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3 years ago

Suppose the moon rotated on its axis just as quickly as Earth. Would we still always see the same side of the moon from Earth?

Physics

1 answer:

jasenka [17]3 years ago

3 0

Answer:

No, The Moon, on the other hand, rotates once around its every 28 days, and once around the Earth in that same 28 days. The result of this combination is that the same side of the Moon is always facing the Earth.

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A proton initially moves left to right along the x-axis at a speed of 2.00 x 103 m/s. It moves into an uniform electric field, w

FinnZ [79.3K]

Answer:

E = 1.04*10⁻¹ N/C

Explanation:

Assuming no other forces acting on the proton than the electric field, as this is uniform, we can calculate the acceleration of the proton, with the following kinematic equation:

$vf^{2} -vo^{2} = 2*a*x$

As the proton is coming at rest after travelling 0.200 m to the right, vf = 0, and x = 0.200 m.

Replacing this values in the equation above, we can solve for a, as follows:

$a = \frac{vo^{2}*mp}{2*x} = \frac{(2.00e3m/s)^{2}}{2*0.2m} = 1e7 m/s2$

According to Newton´s 2nd Law, and applying the definition of an electric field, we can say the following:

F = mp*a = q*E

For a proton, we have the following values:

mp = 1.67*10⁻²⁷ kg

q = e = 1.6*10⁻¹⁹ C

So, we can solve for E (in magnitude) , as follows:

$E = \frac{mp*a}{e} =\frac{1.67e-27kg*1e7m/s2}{1.6e-19C} = 1.04e-1 N/C$

⇒ E = 1.04*10⁻¹ N/C

5 0

3 years ago

What is it plz help me!!

Darina [25.2K]

Answer:

second one

Explanation:

7 0

3 years ago

Read 2 more answers

Determine the critical crack length for a through crack contained within a thick plate of 7150-T651 aluminum alloy that is in un

Mila [183]

Explanation:

Formula to determine the critical crack is as follows.

$K_{IC} = \gamma \sigma_{f} \sqrt{\pi \times a}$

$\gamma$ = 1, $K_{IC}$ = 24.1

[/tex]\sigma_{y}[/tex] = 570

and, $\sigma_{f} = 570 \times \frac{3}{4}$

= 427.5

Hence, we will calculate the critical crack length as follows.

a = $\frac{1}{\pi} \times (\frac{K_{IC}}{\sigma_{f}})^{2}$

= $\frac{1}{3.14} \times (\frac{24.1}{427.5})^{2}$

= $10.13 \times 10^{-4}$

Therefore, largest size is as follows.

Largest size = 2a

= $2 \times 10.13 \times 10^{-4}$

= $20.26 \times 10^{-4}$

Thus, we can conclude that the critical crack length for a through crack contained within the given plate is $20.26 \times 10^{-4}$ .

4 0

3 years ago

PLEASE HELP FOR BRAINLIEST ANSWER!

Alex787 [66]

I believe the answer is the fourth one, hope this helps

3 0

4 years ago

Read 2 more answers

The coils in a motor have a resistance of 0.14 Ω and are driven by an EMF of 62 V.

Brilliant_brown [7]

The difference in the power required by the motor at start up and at operating speed is 9778.57

<h3>What is power? </h3>

This is defined as the rate in which energy is consumed. Electrical power is expressed mathematically as:

Power (P) = voltage (V) × current (I)

P = IV

Power = square voltage (V²) / resistance ®

P = V² / R

<h3>How to determine the power by the EMF</h3>

Voltage (V) = 37 V
Resistance (R) = 0.14 Ω
Power by EMF (P₁) =?

P = V² / R

P₁ = 62² / 0.14

P₁ = 27457.14 watts

<h3>How to determine the power by the back EMF</h3>

Voltage (V) = 62 V
Resistance (R) = 0.14 Ω
Power by back EMF (P₂) =?

P = V² / R

P₂ = 37² / 0.14

P₂ = 9778.57 watts

<h3>How to determine the power difference</h3>

Power by EMF (P₁) = 27457.14 watts
Power by back EMF (P₂) = 9778.57 watts
Difference in power =?

Difference = P₁ - P₂

Difference in power = 27457.14 - 9778.57

Difference in power = 17678.57 watts

Learn more about electrical power:

brainly.com/question/64224

#SPJ1

7 0

2 years ago