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3 years ago

At what speed must a 950-kg subcompact car be moving to have the same kinetic energy as a 3.2×104−kg truck going 20 km/h?

Physics

1 answer:

Ostrovityanka [42]3 years ago

6 0

<u>Answer:</u>

950-kg subcompact car be moving to have the same kinetic energy of truck at 116.08 km/hr

<u>Explanation:</u>

Kinetic energy of body with mass m and moving at velocity v is given by

$KE=\frac{1}{2} mv^2$

KE of truck $= \frac{1}{2}*3.2*10^4*20^2$

KE of subcompact car $= \frac{1}{2}*950*v^2$

Equating

$\frac{1}{2}*3.2*10^4*20^2= \frac{1}{2}*950*v^2\\ \\ v=116.08 km/hr$

So 950-kg subcompact car be moving to have the same kinetic energy of truck at 116.08 km/hr

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Answer:

They can join a gym or a class to help keep themselves motivated to exercise.

Explanation:

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What are the conditions of the thermonuclear fusion reaction in the sun?

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1 year ago

- A fridge has a maximum static friction force

Dennis_Churaev [7]

Answer:

They will move the fridge if they all push in the same direction, but it will not move with constant velocity

Explanation:

The maximum static friction force is

$F_f = -250 N$ (negative sign since its direction is opposite to the push applied by the people)

Sam can apply a force of 130 N, while Amir and Andre can apply a push of 65 N each, so the total force that they can apply, if they push in the same direction, will be:

$F=130 + 65 +65=260 N$

This force is larger than the frictional force, so the fridge will start moving.

However, the net force on the fridge will be:

$\sum F = 260 N - 250 N = 10 N$

And according to Newton's second law,

$\sum F = ma$

where m is the mass of the fridge and a its acceleration, since the net force is not zero, then the fridge will have a non-zero acceleration, so it will not move with constant velocity.

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3 years ago

A spring on a horizontal surface can be stretched and held 0.5 m from its equilibrium position with a force of 60 N. a. How much

Lostsunrise [7]

Answer:

a)1815Joules b) 185Joules

Explanation:

Hooke's law states that the extension of a material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically;

F = ke where;

F is the applied force

k is the elastic constant

e is the extension of the material

From the formula, k = F/e

F1/e1 = F2/e2

If a force of 60N causes an extension of 0.5m of the string from its equilibrium position, the elastic constant of the spring will be ;

k = 60/0.5

k = 120N/m

a) To get the work done in stretching the spring 5.5m from its position,

Work done by the spring = 1/2ke²

Given k = 120N/m, e = 5.5m

Work done = 1/2×120×5.5²

Work done = 60× 5.5²

Work done = 1815Joules

b) work done in compressing the spring 1.5m from its equilibrium position will be gotten using the same formula;

Work done = 1/2ke²

Work done =1/2× 120×1.5²

Works done = 60×1.5²

Work done = 135Joules

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2 years ago

Read 2 more answers

You have four fixed-volume containers at STP . Container A has 0.5 mol of gas in 11.2 L. Container B has 2 mol of gas in 22.4 L.

Colt1911 [192]

Answer:

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ideal gas equation gives P=nRT/V

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Container A and C after dividing number of moles and Volume, are found to be the same=0.0446

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2 years ago