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2 years ago

What causes the erosion of rocks on the moon’s surface?

Physics

2 answers:

liberstina [14]2 years ago

7 0

<span> Lack of an atmosphere on the Moon greatly reduces the erosion of the surface. </span>

tia_tia [17]2 years ago

4 0

impact of particles from space.

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As a new electrical technician, you are designing a large solenoid to produce a uniform 0.170 T magnetic field near the center o

MrRa [10]

Answer:

18.6012339739 A

Explanation:

$\mu_0$ = Vacuum permeability = $4\pi \times 10^{-7}\ H/m$

L = Length of wire = 55 cm

N = Number of turns = 4000

I = Current

Magnetic field is given by

$B=\dfrac{\mu_0NI}{L}\\\Rightarrow I=\dfrac{BL}{\mu_0N}\\\Rightarrow I=\dfrac{0.17\times 0.55}{4\pi \times 10^{-7}\times 4000}\\\Rightarrow I=18.6012339739\ A$

The current necessary to produce this field is 18.6012339739 A

7 0

3 years ago

Select the correct answer

DiKsa [7]

Answer:

C or 3

Explanation:

A: no they are called sources of sound. A is incorrect.

B: It does. Many people attest to this. But this is not a property of physics.

C: A media is required is the correct answer.

D: Dogs might. In general we don't.

I would pick C

7 0

2 years ago

A mars prototype carrying scientific instruments has a mass of 1060kg. a net force of 52000 n is applied to this roverat a test

Nitella [24]

F = 52000 N

m = 1060 kg

a= F/m = 52000 N/1060 kg = 49.0566 m/s^2

8 0

3 years ago

X-rays cannot pass through Earth's atmosphere. Which of these is the best location to place a telescope used to observe x-rays f

Amiraneli [1.4K]

Mountains, tops of buildings, and high-flying aircraft are all part of Earth's atmosphere, no matter how high they are. On the other hand, space doesn't belong to our atmosphere, it is outside of it. Having this in mind, the best location to place a telescope used to observe x-rays from stars is in space.

7 0

2 years ago

Read 2 more answers

Options: A.) 10 N B.) 15 N C.) 25 N D.) 35N

Lady_Fox [76]

Given:

F_gravity = 10 N

F_tension = 25 N

Let's find the net centripetal force exterted on the ball.

Apply the formula:

$\sum ^{}_{}F_{\text{net}}=F_1+F_2=F_{centripetal}$

From the given figure, the force acting towards the circular path will be positive, while the force which points directly away from the center is negative.

Hence, the tensional force is positive while the gravitational force is negative.

Thus, we have:

$F_{\text{net}}=F_{\text{centripetal}}=F_{tension}-F_{gravity}=25N-10N=15N$

Therefore, the net centripetal force exterted on the ball is 15 N.

ANSWER:

15 N

7 0

8 months ago