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kenny6666 [7]
3 years ago
14

Your friend tells you “I no the moon does not rotate because we always see the same side.” do you agree or disagree with your fr

iend? Write up what you would say to your friend to show why you agree or disagree with them. (THIS IS URGENT I REALLY NEED HELP WITH THIS BECAUSE I DONT KNOW HOW TO EXPLAIN)
Physics
1 answer:
nydimaria [60]3 years ago
3 0

Answer:

no the moon does not rotate it only goes in circle just like the sun so I disagree with your friend

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Which of the following is not true about the Law of Conservation of Energy?
SVEN [57.7K]

By definition, the law of conservation of energy states that:

Ei = Ef

Where,

Ei: initial energy

Ef: final energy

Therefore, no matter the type of energy, always the final energy is equal to the final energy.

Energy can be transformed into another type of energy. For example, the potential energy can be transformed into kinetic energy.

Also, energy is not created, nor destroyed.

Answer:

The following is not true about the Law of Conservation of Energy:

A. It states that the total energy in the universe keeps increasing.

8 0
3 years ago
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PLEASE HELP ME
Monica [59]

Answer:

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Explanation:

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4 years ago
Find the position vector of a particle that has the given acceleration and the specified initial velocity and position. a(t) = 1
kondor19780726 [428]

Answer:

Explanation:

Given

Acceleration a(t)=14t\hat{i]+\sin (t)\hat{j}+\cos (2t)\hat{k}[/tex]

and v(0)=\hat{i}

r(0)=\hat{j}

we know a=\frac{\mathrm{d} v}{\mathrm{d} t}

\int dv=\int adt

v(t)=\int (14t\hat{i}+\sin (t)\hat{j}+\cos (2t)\hat{k})dt

v(t)=7t^2\hat{i}-\cos t\hat{j}+\frac{\sin (2t)\hat{k}}{2}+c

at t=0

v(0)=0-1\cdot \hat{j}+0+c

c=\hat{i}+\hat{j}

v(t)=(7t^2+1)\hat{i}+(1-\cos t)\hat{j}+\frac{\sin (2t)\hat{k}}{2}

and \frac{\mathrm{d} r}{\mathrm{d} t}=v(t)

\int dr=\int vdt

r(t)=\int ((7t^2+1)\hat{i}+(1-\cos t)\hat{j}+\frac{\sin (2t)\hat{k}}{2})dt

r(t)=(\frac{7}{2}t^3+t)\hat{i}+(t-\sin (t))\hat{j}+\frac{1}{2}\times (-\frac{1}{2}\cos 2t)\hat{k}+c_2

at t=0

r(0)=\hat{j}

r(t)=(\frac{7}{3}t^3+t)\hat{i}+(1+t-\sin t)\hat{j}+\frac{1}{4}(1-\cos 2t)\hat{k}

       

4 0
3 years ago
Simple curiosity is not a legitimate reason to complete a scientific investigation.
jok3333 [9.3K]
True......I would say...
7 0
3 years ago
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A person holds a rifle horizontally and fires at a target. The bullet leaves the muzzle of the rifle with a velocity of 460 m/s.
Trava [24]

Answer:

the distance travelled from the bullet to the target  is 391m

Explanation:

Hello! To solve this exercise we must follow the following steps.

1. the bullet travels with constant speed which means that the distance traveled to the target is given by the following equation

X=(V1)(T1)

T1=\frac{X}{V1} =\frac{x}{460}

where

X=target distance

V1=bullet speed=460m/s

T1=

time it takes for the bullet to reach the target

2. The distance the sound travels is given by the following equation (it is the same as the distance from the person to the target)

X=(V2)(T2)

T2=\frac{X}{V2} =\frac{x}{340}

X=

target distance

V2= speed of sound=340m/s

T2=   time it takes the sound of the Bullet to return.

3. The total time it takes for the person to hear the bullet(T=2s) is the sum of the time it takes for the bullet to reach the target, plus the time it takes for the sound to reach the person, with the above we infer the following equation

T=T1+T2

2=T1+T2

4. Finally we use the equations found in step 1 and 2 to find the distance traveled using algebra.

2=\frac{x}{340}+\frac{x}{460} \\x(\frac{1}{340} +\frac{1}{460} )=2\\\ X= \frac{2}{(\frac{1}{340} +\frac{1}{460} )} \\\\x=391m

the distance travelled from the bullet to the target  is 391m

3 0
3 years ago
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