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drek231 [11]
3 years ago
5

A rocket's acceleration is 6.0 m/s2. Assuming it starts at 0 m/s, how long will it take for the rocket to reach a velocity of 42

m/s?
250 s
0.14s
7.0 s
290 s
Physics
1 answer:
Over [174]3 years ago
4 0
There are few equations that link acceleration and velocity.

One of them is derived from the definition of acceleration - is the rate of change of velocity in respect with time

a = v-u/t
at=v-u

a = acceleration
v=velocity at the end
u=velocity at the start
t=time

Now put everything you know and solve for time.

6t = 42 -0
6t = 42
t = 42 / 6
t = 7 second

It takes 7 seconds for the roccet to reach a velocity of 42 m/s

Hope it helps :).

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Equipotential surface A has a potential of 5650 V, while equipotential surface B has a potential of 7850 V. A particle has a mas
timurjin [86]

Answer:

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Explanation:

From the principle of conservation of energy, the workdone by the applied force, W = kinetic energy change + electric potential energy change.

So, W = ΔK + ΔU =1/2m(v₂² - v₁²) + q(V₂ - V₁) where m = mass of particle = 5.4 × 10⁻² kg, q = charge of particle = 5.10 × 10⁻⁵ C, v₁ = initial speed of particle = 2.00 m/s, v₂ = final speed of particle = 3.00 m/s, V₁ = potential at surface A = 5650 V, V₂ = potential at surface B = 7850 V.

So, W = ΔK + ΔU =1/2m(v₂² - v₁²) + q(V₂ - V₁)

          = 1/2 × 5.4 × 10⁻²kg × ((3m/s)² - (2 m/s)²) + 5.10 × 10⁻⁵ C(7850 - 5650)

          = 0.135 J + 0.11220 J

          = 0.2472 J

          ≅ 0.247 J = 247 mJ

5 0
3 years ago
what is the magnitude of the compression forces (assumed to be horizontal) acting on both sides of the center board that is sand
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