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Kay [80]
3 years ago
10

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?
Physics
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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You are standing a distance of 17.0 meters from the center of a merry-go-round. The merry-go-round takes 9.50 seconds to go comp
GenaCL600 [577]

Answer:

(a)11.24 m/s

(b)7.44 m/s

(c)409 N

(d)539.55\mu

(e) 0

Explanation:

The period for 1 circle 2\pi of the merry go around is 9.5s. It means the angular speed is:

\omega = \theta / t = 2\pi / 9.5 \approx 0.661 rad/s

(a)The speed is

v = \omega * R = 0.661 * 17 = 11.24 m/s

(b) Centripetal acceleration:

a = \frac{v^2}{R} = \frac{11.24^2}{17} = 7.44 m/s^2

(c) Magnitude of the force that keeps you go around at this acceleration

F = ma = 55 * 7.44 = 409 N

(d) let the coefficient of friction by \mu. The frictional force shall be this coefficient multiplied by normal force reverting gravity of the man

F_f = mg\mu = 55*9.81\mu = 539.55\mu

3 0
3 years ago
What is the total energy of a particle with a rest mass of 1 gram moving with half the speed of light? 1 eV = 1.6 x 10^-19 J. An
GREYUIT [131]

Answer:

6.5 x 10^32 eV

Explanation:

mass of particle, mo = 1 g = 0.001 kg

velocity of particle, v = half of velocity of light = c / 2

c = 3 x 10^8 m/s

Energy associated to the particle

E = γ mo c^2

E=\frac{m_{0}c^}2}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

E=\frac{m_{0}c^}2}{\sqrt{1-\frac{c^{2}}{4c^{2}}}}

E=\frac{2m_{0}c^}2}{\sqrt{3}}

E=\frac{2\times0.001\times9\times10^{16}}{1.732}

E=1.04\times10^{14}J

Convert Joule into eV

1 eV = 1.6 x 10^-19 J

So, E=\frac{1.04\times10^{14}}{1.6\times10^{-19}}=6.5\times10^{32}eV

4 0
3 years ago
A falcon can descend with a speed 250 km/h. If a falcon flies at this speed for 2.0 s and then flies a 100 m in 2.5 s, what is t
Vadim26 [7]

Answer:

v= s/t

Explanation:

250 km/ h =69.44m/s

S1=2 times 69.44 ≈ 139m

Next 2.5 seconds:

S2 = 100m

Average speed:

v=139m+100m/2s+2.5s = 239/4.5s = 53.2 m/s=192km/h

3 0
3 years ago
What can you use to determine whether the SATA port no which you are connecting the drive will also run at 6.0Gbps?
kolbaska11 [484]

Answer:

The answer is "use manual motherboard".

Explanation:

The motherboard is also known as the mainboard, it an electronic circuit board, that can connect with the CPU, RAM, and other networking equipment parts. It is also is known as a chipset, that differ widely in style, context, power source, height and performance (Form Factor).

All the data of the computer is stored memory, which checks into the motherboard, that the SATA port which you are connected to is still going to run at 6.0Gbps or not.

8 0
3 years ago
The diagram shows forces acting on a boat.
Greeley [361]
A

I hope this helps!!:)
7 0
3 years ago
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