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julia-pushkina [17]
3 years ago
7

A 0.71 kg spike is hammered into a railroad tie. The initial speed of the spike is equal to 3.8 m/s. If the tie and spike togeth

er absorb 38.2 percent of the spike’s initial kinetic energy as internal energy, calculate the increase in internal energy of the tie and spike. Answer in units of J.
Physics
1 answer:
Alexeev081 [22]3 years ago
6 0

Answer:

1.96 J

Explanation:

From the law of conservation of energy

ΔU + ΔK = 0  where ΔU = internal energy change and ΔK = kinetic energy change. We neglect potential energy change since we are not given any information about it.

ΔU = -ΔK

ΔK = K₂ - K₁ where  K₁ = initial kinetic energy and K₂ = final kinetic energy = 0 where  ΔU = 0.382K₁

     = 0.382mv²/2  where m = mass of spike = 0.71 kg and v = initial speed of spike = 3.8 m/s

     = 0.382 × 0.71 kg × (3.8 m/s)²/2

     = 1.96 J

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The lowest note on a grand piano has a frequency of 27.5 Hz. The entire string is 2.00 m long and has a mass of 400 g. The vibra
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Answer:

1456 N

Explanation:

Given that

Frequency of the piano, f = 27.5 Hz

Entire length of the string, l = 2 m

Mass of the piano, m = 400 g

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The formula for the tension is represented as

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T = tension

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F = frequency

l = length of the whole part

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T = 1456 N

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4 0
3 years ago
A metal rod has a moves with a constant velocity of 40 cm/s along two parallel metal rails through a magnetic field of 0.575 T.
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Answer:

2.12/R mW

Explanation:

The electrical power, P generated by the rod is

P = B²L²v²/R where B = magnetic field = 0.575 T, L = length of metal rod = separation of metal rails = 20 cm = 0.2 m, v = velocity of metal rod = 40 cm/s = 0.4 m/s and R = resistance of rod = ?

So, the induced emf on the conductor is

E = BLv

= 0.575 T × 0.2 m × 0.4 m/s

= 0.046 V

= 46 mV

The electrical power, P generated by the rod is

P = B²L²v²/R

=  B²L²v²/R

So, P = (0.575 T)² × (0.2 m)² × (0.4 m/s)²

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Two equal forces are applied to a door. The first force is applied at the midpoint of the door, the second force is applied at t
Orlov [11]

Answer:

D) the second at the doorknob

Explanation:

The torque exerted by a force is given by:

\tau = Fdsin \theta

where

F is the magnitude of the force

d is the distance between the point of application of the force and the centre of rotation

\theta is the angle between the direction of the force and d

In this problem, we have:

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