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Paraphin [41]
3 years ago
4

A constant horizontal pull acts on a sled on a horizontal frictionless ice pond. The sled starts from rest. When the pull acts o

ver a distance x, the sled acquires a speed v and a kinetic energy K. If the same pull instead acts over twice this distance, A) The sled’s speed will be 2v and its kinetic energy will be 2K.
B) The sled’s speed will be 2v and its kinetic energy will be K 2 .
C) The sled’s speed will be v 2 and its kinetic energy will be 2K.
D) The sled's speed will be v 2 and its kinetic energy will be K 2 .
E) The sled's speed will be 4v and its kinetic energy will be 2K.
Physics
1 answer:
Kamila [148]3 years ago
4 0

Answer:

K'= 2 K

v'= v√2

Explanation:

As we know that

From work power energy

Work done by all the force = Change in kinetic energy

F . x = K

When distance become 2 x

Then

F.2 x = K'

K' = 2 F . x

K'= 2 K

We know that

K=\dfrac{mv^2}{2}

K'=\dfrac{mv'^2}{2}

K'= 2 K

2\dfrac{mv^2}{2}=\dfrac{mv'^2}{2}

v'²= 2 v²

v'= v√2

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A 1100 kg car rounds a curve of radius 68 m banked at an angle of 16 degrees. If the car is traveling at 95 km/h, will a frictio
Mariulka [41]

Answer:

Yes. Towards the center. 8210 N.

Explanation:

Let's first investigate the free-body diagram of the car. The weight of the car has two components: x-direction: towards the center of the curve and y-direction: towards the ground. Note that the ground is not perpendicular to the surface of the Earth is inclined 16 degrees.

In order to find whether the car slides off the road, we should use Newton's Second Law in the direction of x: F = ma.

The net force is equal to F = \frac{mv^2}{R} = \frac{1100\times (26.3)^2}{68} = 1.1\times 10^4~N

Note that 95 km/h is equal to 26.3 m/s.

This is the centripetal force and equal to the x-component of the applied force.

F = mg\sin(16) = 1100(9.8)\sin(16) = 2.97\times10^3

As can be seen from above, the two forces are not equal to each other. This means that a friction force is needed towards the center of the curve.

The amount of the friction force should be 8.21\times 10^3~N

Qualitatively, on a banked curve, a car is thrown off the road if it is moving fast. However, if the road has enough friction, then the car stays on the road and move safely. Since the car intends to slide off the road, then the static friction between the tires and the road must be towards the center in order to keep the car in the road.

5 0
4 years ago
A copper wire has a diameter of 4.00 x 10-2 inches and is originally 10.0 ft long. What is the greatest load that can be support
jek_recluse [69]

Complete question is;

A copper wire has a diameter of 4.00 × 10^(-2) inches and is originally 10.0 ft long. What is the greatest load that can be supported by this wire without exceeding its elastic limit? Use the value of 2.30 × 10⁴ lb/in² for the elastic limit of copper.

Answer:

F_max = 28.9 lbf

Explanation:

Elastic limit is simply the maximum amount of stress that can be applied to the wire before it permanently deform.

Thus;

Elastic limit = Max stress

Formula for max stress is;

Max stress = F_max/A

Thus;

Elastic limit = F_max/A

F_max is maximum load

A is area = πr²

We have diameter; d = 4 × 10^(-2) inches = 0.04 in

Radius; r = d/2 = 0.04/2 = 0.02

Plugging in the relevant values into the elastic limit equation, we have;

2.30 × 10⁴ = F_max/(π × 0.02²)

F_max = 2.30 × 10⁴ × (π × 0.02²)

F_max = 28.9 lbf

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