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Olenka [21]
3 years ago
6

Scientists are experimenting with a kind of gun that may eventually be used to fire payloads directly into orbit. In one test, t

his gun accelerates a 3.8-kg projectile from rest to a speed of 9.3 × 103 m/s. The net force accelerating the projectile is 9.3 × 105 N. How much time is required for the projectile to come up to speed?
Physics
1 answer:
kkurt [141]3 years ago
5 0

Answer:

t=0.038s

Explanation:

Project mass m=3.8 kg

Initial speed vi= 0m/s

Final speed vf= 9.3×10³ m/s

Force F=9.3×10⁵N

To find

Time t

Solution

From Newtons second law we know that

∑F=ma

Where m is mass

a is acceleration

We can write this equation as

∑F=m(Δv/Δt)

=m\frac{v_{f}-v_{i}}{t}

Rearrange this equation to find time t

So

t=m\frac{v_{f}-v_{i}}{F}

Substitute the given values

t=3.8kg\frac{9.3*10^3m/s-0}{9.3*10^5N} \\t=0.038s  

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Answer:

the <em>ratio F1/F2 = 1/2</em>

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Explanation:

The force that both satellites experience is:

F1 = G M_e m1 / r²       and

F2 = G M_e m2 / r²

where

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Therefore,

F1/F2 = [G M_e m1 / r²] / [G M_e m2 / r²]

F1/F2 = [G M_e m1 / r²] × [r² / G M_e m2]

F1/F2 = m1/m2

F1/F2 = 1000/2000

<em>F1/F2 = 1/2</em>

The other force that the two satellites experience is the centripetal force. Therefore,

F1c = m1 v² / r    and

F2c = m2 v² / r

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  • m2 is the mass of satellite 2
  • v is the orbital velocity
  • r is the orbital velocity

Thus,

a1 = v² / r ⇒ v² = r a1    and

a2 = v² / r ⇒ v² = r a2

Therefore,

F1c = m1 a1 r / r = m1 a1

F2c = m2 a2 r / r = m2 a2

In order for the satellites to stay in orbit, the gravitational force must equal the centripetal force. Thus,

F1 = F1c

G M_e m1 / r² = m1 a1

a1 = G M_e / r²

also

a2 = G M_e / r²

Thus,

a1/a2 = [G M_e / r²] / [G M_e / r²]

<em>a1/a2 = 1</em>

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According to the saving of the momentum, the total momentum before collision is equal to the total momentum after collision

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Answer:

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