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never [62]
2 years ago
12

A 3.50-g bullet has a muzzle velocity of 250 m/s when fired by a rifle with a weight of 25.0 N. (a) Determine the recoil speed o

f the rifle. m/s (b) If a marksman with a weight of 650 N holds the rifle firmly against his shoulder, determine the recoil speed of the shooter and rifle.
Physics
1 answer:
enyata [817]2 years ago
3 0

Answer:

(a) 0.343 m/s

(b) 0.012 m/s

Explanation:

(a) From the question above,

MV = mv............................... Equation 1

Where M = mass of the rifle, V = recoiling speed of the rifle, m = mass of the bullet, v = velocity of the bullet.

make V the subject of the equation

V = mv/M........................... Equation 2

Given: m = 3.5 g = 0.0035 kg, v = 250 m/s, M = 25 N = 25/9.8 = 2.55 kg.

Substitute into equation 2

V = (0.0035×250)/2.55

V = 0.343 m/s.

(b) Similarly,

(M'+M)V' = mv....................... Equation 3

Where M' = mass of the marksman, V' = recoiling speed of the shooter and rifle

make V' the subject of the equation

V' = mv/(M'+M)................... Equation 4

Given: m = 3.5 g = 0.0035 kg, v = 250 m/s, M = 25 N = 25/9.8 = 2.55 kg, M' = 650 N = 650/9.8 = 66.33 N

Substitute into equation 4

V' = (0.0035×250)/(66.33+2.55)

V' = 0.8125/68.88

V' = 0.012 m/s

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Answer:

Distance, d = 192 meters

Explanation:

We have,

Initial velocity of an object is 10 m/s

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We need to find the distance travelled by the object during that time. Second equation of motion gives the distance travelled by the object. It is given by :

d=ut+\dfrac{1}{2}at^2

d=10\times 8+\dfrac{1}{2}\times 3.5\times 8^2\\\\d=192\ m

So, the distance travelled by the object is 192 meters.

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Answer:

Explained below

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