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3 years ago

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A rifle with a mass of 0.9 kg fires a bullet with a mass of 6.0 g (0.006 kg). The bullet moves with a muzzle velocity of 750 m/s

after the rifle is fired a. What is the momentum of the bullet after the rifle is fired? b. If external forces acting on the rifle can be ignored, what is the recoil velocity of the rifle?

1 answer:

SCORPION-xisa [38]3 years ago

5 0

Answer:

a ) 4.5 N.s

b) V =5 m/s

Explanation:

given,

mass of rifle(M) = 0.9 kg

mass of bullet(m) = 6 g = 0.006 kg

velocity of the bullet(v) = 750 m/s

a) momentum of bullet = m × v

= 750 × 0.006

= 4.5 N.s

b) recoil velocity

m × u + M × U = m × v + M × V

0 + 0 = 0.006 × 750 - 0.9 × V

V = $\dfrac{4.5}{0.9}$

V =5 m/s

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1 year ago

A helicopter lifts a 72 kg astronaut 15 m vertically from the ocean by means of a cable. The acceleration of the astronaut is g/

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a) $F_H=776.952\ N$

b) $F_g=706.32\ N$

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Explanation:

Given:

mass of the astronaut, $m=72\ kg$

vertical displacement of the astronaut, $h=15\ m$

acceleration of the astronaut while the lift, $a=\frac{g}{10} =0.981\ m.s^{-2}$

a)

<u>Now the force of lift by the helicopter:</u>

Here the lift force is the resultant of the force of gravity being overcome by the force of helicopter.

$F_H-F_g=m.a$

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$F_H=$ force by the helicopter

$F_g=$ force of gravity

$F_H=72\times 0.981+72\times9.81$

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b)

The gravitational force on the astronaut:

$F_g=m.g$

$F_g=72\times 9.81$

$F_g=706.32\ N$

d)

Since the astronaut has been picked from an ocean we assume her initial velocity to be zero, $u=0\ m.s^{-1}$

using equation of motion:

$v^2=u^2+2a.h$

$v^2=0^2+2\times 0.981\times 15$

$v=5.4249\ m.s^{-1}$

c)

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3 years ago

Read 2 more answers

A stone is thrown from the top of a building 30m high. If it strikes the ground at an angle 45, with what speed was it thrown?

sasho [114]

Answer:

24.2 m/s

Explanation:

The stone strikes the ground at an angle of 45 degrees: this means that its vertical velocity is equal (in magnitude) to its horizontal velocity, in fact:

$tan \theta = \frac{|v_y|}{v_x}\\tan 45^{\circ} = 1 \rightarrow |v_y| = v_x$

The motion along the vertical direction is a uniformly accelerated motion, so we can find the final vertical velocity using the following suvat equation

$v_y^2 -u_y^2 = 2as$

where

$v_y$ is the final vertical velocity

$u_y = 0$ is the initial vertical velocity (zero because the stone is thrown horizontally)

$a=g=9.8 m/s^2$ is the acceleration of gravity (we take downward as positive direction)

s = 30 m is the vertical displacement

Solving for vy,

$v_y = \sqrt{u_y^2+2as}=\sqrt{0+2(9.8)(30)}=24.2 m/s$

This means that the horizontal velocity is also 24.2 m/s: and since the horizontal velocity is constant during the whole motion (there is no acceleration in the horizontal direction), this means that the stone was thrown exactly at 24.2 m/s.

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