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Slav-nsk [51]
3 years ago
15

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 rifleshoulder 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:
BlackZzzverrR [31]3 years ago
7 0

Answer:

(a) 4.58 m/s

(b) 31.5 J

(c) 0.491 m/s

(d) 0.362 J

(e) Momentum of player = 880 kg m/s

Momentum of ball = 10.3 kg m/s

Explanation:

(a) By Newton's third law of motion, action and reaction are equal and opposite.

The momentum action of bullet = (0.0250 kg)(550 m/s)

The momentum reaction of rifle = (3.00 kg)(v m/s)

Hence,

(0.0250 kg)(550 m/s) = (3.00 kg)(v m/s)

v =(0.0250 kg)(550 m/s)/(3.00 kg) = 4.58 m/s

(b) Kinetic energy = \frac{1}{2}mv^2 = \frac{1}{2}\times3.00\text{ kg}\times (4.58\text{ m/s})^2 = 31.5\text{ J}

(c) If the effective mass of the rifle is now 28.0 kg, then it's recoil speed is

v =(0.0250 kg)(550 m/s)/(28.0 kg) = 0.491 m/s

(d) With the new effective mass, kinetic energy is

\frac{1}{2}mv^2 = \frac{1}{2}\times3.00\text{ kg}\times (0.491\text{ m/s})^2 = 0.362\text{ J}

(e) The momentum of the player =(110 kg)(8.00 m/s) = 880 kg m/s

Momentum of ball = (0.41 kg)(25.0 m/s) = 10.3 kg m/s

The momentum of the bullet in the question is = (0.0250 kg)(550 m/s) = 13.8 kg m/s.

This momentum is much smaller than that of the player. Hence, an object or body will feel more 'pain' or impact when hit by the running player than by the bullet.

The impact of the bullet is only slightly more than that of the thrown ball.

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As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff springs arr
zubka84 [21]

Answer:

894.12\ \text{N}

284.013\ \text{J}

Explanation:

U = Energy = 76 J

x = Displacement of spring = 0.17 m

k = Spring constant

Energy is given by

U=\dfrac{1}{2}kx^2\\\Rightarrow k=\dfrac{2U}{x^2}\\\Rightarrow k=\dfrac{2\times 76}{0.17^2}\\\Rightarrow k=5259.51\ \text{Nm}

Force is given by

F=kx\\\Rightarrow F=5259.51\times 0.17\\\Rightarrow F=894.12\ \text{N}

The magnitude of force must you apply to hold the platform is 894.12\ \text{N}.

Now x=0.17+0.2=0.37\ \text{m}

U=\dfrac{1}{2}kx^2\\\Rightarrow U=\dfrac{1}{2}\times 5259.51\times 0.37^2\\\Rightarrow U=360.013\ \text{J}

Additional energy

360.013-76=284.013\ \text{J}

8 0
3 years ago
3. A 0.145 kg ball moving horizontally at 20 m/s is struck by a bat that causes the ball to move in the
kumpel [21]

Answer:

F = 1015 N

Explanation:

Given that,

Mass of a ball, m = 0.145 kg

Initial speed, u = 20 m/s

Final speed, v = -15 m/s (as it moves in opposite direction)

The ball and bat were in contact for 5 ms

We need to find the force the bat applies to the bat. We know that,

F = ma

F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{0.145\times (-15-20)}{5\times 10^{-3}}\\\\F=-1015\ N

So, the bat will apply 1015 N force to the ball.

7 0
3 years ago
How much time does it take for a bird flying at a speed of 45 miles per hour to travel a distance
Marina CMI [18]

Answer:

40

Explanation:

5 0
3 years ago
You are holding a positive charge and there are positive charges of equal magnitude 1 m to your north and 1 m to your east. what
pychu [463]

By holding a positive charge and there are positive charges of equal magnitude 1 m to your north and 1 m to your east. Therefore, the direction of the force on the charge you are holding will be to the southwest.

Let I hold the charge , q at the centre of given co-ordinate system and two positive charge of equal magnitude Q are placed 1 m to my North and 1 m to my South .

now, both the charge are same nature e.g., positive . Let my charge is also positive (well, you can assume negative too , I am considering positive because it makes me easy to solve) then, both charge repel to my charge.

charge Q placed on east is repelling my charge q toward west . similarly charge Q placed on North is repelling my charge q toward south.

Now , use vector for solve it.

vector F_{net} = vector Fe + vector Fn,

⇒ |F_{net}| = \sqrt{}  F^{2} _{e } + F^{2}_n

⇒ Fe = Fs = KqQ/(1m)² = KqQ

⇒ F_{net} = √{Fe² + Fs²} = √{(kqQ)²+(KqQ)²}

⇒ F_{net}= √2KqQ

Hence, net force act on q {my charge } is √2KqQ and the direction of force is S - W (southwest )direction.

To learn more about positive charges here

brainly.com/question/2903220

#SPJ4

8 0
2 years ago
HELP!!
tresset_1 [31]

Answer:

So do 2400 divided by 70. I got 34.285714 and the numbers behind the decimal are repeating. If you round it you get 34.3

3 0
3 years ago
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