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3 years ago

7

This is a small rocky body that orbits the sun, and when close to the sun exhibit a small tail of ice and gas.

1 answer:

Over [174]3 years ago

3 0

This is a comet. It is also called a dirty snowballs.

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A car has an engine which delivers a constant power. It accelerates from rest at time t = 0, and at t = t0 its acceleration is a

olga2289 [7]

Here is the answer of the given problem above.
Use this formula: <span>P = FV = ma*at = ma^2 t
</span><span>Substitute the values, and therefore, we got m(a0)^2t = m(x)^2 (2t)
then, solve for x which is the acceleration at 2t.
</span>The <span>answer would be a0/sqrt(2).
Hope this answers your question. Thanks for posting.
</span>

7 0

3 years ago

In a building with 10.000 cubic feet where the air changes every two hours, what the rate of air change? A. 167.7 cfm B. 83.3 cf

Naya [18.7K]

Answer:

Flow rate of air is given as 83.33 cubic feet per minute

Explanation:

As we know that total volume of the air flow is given as

$V = 10,000 cubic feet$

also we know that total time is

$t = 2 hours = 120 min$

now we have flow rate given as

$Q = \frac{V}{t}$

$Q = \frac{10000}{120}$

$Q = 83.3 cf/m$

3 0

3 years ago

What is the answer be fast

svetlana [45]

Answer:

Explanation:

Same numbers of protons but different number of neutron so i would go for A same atomic number different number of neutrons

5 0

3 years ago

Another name for continuous uniform circular motion is___?

VikaD [51]

I think the answer is constant

5 0

3 years ago

two pendulums of lengths 100cm and 110.25cm start oscillating in phase. after how many oscillations will they again be in same p

goldfiish [28.3K]

Angular frequency of pendulum is given by

$\omega = \sqrt{\frac{g}{l}}$

for both pendulum we have

$\omega_1 = \sqrt{\frac{9.81}{1.00}}$

$\omega_1 = 3.13 rad/s$

For other pendulum

$\omega_2 = \sqrt{\frac{9.81}{1.1025}}$

$\omega_2 = 2.98 rad/s$

now we have relate angular frequency given as

[tex\omega_1 - \omega_2 = 3.13 - 2.98 = 0.15 rad/s[/tex]

now time taken to become in phase again is given as

$t = \frac{2\pi}{\omega_1 - \omega_2}$

$t = \frac{2\pi}{0.15} = 41.88 s$

now number of oscillations complete in above time

$N = \frac{t}{\frac{2\pi}{\omega_1}}$

$N = \frac{41.88}{\frac{2\pi}{3.13}}$

$N = 21 oscillation$

3 0

3 years ago