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Feliz [49]
3 years ago
13

What's the kinetic energy of an object that has a mass of 30 kilograms and moves with a velocity of 20m/s?

Physics
1 answer:
icang [17]3 years ago
4 0

Ek = 6KJ.

In physics, the kinetic energy of a body or object is the one that owns due to its movement and is given by the equation E_{k} = \frac{1}{2} mv^{2}, where m is the mass of the object in kilograms and v is the velocity in m/s.

An object that it has a mass of 30 kilograms and moves with a velocity of 20m/s, its kinetic energy is given by:

E_{k} = \frac{1}{2} (30kg)(20m/s)^{2}=6000J=6KJ

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A grocery cart with a mass of 15 kg is pushed at constant speed along an aisle by a force fp = 12 n which acts at an angle of 17
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<span>Given:

Mass of cart: 15kg</span>

Aisle length = 14m

Angle = 17° below the horizontal

Force fp = 12 N

 

So, the solution would be like this for this specific problem:

 

<span><span>1)    </span>W(by applied force) = F(applied) x s x cosθ <span>
=>W(a) = 12 x 14 x cos17* = 160.66 J </span></span>
<span><span>
2)    </span>By F(net) = F(applied) - F(friction) <span>
=>As v = constant => a = 0 => F(net) = 0 
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=>W(friction) = -F(applied) x s 
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3)    </span>0, as the displacement is perpendicular to Force </span>
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4)    </span>0, as the displacement is perpendicular to Force</span>  
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To add, the force that is applied<span> to an object by a person or another object is called the applied force.</span></span>
8 0
3 years ago
One charge is decreased to one-third of its original value, and a second charge is decreased to one-half of its original value.
Nadya [2.5K]
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Fc=(k*Q₁*Q₂)/r², where k is the Coulomb constant and k=9*10⁹ N m² C⁻², Q₁ is the first charge, Q₂ is the second charge and r is the distance between the charges.

The magnitude of the force is independent of the sign of the charge so I can simply say they are both positive. 

Q₁ is decreased to Q₁₁=(1/3)*Q₁=Q₁/3 and
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New force:

Fc₁=(k*Q₁₁*Q₂₂)r², now we input the decreased values of the charge

Fc₁=(k*{Q₁/3}*{Q₂/2})/r², that is equal to:

Fc₁=(k*(1/3)*(1/2)*Q₁*Q₂)/r²,

Fc₁=(k*(1/6)*Q₁*Q₂)/r²

Fc₁=(1/6)*(k*Q₁*Q₂)/r², and since the original force is: Fc=(k*Q₁*Q₂)/r² we get:

Fc₁=(1/6)*Fc

So the magnitude of the new force Fc₁ with decreased charges is 6 times smaller than the original force Fc. 
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