I need help with my homework really FAAAST

A satellite of mass m is moving in a circular orbit around the earth at a constant speed v and at an altitude h above the earth'

qwelly [4]

The satellite executes a rotation motion around the earth, because Earth's force of attraction plays the role of centripetal force:

Fa=Fcp=>k*Mp*m/(Rp+r)²=mv²/(Rp+r)=>v=√(k*Mp/(Rp+r))=√(6.67*10⁻¹¹*5.98*10²⁴/(6371*10³+1000*10³))=√(39.88*10¹³/(7371*10³))=√(5.41*10⁷)=7355.53 m/s

Check the calculations again !

7 0

3 years ago

An incident on route 12 summary?I need to describe what happens in the book in a short summary can anyone help?

Serga [27]

Answer: An Incident on Route 12 is presented here in a high quality paperback edition. This popular classic work by James H. Schmitz is in the English language, and may not include graphics or images from the original edition.

Explanation: I HOPE THAT HELPED

5 0

3 years ago

A ball is projected with an initial velocity of 40 meter per second and reached maximum height of 160 meters calculate tge angle

Andru [333]

There's a problem with the question as given. Even with a maximum projection angle of θ = 90°, the initial velocity is not large enough to get the ball up in the air 160 m. With angle 90°, the ball's height y at time t would be

y = (40 m/s) t - 1/2 g t ²

Set y = 160 m, and you'll find that there is no (real) solution for t, so the ball never attains the given maximum height.

From another perspective: recall that

v ² - v₀² = 2a ∆y

where

• v₀ = initial velocity

• v = final velocity

• a = acceleration

• ∆y = displacement

At its maximum height, the ball has zero vertical velocity, and ∆y = maximum height = 160 m. The ball is in free fall once it's launched, so a = -g.

So we have

0² - (40 m/s)² = -2g (160 m)

but this reduces to

(40 m/s)² = 2 (9.8 m/s²) (160 m)

1600 m²/s² ≠ 3136 m²/s²

7 0

3 years ago

A child pulls a sled up a snow covered hill. If the child does 504J of work on the sled while pulling the sled 23m up the hill t

zhenek [66]

Explanation:

ans is equal to 504j* 23 m* 10 ms

7 0

3 years ago

Choose the scenario in which the sound frequency of the waves is higher.

mrs_skeptik [129]

Answer:

B) the sound source moves towards you at 100 m/sec

Explanation:

The Dopper Effect is a phenomenon that occur when there is relative motion between an observer and a source of a wave. When this situation occurs, there is an apparent shift in frequency of the wave, as observed by the observer.

The apparent frequency observed by the observer is given by

$f'=\frac{v\pm v_o}{v\pm v_s}f$

where

f is the original frequency of the wave

f' is the apparent frequency

v is the speed of the wave

$v_o$ is the velocity of the observer (positive if moving towards the source of the wave, negative otherwise)

$v_s$ is the velocity of the source (negative if moving towards from the observer, positive otherwise)

In this problem, we want to find the scenario in which the sound frequency is higher.

We see that in all 4 scenarios, the sound source is moving: this means we have to find the scenario in which the denominator of the equation is smaller.

First of all, we notice the sound source moves towards the observer, $v_s$ is negative, so the denominator is higher: this means that the correct option must be either A or B.

Also, we notice that since $v_s$ is negative, a value larger in magnitude will mean a smaller denominator: therefore, the correct answer will be

B) the sound source moves towards you at 100 m/sec

Since this situation will make the denominator of the formula the smallest possible.

5 0

3 years ago

I need help with my homework really FAAAST​

I need help with my homework really FAAAST