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vlabodo [156]
4 years ago
15

A particular gun (10kg) is able to fire 20 gram bullets at a speed of 350 m/s. From this information, calculate roughly how much

energy contained in the powder in the bullet (Hint: you will need to calculate the recoil speed of the gun.) A. 2455 J B. 620 J C. Cannot be determined. D. 1230 J
Physics
1 answer:
lakkis [162]4 years ago
7 0

Answer:

The answer is

D. 1230j

Explanation:

When a bullet is shot out of a gun the person firing experiences a backward impact, which is the recoil force, while the force propelling the bullet out of the gun is the propulsive force

given data

Mass of gun M=10kg

Mass of bullet m=20g----kg=20/1000 =0.02kg

Propulsive speed of bullet = 350m/s

Hence the moment of the bullet will be equal and opposite to that of the gun

mv=MV

where V is the recoil velocity which we are solving for

V=mv/M

V=0.02*350/10

V=7/10

V=0.7m/s

The energy contained in the bullet can be gotten using

KE=1/2m(v-V)²

KE=1/2*0.02(350-0.7)²

KE=1/2*0.02(349.3)²

KE=1/2*0.02*122010.49

KE=1/2*2440.20

KE=1220.1J

roughly the energy is 1230J

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Find the ratio of the diameter of aluminium to copper wire, if they have the same
kicyunya [14]

Answer:

1.24

Explanation:

The resistivity of copper\rho_1=2.65\times 10^{-8}\ \Omega-m

The resistivity of Aluminum,\rho_2=1.72\times 10^{-8}\ \Omega-m

The wires have same resistance per unit length.

The resistance of a wire is given by :

R=\rho \dfrac{l}{A}\\\\R=\rho \dfrac{l}{\pi (\dfrac{d}{2})^2}\\\\\dfrac{R}{l}=\rho \dfrac{1}{\pi (\dfrac{d}{2})^2}

According to given condition,

\rho_1 \dfrac{1}{\pi (\dfrac{d_1}{2})^2}=\rho_2 \dfrac{1}{\pi (\dfrac{d_2}{2})^2}\\\\\rho_1 \dfrac{1}{{d_1}^2}=\rho_2 \dfrac{1}{{d_2}^2}\\\\(\dfrac{d_2}{d_1})^2=\dfrac{\rho_1}{\rho_2}\\\\\dfrac{d_2}{d_1}=\sqrt{\dfrac{\rho_1}{\rho_2}}\\\\\dfrac{d_2}{d_1}=\sqrt{\dfrac{2.65\times 10^{-8}}{1.72\times 10^{-8}}}\\\\=1.24

So, the required ratio of the diameter of Aluminum to Copper wire is 1.24.

3 0
3 years ago
An electron is in a vacuum near Earth's surface and located at y = 0 m on a vertical y axis. At what value of y should a group o
Bingel [31]

Answer:

the bunch of 23 electrons must be placed on y axis at coordinate y = - 24.35 m.

Explanation:

As we know that the gravitational force on electron at y = 0 is counter-balanced by the weight of the electron

So we have

\frac{kq_1q_2}{r^2} = mg

here we have

q_1 = e

q_2 = 23 e

m = 9.11 \times 10^{-31} kg

also we know that

e = 1.6 \times 10^{-19} C

so we will have

\frac{(9\times 10^9)(1.6 \times 10^{-19})(23\times 1.6 \times 10^{-19})}{r^2} = (9.11 \times 10^{-31})(9.81)

\frac{5.3 \times 10^{-27}}{r^2} = 8.94 \times 10^{-30}

r = 24.35 m

so the bunch of 23 electrons must be placed on y axis at coordinate y = - 24.35 m.

7 0
3 years ago
A thin uniform rod of mass M and length L is bent at its center so that the two segments are perpendicular to each other. Find i
serg [7]

Answer:

\frac{1}{12}ML^2

Explanation:

The moments of the whole object is the sum of the moments of the 2 segments of rod at their ends of which length is L/2 and mass M/2:

I = 2I_{end} = 2\frac{1}{3}\frac{M}{2}\left(\frac{L}{2}\right)^2

I = \frac{1}{3}M\frac{L^2}{4}

I = \frac{1}{12}ML^2

5 0
3 years ago
How are meteors and meteorites different?
konstantin123 [22]

A meteor is the flash of light that we see in the night sky when a small chunk of interplanetary debris burns up as it passes through our atmosphere. "Meteor" refers to the flash of light caused by the debris, not the debris itself.

If any part of a meteoroid survives the fall through the atmosphere and lands on Earth, it is called a meteorite.

5 0
3 years ago
Read 2 more answers
A student rings a brass bell with a frequency of 300 Hz. The sound wave
Citrus2011 [14]

Answer:

16 m

Explanation:

i got is right on my ap3x test :)

3 0
3 years ago
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