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Vladimir79 [104]
3 years ago
13

Fast Bullets A rifle with that shoots bullets at 420 m/s shoots a bullet at a target 49.8 m away. How high above the target must

the rifle barrel be pointed to ensure that the bullet will hit the target?
Physics
1 answer:
Nezavi [6.7K]3 years ago
8 0

Answer:

0.08°

Explanation:

velocity of projection, u = 420 m/s

Horizontal range, R = 49.8 m

Let the angle of projection is θ.

Use the formula for the horizontal range.

R=\frac{u^{2}Sin2\theta }{g}

49.8=\frac{420^{2}Sin2\theta }{9.8}

Sin 2θ = .0027

2θ = 0.1585

θ = 0.08°

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You observe that a negatively charged plastic pen repels a charged piece of magic tape. You then observe that the same piece of
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Answer:

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Can a object have have zero velocity and nonzero acceleration?
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A 15 kg bucket of water is suspended by a very light ropewrapped around a solid uniform cylinder 0.300 m in diamter withmass 12.
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Answer:

Part a)

T = 42 N

Part b)

v_f = 11.8 m/s

Part c)

t = 1.7 s

Part d)

F = 159.7 N

Explanation:

Part a)

While bucket is falling downwards we have force equation of the bucket given as

mg - T = ma

for uniform cylinder we will have

TR = I\alpha

so we have

T = \frac{1}{2}MR^2(\frac{a}{R^2})

T = \frac{1}{2}Ma

now we have

mg = (\frac{M}{2} + m)a

a = \frac{mg}{(\frac{M}{2} + m)}

a = \frac{15 \times 9.81}{(6 + 15)}

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now we have

T = \frac{12 \times 7}{2}

T = 42 N

Part b)

speed of the bucket can be found using kinematics

so we have

v_f^2 - v_i^2 = 2 a d

v_f^2 - 0 = 2(7)(10)

v_f = 11.8 m/s

Part c)

now in order to find the time of fall we can use another equation

v_f - v_i = at

11.8 - 0 = 7 t

t = 1.7 s

Part d)

as we know that cylinder is at rest and not moving downwards

so here we can use force balance

F = T + Mg

F = 42 + (12 \times 9.81)

F = 159.7 N

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