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Ad libitum [116K]

3 years ago

12

The moon completes one (circular) orbit of the earth in 27.3 days. The distance from the earth to the moon is 3.84×108 m. What i

s the moon’s centripetal acceleration?

1 answer:

I am Lyosha [343]3 years ago

4 0

Answer:

$2.72\cdot 10^{-3} m/s^2$

Explanation:

Let's start by calculating the angular velocity of the Moon. We know that the period is:

$T=27.3 d \cdot 24 \cdot 60 \cdot 60 =2.36\cdot 10^6 s$

So now we can calculate its angular velocity:

$\omega=\frac{2\pi}{T}=\frac{2\pi}{(2.36\cdot 10^6)}=2.66 \cdot 10^{-6} rad/s$

The centripetal acceleration is given by

$a=\omega^2 r$

where

$\omega=2.66\cdot 10^{-6}rad/s$

$r=3.84\cdot 10^8 m$ is the radius of the orbit

Substituting,

$a=(2.66\cdot 10^{-6})^2(3.84\cdot 10^8)=2.72\cdot 10^{-3} m/s^2$

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a spacecraft that seems to be motionless in deep space is given some type of quick push. which statement describes what will hap

bogdanovich [222]

An applied force applied causes a body such as the spacecraft to move. The magnitude of the force determines the change in the velocity

What will happen to the spacecraft is given by option C. from among the possible question options.

C. The spacecraft will begin to move and it will continue moving until it is stopped by an equal and opposite force

Reason:

<em>The possible question options obtained from a similar question includes;</em>

<em>A. The spacecraft will move for some time then stop slowly</em>

<em>B. Air resistance will prevent the spacecraft from moving</em>

<em>C. The spacecraft will begin to move and the motion will continue until a force equal and opposite to the applied force stops it</em>

<em>D. The quick push will not cause the spacecraft to move, as a quick push works on Earth only</em>

The state of the space craft = Motionless

Location of the spacecraft = Deep space

Type of force applied = Quick push

The statement that describes what will happen = Required

Solution;

Let <em>F</em>, represent the force applied, and let Δt be the duration of the applied force, we have;

The impulse of the force, F × Δt = m·(v₂ - v₁)

Where;

m = The mass of the spacecraft

v₁ = The initial velocity of the spacecraft = 0

v₂ = Final velocity of the spacecraft

Plugging in v₁ = 0, gives;

F × Δt = m·v₂

The space craft is given a velocity, <em>v₂</em>, and according to Newton's First Law of Motion, it continues moving in a straight line until another force acts on it

Therefore, the correct option is option C. <u>The spacecraft will begin to move and the motion will continue until it is stopped by an equal and opposite force</u>

<u />

Learn more about Newton's First Law of Motion here:

brainly.com/question/20841616

8 0

3 years ago

You might say that this experiment was an attempt to build a scale, and then calibrate it against a scale that we trust (the ele

Allushta [10]

No.

Since repeated measurements are taken and the average and 95% confidence interval are calculated, the possibility of the lack of agreement being a random error has been minimized or even eliminated.

<h3>What is a random error?</h3>

Random error is defined as the deviation of the total error from its mean value due to chance.

Random errors can result from the instrument not being precise or from mistakes by the researcher.

Random errors can be minimized by taking multiple readings and averaging the results.

Since repeated measurements are taken and the average and 95% confidence interval are calculated, the possibility of the lack of agreement being a ransom error has been minimized.

Learn more about random errors at: brainly.com/question/22041172

3 0

3 years ago

Read 2 more answers

Bart runs up a 2.91 meters high flight of stairs at a constant speed in 2.15 seconds. If barts mass is 65.9kg determine the work

Tanzania [10]

Answer:

1879.33J

874.1W

Explanation:

Given parameters:

Distance covered = 2.91m

Time taken = 2.15s

Mass of Bart = 65.9kg

Unknown:

Work done = ?

Power rating = ?

Solution:

Here, the work done is related to the the potential energy in climbing this flight of stairs.

Work done = Potential energy = mgH

where m is the mass

g is the acceleration due to gravity

H is the height

Work done = 65.9 x 9.8 x 2.91 = 1879.33J

Power is defined as the rate at which work is being done.

Power = $\frac{Work done }{time taken}$

= $\frac{1879.33}{2.15}$

= 874.1W

6 0

3 years ago

Fossil fuels like gasoline or coal are examples of

disa [49]

Answer:

Nonrenewable Resources

8 0

3 years ago

Can someone help meeeeee... show how to solve it plzzzzzzzz

liubo4ka [24]

<h2>Right answer: 64 units</h2><h2></h2>

According to the law of universal gravitation, which is a classical physical law that describes the gravitational interaction between different bodies with mass:

$F=G\frac{m_{1}m_{2}}{r^2}$

Where:

$F$ is the module of the force exerted between both bodies

$G$ is the universal gravitation constant.

$m_{1}$ and $m_{2}$ are the masses of both bodies.

$r$ is the distance between both bodies

In this case we have a gravitation force $F_{1}=16units$ , given by the formula written at the beginning. Let’s rename the distance $r$ as $d$ :

$F_{1}=G\frac{m_{1}m_{2}}{d^2}$ (1)

And we are asked to find the gravitation force $F_{2}$ with a given distance of $\frac{d}{2}$ :

$F_{2}=G\frac{m_{1}m_{2}}{({\frac{d}{2})}^{2}}$

$F_{2}=G\frac{m_{1}m_{2}}{{\frac{d^{2}}{4}}}$ (2)

The gravity constant is the same for both equations, and we are assuming both masses are constants, as well. So, let’s isolate $G m_{1}m_{2}$ in both equations:

From (1):

$Gm_{1}m_{2}=F_{1}{d}^{2}$ (3)

From (2):

$Gm_{1}m_{2}=F_{2}\frac{{d}^{2}}{4}$ (4)

If (3)=(4):

$F_{1}{d}^{2}=F_{2}\frac{{d}^{2}}{4}$ (5)

Now we have to find $F_{2}$ :

$F_{2}=F_{1}{d}^{2}\frac{4}{{d}^{2}}$

$F_{2}=4F_{1}$ (6)

If $F_{1}=16 units$ :

$F_{2}=(4)(16 units)$

$F_{2}=64 units$ >>>>This is the new force of attraction

3 0

3 years ago