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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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Which of the following statements is TRUE? The side of the Moon facing away from the Earth is in perpetual darkness. The Moon is
Ivanshal [37]

Answer:

The Moon's distance from the Earth varies during its orbit

Explanation:

The correct statement is ,The Moon's distance from the Earth varies during its orbit.

Important point regarding moon:

1 .Moon is a natural satellite of the earth.

2. Moon is the fifth largest satellite in solar system.

3.Second densest  satellite in solar system.

4.Moon rotates about earth.

5.Moon is an astronomical body .

5 0
3 years ago
a foul ball is hit into the stands at a baseball game. the ball rises to a height of 38 meters and is caught on its way down by
lisov135 [29]

The velocity of the ball when it was caught is 12.52 m/s.

<em>"Your question is not complete it seems to be missing the following, information"</em>,

find the velocity of the ball when it was caught.

The given parameters;

maximum height above the ground reached by the ball, H = 38 m

height above the ground where the ball was caught, h = 30 m

The height traveled by the ball when it was caught is calculated as follows;

y = H - h

y = 38 - 30 = 8 m

The velocity of the ball when it was caught is calculated as;

v_f^2 = v_0 + 2gh\\\\v_f^2 = 0 + (2\times 9.8 \times 8)\\\\v_f^2 = 156.8\\\\v_f = \sqrt{156.8} \\\\v_f = 12.52 \ m/s

Thus, the velocity of the ball when it was caught is 12.52 m/s.

Learn more here: brainly.com/question/14582703

4 0
2 years ago
6. A 2-kg ball B is traveling around in a circle of radius r1 = 1 m with a speed (vB)1 = 2 m/s. If the attached cord is pulled d
kipiarov [429]

Answer:

Explanation:

Given that,

Mass of ball m = 2kg

Ball traveling a radius of r1= 1m.

Speed of ball is Vb = 2m/s

Attached cord pulled down at a speed of Vr = 0.5m/s

Final speed V = 4m/s

Let find the transverse component of the final speed using

V² = Vr²+ Vθ²

4² = 0.5² + Vθ²

Vθ² = 4²—0.5²

Vθ² = 15.75

Vθ =√15.75

Vθ = 3.97 m/s.

Using the conservation of angular momentum,

(HA)1 = (HA)2

Mb • Vb • r1 = Mb • Vθ • r2

Mb cancels out

Vb • r1 = Vθ • r2

2 × 1 = 3.97 × r2

r2 = 2/3.97

r2 = 0.504m

The distance r2 to the hole for the ball to reach a maximum speed of 4m/s is 0.504m

The required time,

Using equation of motion

V = ∆r/t

Then,

t = ∆r/Vr

t = (r1—r2) / Vr

t = (1—0.504) / 0.5

t = 0.496/0.5

t = 0.992 second

7 0
3 years ago
If a puddle of water is frozen, do the particles in the ice have kinetic energy? Explain.
TEA [102]
They have some but not very much, the particles in the ice are still vibrating just not as much as in water. the only time a substance would have 0 kinetic energy is when that substance is at 0 degrees kelvin(absolute zero) so far no place in the universe has been recorded at absolute zero though
8 0
3 years ago
he adventurous robot M.A.N.D.I. is orbiting Saturn’s moon Dione. She wants to cause an impact with themoon to kick up some of th
GuDViN [60]

Answer:

v = 2.928 10³ m / s

Explanation:

For this exercise we use Newton's second law where the force is the gravitational pull force

         F = ma

         a = F / m

Acceleration is

        a = dv / dt

        a = dv / dr dr / dt

        a = dv / dr v

        v dv = a dr

We substitute

       v dv = a dr

       ∫ v dv = 1 / m G m M ∫ 1 / r² dr

We integrate

       ½ v² = G M (-1 / r)

We evaluate from the lower limit v = 0 for r = R m to the upper limit v = v for r = R + 2.73 10³, where R is the radius of Saturn's moon

       v² = 2G M (- 1 / R +2.73 10³+ 1 / R)

         

We calculate

       v² = 2 6,674 10⁻¹¹ 1.10 10²¹ (10⁻³ / 5.61  - 10⁻³ /(5.61 + 2.73))

       v² = 14.6828 10⁷ (0.1783 -0.1199)

       v = √8.5748 10⁶

       v = 2.928 10³ m / s

5 0
3 years ago
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