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3 years ago

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A satellite is in a circular orbit 21000 km above the Earth’s surface; i.e., it moves on a circular path under the influence of

nothing but the Earth’s gravity. Find the speed of the satellite. The radius of the Earth is 6.37 × 106 m, and the acceleration of gravity at the satellite’s altitude is 0.532655 m/s 2. Find the time it takes to complete one orbit around the Earth.

1 answer:

mina [271]3 years ago

6 0

Answer:

(orbital speed of the satellite) V₀ = 3.818 km

Time (t) = 4.5 × 10⁴s

Explanation:

Given that:

The radius of the Earth is 6.37 × 10⁶ m; &

the acceleration of gravity at the satellite’s altitude is 0.532655 m/s

We can calculate the orbital speed of the satellite by using the formula:

Orbital Speed (V₀) = √(r × g)

radius of the orbit (r) = 21000 km + 6.37 × 10⁶ m

= (2.1 × 10⁷ + 6.37 × 10⁶) m

= 27370000

= 2.737 × 10⁷m

Orbital Speed (V₀) = √(r × g)

Orbital Speed (V₀) = √(2.737 × 10⁷ × 0.532655 )

= 3818.215

= 3.818 × 10³

= 3.818 Km

To find the time it takes to complete one orbit around the Earth; we use the formula:

Time (t) = 2 π × $\frac{r}{V_o}$

= 2 × 3.14 × $\frac{2.737*10^7}{3.818*10^3}$

= 45019.28

= 4.5 × 10 ⁴ s

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Which of the following statements best describe destructive forces? A. Forces that build up, create, landmasses. B. Forces that

MissTica

<span>An event that breaks objects into smaller objects or pieces is called destructive force
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4. How much work is done when you lift a 25 N box up to a 2 m shelf?

Semmy [17]

Answer:

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The work done by an object can be found by using the formula

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Hope this helps you

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3 years ago

If 35 J of work was performed in 70 seconds, how much power was used to do this task?

Irina18 [472]

$P = w \times delta \: T$
P = Power
W = Work
Delta T = Change of Time

So:
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Delta T = 70 secs
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3 years ago

a bullet with a mass of 4.0g and a speed of 650m/s is fired at a block of wood with a mass of 0.095kg. the block rests on a fric

n200080 [17]

Part a)

Here in this we can use momentum conservation as there is no external force on it

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} +m_2v_{2f}$

here we know that

$m_1 = 0.004 kg$

$v_{1i} = 650 m/s$

$m_2 = 0.095$

$v_{2i} = 0$

$v_{2f} = 23 m/s$

now by above equation

$0.004*650 + 0.095* 0 = 0.004*v + 0.095*23$

$2.6 + 0 = 0.004*v + 2.185$

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Part b)

Final kinetic energy of the system

$KE_f = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2$

$KE_f = \frac{1}{2}*0.004*(103.75^2) + \frac{1}{2}*(0.095)*23^2$

$KE_f = 46.65 J$

Initial Kinetic energy of the system will be

$KE_f = \frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2$

$KE_f = \frac{1}{2}*0.004*(650^2) + \frac{1}{2}*(0.095)*0^2$

$KE_f = 845 J$

So here kinetic energy is decreased for this system

final energy is less than initial energy

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