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Ulleksa [173]
3 years ago
12

Satellite in a circular orbit of radius r around planet x has an orbital period t. if planet x had one-fourth as much mass, the

orbital period of this satellite in an orbit of the same radius would be
Physics
1 answer:
postnew [5]3 years ago
7 0
<span>Satellite in a circular orbit of radius r around planet x has an orbital period t. if planet x had one-fourth as much mass, the orbital period of this satellite in an orbit of the same radius would be 2t</span>
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A wave has a wavelength of 30 mm and a frequency of 5.0 hertz. What is its speed?
Alenkinab [10]
                 Wave speed  =  (frequency) x (wavelength)

                                       =  (5 / second)  x  (30 mm) 

                                       =      150 mm/second  =  0.15 meter/second .
8 0
3 years ago
You push a 2.3 kg block against a horizontal spring, compressing the spring by 17 cm. then you release the block, and the spring
horsena [70]
Some dogs may inherit a susceptibility to epilepsy.

6 0
3 years ago
The radius of a circle of area A and circumference C is doubled. find the new circumference of the circle in terms of C.
Fudgin [204]

If the radius of a circle of area A and circumference C is doubled, <u>the new circumference</u> of the circle, in terms of C, <u>will be</u> equal to 2C.

The area (A) of a circle is given by:

A = \pi r^{2}   (1)

And a circumference (C) is:

C = 2\pi r   (2)

Where:

r: is the radius

<u>If the radius</u> of a circle and circumference <u>is doubled</u>, the new circumference of the circle is:

C_{2} = 2 (2 \pi r) = 2C

Therefore, the new circumference of the circle in terms of C is equal to 2C.

To find more about circumference, go here: brainly.com/question/4268218?referrer=searchResults

I hope it helps you!

3 0
3 years ago
A ball thrown horizontally from the top of a building hits the ground in 0.600 s. If it had been thrown with twice the speed in
ludmilkaskok [199]

Answer:

none of the answers is correct, the time  is the same  t₁ = t₂ = 0.600 s

Explanation:

This is a kinematics exercise, analyze the situation a bit. The vertical speed in both cases is the same is zero, the horizontal speed in the second case is double (vₓ₂ = 2 vₓ₁)

let's find the time to hit the ground

     y = y₀ + I go t - ½ g t²

     0 = y₀ - ½ g t²

     t = √ 2y₀ / g

with the data from the first launch

     y₀i = ½ g t²

     y₀ = ½  9.8  0.6²

     y₀ = 1,764 m

with this is the same height the time to descend in the second case is the same

    t₂ = 0.600 s

this is because the horizontal velocity change changes the offset on the x axis, but does not affect the offset on the y axis

Therefore, none of the answers is correct, the time  is the same

t₁ = t₂ = 0.600 s

5 0
2 years ago
Select the correct answer
Flauer [41]

Answer: The correct answer is 1800 meters

Explanation: The velocity is equal to distance over time. Thus the formula for distance is d=vt

And 1minute = 60 seconds

d= 30m/s×60s

d= 1800m

5 0
3 years ago
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