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Natasha2012 [34]
2 years ago
7

7. A 1.0 kg metal head of a geology hammer strikes a solid rock with a velocity of 5.0 m/s. Assuming all the energy is retained

by the hammer head, how much will its temperature increase
Physics
1 answer:
marta [7]2 years ago
8 0

The increase in temperature of the metal hammer is 0.028 ⁰C.

The given parameters:

  • <em>mass of the metal hammer, m = 1.0 kg</em>
  • <em>speed of the hammer, v = 5.0 m/s</em>
  • <em>specific heat capacity of iron, 450 J/kg⁰C</em>

The increase in temperature of the metal hammer is calculated as follows;

Q = K.E\\\\mc \Delta T = \frac{1}{2}  mv^2\\\\\Delta T = \frac{v^2}{2 c}

where;

<em>c is the </em><em>specific heat capacity</em><em> of the metal hammer</em>

<em />

Assuming the metal hammer is iron, c = 450 J/kg⁰C

\Delta T = \frac{5^2}{2 \times 450} \\\\\Delta T = 0.028 \ ^0C

Thus, the increase in temperature of the metal hammer is 0.028 ⁰C.

Learn more about heat capacity here: brainly.com/question/16559442

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Which equations represent the relationship between wavelength and frequency for a sound wave? Check all that apply.
Luba_88 [7]

Answer: A

Explanation:

8 0
3 years ago
a lorry travels 3600m on a test track accelerating constantly at 3m/s squared from standstill. what is the final velocity (3 sig
suter [353]

The final velocity of the truck is found as 146.969 m/s.

Explanation:

As it is stated that the lorry was in standstill position before travelling a distance or covering a distance of 3600 m, the initial velocity is considered as zero. Then, it is stated that the lorry travels with constant acceleration. So we can use the equations of motion to determine the final velocity of the lorry when it reaches 3600 m distance.

Thus, a initial velocity (u) = 0, acceleration a = 3 m/s² and the displacement s is 3600 m. The third equation of motion should be used to determine the final velocity as below.

2as =v^{2} - u^{2} \\\\v^{2} = 2as + u^{2}

Then, the final velocity will be

v^{2} = 2 * 3 * 3600 + 0 = 21600\\ \\v=\sqrt{21600}=146.969 m/s

Thus,  the final velocity of the truck is found as 146.969 m/s.

8 0
3 years ago
A Tennis ball falls from a height 40m above the ground the ball rebounds
worty [1.4K]

If the ball is dropped with no initial velocity, then its velocity <em>v</em> at time <em>t</em> before it hits the ground is

<em>v</em> = -<em>g t</em>

where <em>g</em> = 9.80 m/s² is the magnitude of acceleration due to gravity.

Its height <em>y</em> is

<em>y</em> = 40 m - 1/2 <em>g</em> <em>t</em>²

The ball is dropped from a 40 m height, so that it takes

0 = 40 m - 1/2 <em>g</em> <em>t</em>²

==>  <em>t</em> = √(80/<em>g</em>) s ≈ 2.86 s

for it to reach the ground, after which time it attains a velocity of

<em>v</em> = -<em>g</em> (√(80/<em>g</em>) s)

==>  <em>v</em> = -√(80<em>g</em>) m/s ≈ -28.0 m/s

During the next bounce, the ball's speed is halved, so its height is given by

<em>y</em> = (14 m/s) <em>t</em> - 1/2 <em>g</em> <em>t</em>²

Solve <em>y</em> = 0 for <em>t</em> to see how long it's airborne during this bounce:

0 = (14 m/s) <em>t</em> - 1/2 <em>g</em> <em>t</em>²

0 = <em>t</em> (14 m/s - 1/2 <em>g</em> <em>t</em>)

==>  <em>t</em> = 28/<em>g</em> s ≈ 2.86 s

So the ball completes 2 bounces within approximately 5.72 s, which means that after 5 s the ball has a height of

<em>y</em> = (14 m/s) (5 s - 2.86 s) - 1/2 <em>g</em> (5 s - 2.86 s)²

==>  (i) <em>y</em> ≈ 7.5 m

(ii) The ball will technically keep bouncing forever, since the speed of the ball is only getting halved each time it bounces. But <em>y</em> will converge to 0 as <em>t</em> gets arbitrarily larger. We can't realistically answer this question without being given some threshold for deciding when the ball is perfectly still.

During the first bounce, the ball starts with velocity 14 m/s, so the second bounce begins with 7 m/s, and the third with 3.5 m/s. The ball's height during this bounce is

<em>y</em> = (3.5 m/s) <em>t</em> - 1/2 <em>g</em> <em>t</em>²

Solve <em>y</em> = 0 for <em>t</em> :

0 = (3.5 m/s) <em>t</em> - 1/2 <em>g t</em>²

0 = <em>t</em> (3.5 m/s - 1/2 <em>g</em> <em>t</em>)

==>  (iii) <em>t</em> = 7/<em>g</em> m/s ≈ 0.714 s

As we showed earlier, the ball is in the air for 2.86 s before hitting the ground for the first time, then in the air for another 2.86 s (total 5.72 s) before bouncing a second time. At the point, the ball starts with an initial velocity of 7 m/s, so its velocity at time <em>t</em> after 5.72 s (but before reaching the ground again) would be

<em>v</em> = 7 m/s - <em>g t</em>

At 6 s, the ball has velocity

(iv) <em>v</em> = 7 m/s - <em>g</em> (6 s - 5.72 s) ≈ 4.26 m/s

4 0
3 years ago
if you crash your car how could you decrease the damage to you or the car using the concept of impulse
kotykmax [81]

Explanation:

Crumple zones are sections in cars that are designed to crumple up when the car encounters a collision. Crumple zones minimize the effect of the force in an automobile collision in two ways. By crumpling, the car is less likely to rebound upon impact, thus minimizing the momentum change and the impulse.

3 0
3 years ago
Can you make the work output of a machine greater than the work input?
ExtremeBDS [4]

Answer:

It is<em> impossible</em> to construct a machine which produces the <em>work output greater than the work input.</em>

Let us consider the II law of thermodynamics.

According to Kelvin Plank's statement any engine/machine does not give hundred percent efficiency. And violating the PMM-II(Perpetual motion of machine II kind), Always some amount of energy transferred  to the sink or surroundings.

Therefore

                     W(ouput) = Q₁-Q₂

There are many reasons to lower the work output, just for an example friction between the mating parts reduces the work output.

5 0
3 years ago
Read 2 more answers
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