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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 vehicle that goes from 5m/s to 45m/s in 8s. what is its acceleration?
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Answer: 5m/s^2

Explanation:

V= 45m/s

U = 5m/s

t = 8s

a =?

V = u + at

45 = 5 + 8a

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8a = 40

a = 40 / 8

a = 5m/s^2

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Two cars A and B are moving with velocities 20 m/s and 15 m/s in the direction east and west respectively. If
Sphinxa [80]

Answer:

<u>Distance</u><u> </u><u>between</u><u> </u><u>them</u><u> </u><u>is</u><u> </u><u>4</u><u>,</u><u>2</u><u>0</u><u>0</u><u> </u><u>meters</u><u>.</u>

Explanation:

Consinder car A:

{ \bf{distance =  speed \times time }}

substitute:

distance = 20 \times (2 \times 60) \\  = 2400 \: m

Consider car B:

distance = 15 \times (2 \times 60) \\  = 1800 \: m

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distance \: between = { \sum(distance \: of \: each \: car)} \\  = 2400 + 1800 \\  = 4200 \: m

3 0
3 years ago
An avant-garde composer wants to use the Doppler effect in his new opera. As the soprano sings, he wants a large bat to fly towa
Allisa [31]

Answer:

     v’= 9.74 m / s

Explanation:

The Doppler effect is due to the relative movement of the sound source and the receiver of the sound, in this case we must perform the exercise in two steps, the first to find the frequency that the bat hears and then the frequency that the audience hears that also It is sitting.

Frequency shift heard by the murciela, in case the source is still and the observer (bat) moves closer

        f₁ ’= f₀ (v + v₀)/v

         

Frequency shift emitted by the speaker in the bat, in this case the source is moving away from the observer (public sitting) that is at rest

         f₂’= f₁’ v/(v - vs)

           

Note that in this case the bat is observant in one case and emitter in the other, called its velocity v’

            v’= vo = vs

Let's replace

           f₂’= f₀   (v + v’)/v   v/(v -v ’)

           f₂’= f₀   (v + v’) / (v -v ’)

           (v –v’ ) f₂’ / f₀ = v + v ’

           v’ (1+ f₂’ /f₀) = v (f₂’/fo - 1)

           v’ (1 + 1.059) = 340 (1.059 - 1)

           v’= 20.06 / 2.059

           v’= 9.74 m / s

6 0
3 years ago
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