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Step2247 [10]
2 years ago
9

Determine the orbits period of the moon when the distance between the earth and the moon is 3.82 x 10 to the power of 8

Physics
1 answer:
alexandr402 [8]2 years ago
4 0

Answer

From

V=Distance/time

The distance round a circular path or Object is 2πr

so

v=2πr/T

Making T(period) subject

T=2πr/v

where v is the linear velocity.

We don't have the velocity but we can get it.

The Moon and Earth exert gravitational force on each other and the Moon is kept in Orbit by Centripetal force.

Since the Moon doesn't fly out of Orbit... it must mean that the Gravitational force being exerted on it by the Earth is equal to its centripetal Force.

Equating Both (Fg=Fc)

Fg=GMm/R²

Fc=mv²/r

GMm/R² = mv²/R

Canceling out "m" and "r" on both sides

we're left with

V²=GM/R

V=√GM/R

Where M= Mass of Earth (5.98x10^24)

R=Distance between the center of earth and the Moon

G= Gravitational Constant(Value 6.67x10^-11)

V=√6.67x10^-11 x 5.98x10^24/(3.82x10^8)

V=1021.84meters per second.

Now

T=2πr/v

=2π x 3.82x10^8 / 1021.84

T=2.35x10^6seconds

Converting to days by dividing by(24 x 3600)

You have

T=27.2days approx.

✅✅✅

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What is the % error in using g = 10.0 m/s^2? (Take the value ofg as 9.8 m/s^2)
Alenkasestr [34]

Answer:

So percentage error will be 2 %

Explanation:

We have given initial value of acceleration due to gravity g=10m/sec^2

And final value of acceleration due to gravity g=9.8m/sec^2

We have to find the percentage error

We know that percentage error is given by percentage\ error=\frac{initial\ value-final\ value}{initial\ value}\times 100

So percentage\ error=\frac{10-9.8}{10}\times 100=2 %

4 0
3 years ago
24) a plane takes off at 7:00 pm and lands at 1:00 am after travelling a distance of 4818km. what is the minimum number of times
12345 [234]

Answer:

Twice

Step-by-Step Explanation:

Time between 7:00 PM and 1:00 AM: 6 hours

Distance: 4818km

Since the distance is 4818km, and the time is 6 hours, you divide 4818 by 6.

803.0000015999 km/h.

The average speed is 803 km/h

Which considering the ideal case scenario if the plane starts at 0 reaches the speed of 803 and the end reduces its speed from 803 to 0. This means we have come across the value of 800 at least twice. Hence, the plane was travelling at a speed of 800 km/h at least 2 times.

7 0
1 year ago
A 0.500-kg stone is moving in a vertical circular path attached to a string that is 75.0 cm long. The stone is moving around the
Semenov [28]

Answer:

B. 7.07 m/s

Explanation:

The velocity of the stone when it leaves the circular path is its tangential velocity, v, which is given by

v=\omega r

where \omega is the angular speed and r is the radius of the circular path.

\omega is given by

\omega = 2\pi f

where f is the frequency of revolution.

Thus

v=2\pi fr

Using values from the question,

v=2\pi\times 1.50\times0.75

<em>Note the conversion of 75 cm to 0.75 m</em>

v=2\times3.14\times 1.50\times0.75 = 9.42\times0.75 = 7.065=7.07

6 0
3 years ago
Two blocks, A and B, are connected by a rope. A second rope is connected to block B and a steady, horizontal tension force of T
Oksanka [162]

Answer:

40 N

Explanation:

We are given that

Speed of system is constant

Therefore, acceleration=a=0

Tension applied on block B=T=50 N

Friction force=f=10 N

We have to find the friction force acting on block A.

Let T' be the tension in string connecting block A and block B and friction force on block A be f'.

For Block B

T-f-T'=m_Ba

Where m_B=Mass of block B

Substitute the values

50-10-T'=m_B\times 0=0

T'==40 N

For block A

T'-f'=m_Aa

Where m_A=Mass of block A

Substitute the values

40-f'=m_A\times 0=0

f'=40 N

Hence, the friction force acting on block A=40 N

3 0
3 years ago
If the car goes exits a freeway and goes from 65<br> mph to 35 mph is it accelerating?
Zolol [24]

Answer:

No, the car is decelerating  

Explanation:

No the car is decelerating if it exits a freeway and goes from 65

mph to 35 mph since the change in velocity is negative.

change in velocity = final - initial

change in velocity  = 35 - 65

change in velocity = -30mph

Since the change in velocity is negative, hence the car is decelerating. Deceleration is a negative acceleration

8 0
3 years ago
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