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Oduvanchick [21]
3 years ago
5

A 23 g bullet is accelerated in a rifle barrel 62 cm long to a speed of 593 m/s. Use the work-energy theorem to find the average

force exerted on the bullet while it is being accelerated. Answer in units of N.
Physics
1 answer:
Anuta_ua [19.1K]3 years ago
6 0

Answer:

The average force exerted on the bullet while it is being accelerated is 6,522.52 N.

Explanation:

Given;

mass of the bullet, m = 23 g = 0.023 kg

length of the barrel, L = 62 cm = 0.62 m

speed of the bullet, v = 593 m/s

Applying work-energy theorem;

the work done in accelerating the bullet in the riffle = kinetic energy acquired by the bullet.

W = K.E

F x d = ¹/₂mv²

where;

d is the is the distance traveled by the bullet in the riffle = L

F(0.62) = ¹/₂ x 0.023 x (593²)

F(0.62) = 4043.964

F = (4043.964) / (0.62)

F = 6,522.52 N

Therefore, the average force exerted on the bullet while it is being accelerated is 6,522.52 N.

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If a 2 kg object is falling at 3 m/s at what rate is gravity working on the object
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+9.8m/s^2

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The rate of gravity of the object is constant thriughout the surface of the earth.

For falling object, the rate of gravity is positive since the body is coming down (falling)

The rate of gravity is negative if the body is going up

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4 0
3 years ago
A certain automobile is 6.0 m long if at rest. If it is measured to be 4.8 m long while moving, its speed is:
tatuchka [14]

Answer:

1.8*10^8m/s

Explanation:

Using

L= lo√1-v²/c²

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V= 0.6*3E8m/s

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7 0
3 years ago
Automobile A and B are initially 30 m apart travelling in adjacent highway lanes at speeds VA = 14.4 km/hr., VB 23.4 km/hr. at t
marshall27 [118]

Answer:

        x = 240 m

Explanation:

This is a kinematics exercise

Let's fix our frame of reference on car A

           x = x₀ₐ+ v₀ₐ t + ½ aₐ t²

         

the initial position of car a is zero

           x = 0 + v₀ₐ t + ½ 0.8 t²

for car B

          x = x_{ob} + v_{ob} t - ½ a_b t²

     

car B's starting position is 30 m

         x = 30 + v_{ob} t - ½ 0.4 t²

at the point where they meet, the position of the two vehicles is the same

         0 + v₀ₐ t + ½ 0.8 t² = 30 + v_{ob} t - ½ 0.4 t²

let's reduce the speeds to the SI system

        v₀ₐ = 14.4 km / h (1000 m / 1 km) (1h / 3600s) = 4 m / s

        v_{ob} = 23.4 km / h = 6.5 m / s

        4 t + 0.4 t² = 30 + 6.5 t - 0.2 t²

        0.2 t² - 2.5 t - 30 = 0

        t² - 12.5 t - 150 = 0

we solve the quadratic equation

       t = \frac{12.5 \pm \sqrt{12.5^2 + 4 \ 150}  }{2}

       t = \frac{12.5 \  \pm 27.5}{2}

       t₁ = 20 s

       t₂ = -7.5 s

time must be a positive quantity so the correct result is t = 20 s

let's look for the distance

        x = 4 t + ½ 0.8 t²

        x = 4 20 + ½ 0.8 20²

        x = 240 m

8 0
3 years ago
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