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Andrew [12]
3 years ago
12

A bullet is fired from a gun at 30 angle to the horizontal with a muzzle velocity of 600 m/s.

Physics
1 answer:
boyakko [2]3 years ago
6 0

Answer:

Part a)

Time to reach the maximum height is given as

t = 30 s

B)

Maximum height of the bullet is

H = 4500 m

C)

Horizontal range of the bullet is

R = 31177 m

Part d)

Final velocity of the bullet is v = 600 m/s at 30 degree

Explanation:

A) As we know that velocity of the bullet is

v = 600 m/s

\theta = 30^0

So we have

v_y = v sin\theta

v_y = 600 sin30 = 300 m/s

Now time to reach the maximum height is given as

0 = 300 - g t

t = 30 s

B) As we know that the object have zero final speed in y direction

0 - v_y^2 = 2(-g) h

0 - 300^2 = (2)(-10)H

H = 4500 m

C)

Now the horizontal displacement of the object in total time of flight is known as range of projectile

So we have

R = v_x T

R = (600 cos30)(2 \times 30)

R = 31177 m

Part d)

As we know that the final height is same as that of initial So it will strike the ground at same velocity as that of initial velocity

So we have

v = 600 m/s at 30 degree

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A 40 kg girl and an 8.4 kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope, the girl exerts a 5.2 N for
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Answer:

(a) a_s=0.62\frac{m}{s^2}

(b) a_s=0.13\frac{m}{s^2}

(c) x_f=2.6m

Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}

(c) Using the kinematics equation:

x_f=x_0+v_0t \pm  \frac{at^2}{2}

For the girl, we have x_0=0 and v_0=0. So:

x_f_g=\frac{a_gt^2}{2}(1)

For the sled, we have v_0=0. So:

x_f_s=x_0_s-\frac{a_st^2}{2}(2)

When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s

Now, we solve (1) for x_f_g

x_f_g=\frac{0.13\frac{m}{s^2}(6.32s)^2}{2}\\x_f_g=2.6m\\x_f=2.6m

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3 years ago
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In severe head-on automobile accidents, a deceleration of 60 g’s or more (1 g 5 32.2 ft/s2) often results in a fatality. What fo
Musya8 [376]

Answer:

The resulting force on the child is 3000 lbf

Explanation:

To find the force that acts on a child of 50 lb with a deceleration of 60 g's, we can use the formula:

Force = mass * acceleration

To find the force in lbf, we need to use the mass in lb and the acceleration in g (standard unit of gravity).

So we have that:

Force = 50 * 60

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So the resulting force on the child is 3000 lbf.

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