , stands at the top of a 11-m-high, 110-m-long snow-covered slope. His skis have a coefficient of kinet
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Answer:

v = 8.09   m/s

Explanation:

For this exercise we use that the work done by the friction force plus the potential energy equals the change in the body's energy.

Let's calculate the energy

       

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         Em₀ = U = m gh

final point. To go down the slope

         Em_f = K = ½ m v²

The work of the friction force is

         W = fr L cos 180

to find the friction force let's use Newton's second law

Axis y

        N - W_y = 0

        N = W_y

X axis

        Wₓ - fr = ma

let's use trigonometry

        sin  θ = y / L

         sin θ = 11/110 = 0.1

         θ = sin⁻¹  0.1

          θ = 5.74º

         sin 5.74 = Wₓ / W

         cos 5.74 = W_y / W

         Wₓ = W sin 5.74

         W_y = W cos 5.74

the formula for the friction force is

         fr = μ N

         fr = μ W cos θ

Work is friction force is

         W_fr = - μ W L cos θ  

Let's use the relationship of work with energy

        W + ΔU = ΔK

         -μ mg L cos 5.74 + (mgh - 0) = 0  - ½ m v²

        v² = - 2 μ g L cos 5.74 +2 (gh)

        v² = 2gh - 2 μ gL cos 5.74

let's calculate

        v² = 2 9.8 11 - 2 0.07 9.8 110 cos 5.74

        v² = 215.6 -150.16

        v = √65.44

        v = 8.09   m/s

6 0
2 years ago
When magma cools quickly, what kind of texture or
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Answer:

a

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6 0
3 years ago
what will be the moment of inertia of a body if it rotates at a uniform rate of 10rad/sec^2 by a torque of 120Nm?​
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Answer:

12 kgm²

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here angular acceleration = 10rad/sec²

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we know,

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Unpolarized light with an intensity of (25.0 A) units is passed through two successive polarizing filters, the first with its po
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Answer:

the intensity of the light after passing through the two polarizing filters is 4.11 units  

 

Explanation:

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