1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Dominik [7]
2 years ago
14

A 0.0250-kg bullet is accelerated from rest to a speed of 550 m/s in a 3.00-kg rifle. The pain of the rifle’s kick is much worse

if you hold the gun loosely a few centimeters from your shoulder rather than holding it tightly against your shoulder. (a) Calculate the recoil velocity of the rifle if it is held loosely away from the shoulder. (b) How much kinetic energy does the rifle gain? (c) What is the recoil velocity if the rifle is held tightly against the shoulder, making the effective mass 28.0 kg? (d) How much kinetic energy is transferred to the rifle-shoulder combination? The pain is related to the amount of kinetic energy, which is significantly less in this latter situation. (e) Calculate the momentum of a 110-kg football player running at 8.00 m/s. Compare the player’s momentum with the momentum of a hard-thrown 0.410-kg football that has a speed of 25.0 m/s. Discuss its relationship to this problem.
Physics
1 answer:
kondaur [170]2 years ago
4 0

Answer:

a) 4.583 m/s

b) 31.505 J

c) 0.491 m/s

d) 3.375 J

e)

   p_player = (110 kg)(8 m/s) = 880 kg m/s

   p_ball = (0.41 kg)(25 m/s) = 10.25 kg m/s

Explanation:

HI!

a)

We can calculate the recoil velocity by conservation of momentum, remember that p=mv.

The momentum of the bullet is:

p_b = (0.0250 kg)*(550 m/s )

The momentum of the rifle is:

p_r = (3 kg) * v

Since the total initial momentum is zero:

p_b = p_r

That is:

v = (550 m/s ) (0.0250 kg/ 3 kg ) = 4.583 m/s

b)

The kinetic energy gained by the rifle is:

K = (1/2) m v^2 = (1/2) *(3 kg) *(4.583 m/s)^2 = 31.505 J

c)

We use the same formula as in a), but with m=28kg instead of 3 kg

v = (550 m/s ) (0.0250 kg/ 28 kg ) = 0.491 m/s

d)

Again, the same formula as b, but with m=28 and v=0.491 m/s

K = 3.375 J

e)

p_player = (110 kg)(8 m/s) = 880 kg m/s

p_ball = (0.41 kg)(25 m/s) = 10.25 kg m/s

I believe that the kinetic energy is more related to the problem than the momentum. The relation between these two quantities is:

K = p^2/(2m)

usiing this relation, we get:

K_player = 3520 J

K_ball =  128.125 J

Therefore the kinetic energy of the player is around 27 time larger than the kinetic energy of the ball, that being said, the pain of being tackled by that player is around 27 times greater that being hit by the ball!

You might be interested in
describe the physics modeling process in moving a basketball and say what effects you should overlook​
Dahasolnce [82]

Answer:ui9i

Explanation:

jiooooooooooooo

7 0
2 years ago
An athlete needs to lose weight and decides to do it by “pumping iron.” (a) How many times must an 90.0 kg weight be lifted a di
Wittaler [7]

Answer:

230kg would be the best answer

Explanation:

7 0
3 years ago
A very small object carrying -7 μC of charge is attracted to a large, well-anchored, positively charged object. How much kinetic
miss Akunina [59]

Answer:

ΔK.E = 14 nJ

Explanation:

Solution:

- The charge that moves under the influence of an Electric Field produced between a potential difference (V) stores electric potential energy U within that is converted to kinetic energy.

- We will use conservation of energy on the system that contains the charged particle with charge q loses its electric potential energy U as it moves towards positively charged object that converts into a gain in Kinetic energy of the charged particle ΔK.E:

                                 ΔK.E = U

Where,

                                 U = V*q

                                 ΔK.E = V*q

                                 ΔK.E = (7*10^-6)*(2*10^-3)

                                 ΔK.E = 14 nJ

- The gain in kinetic energy is 14 nJ.

7 0
3 years ago
What is someones target heart rate at 70% at age 17​
baherus [9]

Answer: normal

Explanation:

6 0
2 years ago
What are wind and water erosion not likely to affect
zavuch27 [327]

Answer:

Tides, bodies of water

Explanation:

5 0
2 years ago
Read 2 more answers
Other questions:
  • An area of wave-washed sediment along a coast is called a __________.
    11·2 answers
  • Why do astronauts feel weightless in space?
    11·1 answer
  • A rigid container holds 0.30g of hydrogen gas.
    12·1 answer
  • Which lobes of the brain receive the input that enables you to feel someone scratching your back? (2 points)
    8·1 answer
  • In which type of mixture can different parts be seen?
    10·1 answer
  • As the proton approaches the uranium nucleus, the repulsive force slows down the proton until it comes momentarily to rest, afte
    8·1 answer
  • Plan an experiment to measure the ideal mechanical advantage of a three-hole punch. (a) What materials would you need? (b) What
    9·1 answer
  • A 7600 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.35 m/s2 and feels no appreci
    9·1 answer
  • URGENT!!!
    7·1 answer
  • If a battery causes a wire to carry a current of 4 Amps how many coulombs of charge flow past any point in the wire in 3 seconds
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!