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3 years ago

5

The image above was taken by the spacecraft Messenger as it flew by the planet Mercury. The terrain shown in this image is typic

al for the surface of Mercury. There are many circular depressions of various sizes in the scene (the largest one in the upper right part of the image is about 133 km across). These depressions are called _______ and formed when asteroids and comets struck the planet.
A.
volcanoes
B.
mountain ranges
C.
tectonic plates
D.
impact craters

1 answer:

lorasvet [3.4K]3 years ago

7 0

The answer is

C

Hope this helps

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If you push on the wall with a force of 400 N, what is the magnitude of the force at the wall exerts on you?

Inessa05 [86]

Answer:

400 N

Explanation:

Due to action-Reaction, the wall pushes back on the person with a force of equal magnitude and in opposite direction to the force exerted by the person.

So the magnitude must be also 400 N.

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3 years ago

In the blanks below write what the independent variable and the dependent variable would be for each experiment. (what are you t

tiny-mole [99]

Answer:

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9. independent- if you eat no sugar or no carbs. dependent- how much weight you lose

10. independent- which thing you measure. dependent- the measurements

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3 years ago

A circuit contains an EMF source, a resistor R, a capacitor C, and an open switch in series. The capacitor initially carries zer

Flura [38]

Answer:

t = 1.098*RC

Explanation:

In order to calculate the time that the capacitor takes to reach 2/3 of its maximum charge, you use the following formula for the charge of the capacitor:

$Q=Q_{max}[1-e^{-\frac{t}{RC}}]$ (1)

Qmax: maximum charge capacity of the capacitor

t: time

R: resistor of the circuit

C: capacitance of the circuit

When the capacitor has 2/3 of its maximum charge, you have that

Q=(2/3)Qmax

You replace the previous expression for Q in the equation (1), and use properties of logarithms to solve for t:

$Q=\frac{2}{3}Q_{max}=Q_{max}[1-e^{-\frac{t}{RC}}]\\\\\frac{2}{3}=1-e^{-\frac{t}{RC}}\\\\e^{-\frac{t}{RC}}=\frac{1}{3}\\\\-\frac{t}{RC}=ln(\frac{1}{3})\\\\t=-RCln(\frac{1}{3})=1.098RC$

The charge in the capacitor reaches 2/3 of its maximum charge in a time equal to 1.098RC

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4 years ago

A composite load consists of three loads connected in parallel. One draws 100 W at a PF of 0.92 lagging, another takes 250 W at

fenix001 [56]

Answer:

a) $I_{RMS} = 4.79 A$

b) $PF = 0.908$

Explanation:

Get the reactive powers for each of the loads:

Reactive power = Real Power * tanθ

For load 1

Active power, P₁ = 100 W

Power factor, $cos \theta_{1} = 0.92$

$\theta_{1} = cos^{-1} 0.92\\\theta_{1} = 23.074$

$Q_{1}= P_{1} tan \theta_{1} \\Q_{1}= 100tan 23.074\\Q_{1}= 42.60 W$

For load 2

Active power, P₂ = 250 W

Power factor, $cos \theta_{2} = 0.8$

$\theta_{2} = cos^{-1} 0.8\\\theta_{2} = 36.87$

$Q_{2}= P_{1} tan \theta_{2} \\Q_{2}= 250tan 36.87\\Q_{2}= 187.5 W$

For load 3

Active power, P₃ = 250 W

Power factor, $cos \theta_{3} = 1$

$\theta_{3} = cos^{-1} 1\\\theta_{3} =0$

$Q_{2}= P_{1} tan \theta_{3} \\Q_{3}= 150tan 0\\Q_{3}= 0 W$

Calculate the total reactive power, $Q_{net} = 42.6 + 187.5 + 0$

$Q_{net} = 230.1 W$

Calculate the total active power, $P_{net} = 100 + 250 + 150 = 500 W$

$S_{net} = P_{net} + Q_{net} \\S_{net} = 500 + j230.1$

$P_{net} = IVcos \theta_{net}$

$\theta_{net} = tan^{-1} \frac{230.1}{500} \\\theta_{net} = 24.712$

V = 115 $V_{rms}$

$500 = I_{RMS} * 115 cos 24.712\\I_{RMS} = 500/104.47\\ I_{RMS} = 4.79 A$

b) Power factor of the composite load is $cos\theta_{net}$

$\theta_{net} = 24.712\\PF = cos 24.712\\PF = 0.908$

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3 years ago

When an orange rolls off a table, its potential energy is converted into?

Anna007 [38]

Since the orange rolls off the table and is (falling) its potential energy turnes into kinetic energy.

Hope this helped
:D

7 0

3 years ago