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never [62]
3 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]3 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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Ignacio and Velvet walk eastward with a speed of 8 m/s. If it takes them 7.5 minutes to walk to the store, how far in meters hav
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Answer:

They walk 3,600 meters

Explanation:

7.5 minutes = 450 seconds.

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3 years ago
Read 2 more answers
Kyle is flying a helicopter at 125 m/s on a heading of 325 o . If a wind is blowing at 25 m/s toward a direction of 240.0 o , wh
frosja888 [35]

Answer:

The resultant velocity of the helicopter is \vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right).

Explanation:

Physically speaking, the resulting velocity of the helicopter (\vec v_{H}), measured in meters per second, is equal to the absolute velocity of the wind (\vec v_{W}), measured in meters per second, plus the velocity of the helicopter relative to wind (\vec v_{H/W}), also call velocity at still air, measured in meters per second. That is:

\vec v_{H} = \vec v_{W}+\vec v_{H/W} (1)

In addition, vectors in rectangular form are defined by the following expression:

\vec v = \|\vec v\| \cdot (\cos \alpha, \sin \alpha) (2)

Where:

\|\vec v\| - Magnitude, measured in meters per second.

\alpha - Direction angle, measured in sexagesimal degrees.

Then, (1) is expanded by applying (2):

\vec v_{H} = \|\vec v_{W}\| \cdot (\cos \alpha_{W},\sin \alpha_{W}) +\|\vec v_{H/W}\| \cdot (\cos \alpha_{H/W},\sin \alpha_{H/W}) (3)

\vec v_{H} = \left(\|\vec v_{W}\|\cdot \cos \alpha_{W}+\|\vec v_{H/W}\|\cdot \cos \alpha_{H/W}, \|\vec v_{W}\|\cdot \sin \alpha_{W}+\|\vec v_{H/W}\|\cdot \sin \alpha_{H/W} \right)

If we know that \|\vec v_{W}\| = 25\,\frac{m}{s}, \|\vec v_{H/W}\| = 125\,\frac{m}{s}, \alpha_{W} = 240^{\circ} and \alpha_{H/W} = 325^{\circ}, then the resulting velocity of the helicopter is:

\vec v_{H} = \left(\left(25\,\frac{m}{s} \right)\cdot \cos 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \cos 325^{\circ}, \left(25\,\frac{m}{s} \right)\cdot \sin 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \sin 325^{\circ}\right)\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)

The resultant velocity of the helicopter is \vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right).

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Answer:

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Answer:

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Explanation:

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When the freight car is moving at a certain speed, it has a constant momentum.

We will call this M1.

The equation for M1 will be:

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Since conservation of momentum states that the momentum will remain the same. We have:

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Since M1 is constant, if the mass increases, the speed had to decrease to keep the equation true.

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