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BartSMP [9]
3 years ago
14

Satellite A has an orbital radius 3.00 times greater than that of satellite B. Satellite B's orbital period around Earth is 120

minutes. What is satellite A's orbital period?
Physics
1 answer:
igomit [66]3 years ago
7 0

Answer:

To find the circumference (orbit) of an object, you use Pi x Diameter. 

As you have the circumference of B, you divide it by Pi to get the Diameter. 

So 120 divided by 3.141592654 = 38.2 minutes for the Diameter. 

As' radius and Diameter will be 3x greater than B. 

38.2 x 3 = 114.6 

To get back to the orbital period, times 114.6 by Pi, and you will get 360 minutes

HOPE THIS HELPS AND PLS MARK AS BRAINLIEST

THNXX :)

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An elephant kicks a 2.5 kg rock that is initially at rest. The
erma4kov [3.2K]

0.20Answer:

Explanation:

6 0
3 years ago
In the physics lab, a block of mass M slides down a frictionless incline from a height of 35cm. At the bottom of the incline it
bogdanovich [222]

Solution :

Given :

M = 0.35 kg

$m=\frac{M}{2}=0.175 \ kg$

Total mechanical energy = constant

or $K.E._{top}+P.E._{top} = K.E._{bottom}+P.E._{bottom}$

But $K.E._{top} = 0$ and $P.E._{bottom} = 0$

Therefore, potential energy at the top = kinetic energy at the bottom

$\Rightarrow mgh = \frac{1}{2}mv^2$

$\Rightarrow v = \sqrt{2gh}$

      $=\sqrt{2 \times 9.8 \times 0.35}$      (h = 35 cm = 0.35 m)

      = 2.62 m/s

It is the velocity of M just before collision of 'm' at the bottom.

We know that in elastic collision velocity after collision is given by :

$v_1=\frac{m_1-m_2}{m_1+m_2}v_1+ \frac{2m_2v_2}{m_1+m_2}$

here, $m_1=M, m_2 = m, v_1 = 2.62 m/s, v_2 = 0$

∴ $v_1=\frac{0.35-0.175}{0.5250}+\frac{2 \times 0.175 \times 0}{0.525}

      $=\frac{0.175}{0.525}+0$

     = 0.33 m/s

Therefore, velocity after the collision of mass M = 0.33 m/s

 

3 0
3 years ago
Using hooke's law find the elastic constant of a spring that stretches 2 cm when 4newton force is applied to it
Kay [80]

<u>Answer:</u>

2N/cm

<u>Step-by-step explanation:</u>

According to the Hooke's Law, the force required to extend or compress a spring is directly proportional distance you can stretch it, which is represented as:

F=kx

where, F is the force which is stretching or compressing the spring,

k is the spring constant; and

x is the distance the spring is stretched.

Substituting the given values to find the elastic constant  k to get:

F=kx

4=k(2)

k=\frac{4}{2}

k=2

Therefore, the elastic constant is 2 Newton/cm.

3 0
3 years ago
There is a species of bamboo that can grow 36 inches per day. If a plant grew at this rate and was measured at 40 inches initial
LekaFEV [45]

Answer:

It will take the plant 4\frac{4}{9} days or 4.44 days to grow to a height of 200 inches tall.

Explanation:

From the question, the rate at which the species of the bamboo tree grows is 36 inches per day.

To determine how long it would take a plant 40 inches tall initially to grow at this rate (that is, 36 inches per day) to a height of 200 inches.

This means we will calculate the number of days it will take the plant to grow additional 160 inches ( 200 inches - 40 inches) at this rate.

Now,

If the plant grows 36 inches in 1 day

then it will grow 160 inches in x days

x = (160 inches × 1 day) / 36 inches

x = 160 / 36

x = 4\frac{4}{9} days or 4.44 days

Hence, it will take the plant 4\frac{4}{9} days or 4.44 days to grow to a height of 200 inches tall.

8 0
3 years ago
What happens when stress builds up at faults?
Harlamova29_29 [7]

Answer:

The answer is D because when the faults move that is the tectonic plates moving. So earth quakes will be forming when the fault moves.

Explanation:

3 0
3 years ago
Read 2 more answers
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