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arsen [322]
3 years ago
13

A 5.00-g bullet leaves the muzzle of a rifle with a speed of 320 m/s. the expanding gases behind it exert what force on the bull

et while it is traveling down the barrel of the rifle, 0.820 m long? assume constant acceleration and negligible friction.
Physics
1 answer:
muminat3 years ago
3 0

Answer;

= 312 Newtons

Explanation;

The bullet has a mass of 0.005 kg, and a velocity of 320 m/s, so we need to find it's final kinetic energy.  

KE = 1/2*m*v^2

      = 1/2*0.005*320^2

      = 256 Joules.  

Divide this by the distance over which this energy was received and you have the force that provided that energy.  

  = 256/0.820 = 312.195 Newtons  

Rounded off, this is 312 N

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 a substance's density is the same at a certain pressure and temperature, and the density of one substance is usually different than another substance.
4 0
3 years ago
A kangaroo jumps straight up to a vertical height of 1.45 m. How long was it in the air before returning to Earth?
dexar [7]

Answer:

1.08 s

Explanation:

From the question given above, the following data were obtained:

Height (h) reached = 1.45 m

Time of flight (T) =?

Next, we shall determine the time taken for the kangaroo to return from the height of 1.45 m. This can be obtained as follow:

Height (h) = 1.45 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

1.45 = ½ × 9.8 × t²

1.45 = 4.9 × t²

Divide both side by 4.9

t² = 1.45/4.9

Take the square root of both side

t = √(1.45/4.9)

t = 0.54 s

Note: the time taken to fall from the height(1.45m) is the same as the time taken for the kangaroo to get to the height(1.45 m).

Finally, we shall determine the total time spent by the kangaroo before returning to the earth. This can be obtained as follow:

Time (t) taken to reach the height = 0.54 s

Time of flight (T) =?

T = 2t

T = 2 × 0.54

T = 1.08 s

Therefore, it will take the kangaroo 1.08 s to return to the earth.

3 0
3 years ago
If a galaxy has an apparent velocity of 2300 km/s, what is its distance if the Hubble constant is assumed to be 70 km/s/Mpc
prohojiy [21]

The distance of the galaxy is 32.86 Mpc.

Using the hubble law, v = H₀D where v = apparent velocity of galaxy = 2300 km/s, H = hubble constant = 70 km/s/Mpc and D = distance of galaxy.

Since we require the distance of the galaxy, we make D subject of the formula in the equation. So, we have

D = v/H₀

Substituting the values of the variables into the equation, we have

D = 2300 km/s ÷ 70 km/s/Mpc

D = 32.86 Mpc

So, the distance of the galaxy is 32.86 Mpc

Learn more about hubble law here:

brainly.com/question/18484687

4 0
2 years ago
A source of sound is kept in a jar in a vacuum. Air is slowly introduced in to the jar. What happens to the sound coming out of
timama [110]
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3 years ago
The planet in our solar system whose orbit actually brings it inside the orbit of another planet is
Ber [7]

Answer:

Pluto

Explanation:

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7 0
4 years ago
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