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2 years ago

If you were to descend 100 m below the surface of the ocean, you would be in _____.

Physics

2 answers:

V125BC [204]2 years ago

8 0

If you descend 100 m in the ocean, you are going to be in colder waters

harina [27]2 years ago

6 0

You would be in deep, dark, cold water, at high pressure.

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A rocket car on a horizontal rail has an initial mass of 2500 kg and an additional fuel mass of 1000 kg. At time t0 the rocket m

slamgirl [31]

Answer: Acceleration of the car at time = 10 sec is 108 $m/s^{2}$ and velocity of the car at time t = 10 sec is 918.34 m/s.

Explanation:

The expression used will be as follows.

$M\frac{dv}{dt} = u\frac{dM}{dt}$

$\int_{t_{o}}^{t_{f}} \frac{dv}{dt} dt = u\int_{t_{o}}^{t_{f}} \frac{1}{M} \frac{dM}{dt} dt$

= $u\int_{M_{o}}^{M_{f}} \frac{dM}{M}$

$v_{f} - v_{o} = u ln \frac{M_{f}}{M_{o}}$

$v_{o} = 0$

As, $v_{f} = u ln (\frac{M_{f}}{M_{o}})$

u = -2900 m/s

$M_{f} = M_{o} - m \times t_{f}$

= $2500 kg + 1000 kg - 95 kg \times t_{f}s$

= $(3500 - 95t_{f})s$

$v_{f} = -2900 ln(\frac{3500 - 95 t_{f}}{3500}) m/s$

Also, we know that

a = $\frac{dv_{f}}{dt_{f}} = \frac{u}{M} \frac{dM}{dt}$

= $\frac{u}{3500 - 95 t} \times (-95) m/s^{2}$

= $\frac{95 \times 2900}{3500 - 95t} m/s^{2}$

At t = 10 sec,

$v_{f}$ = 918.34 m/s

and, a = 108 $m/s^{2}$

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nasty-shy [4]

Explanation:

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Answer:

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