4 years ago

Which of the following planets rotates on its side?

Physics

2 answers:

Vesna [10]4 years ago

5 0

Im pretty sure the answer is Uranus :)

liraira [26]4 years ago

4 0

The answer for k12 students would be, also here is a helpful attachment of a chart from the lesson, feel free to copy it its really useful!

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5) When a substance changes its state, all BUT ONE of these statements is true That is A) Mass remains constant B) Only the form

expeople1 [14]

C) kinetic energy changes. Gas>liquid>solid

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3 years ago

What is true for ALL of the examples of electromagnetic waves? A) They all move at the same speed in a vacuum. B) They all have

gizmo_the_mogwai [7]

<em>A statement that is true for ALL of the examples of electromagnetic waves is that;</em>

A) They all move at the same speed in a vacuum

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3 years ago

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A 12 N force is used to pull a wagon 10 m to the left in 15 seconds. Which of the following represents the magnitude of the forc

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3. - 0.5 m/s/s

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3 years ago

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You have been hired by a "storm chaser" as an assistant. This individual loves to find locations at which tornadoes and violent

Studentka2010 [4]

Answer:

58.44 C

Explanation:

Electric field is found by

$E=\frac {\sigma}{\epsilon_o}=\frac {Q}{A\epsilon_o}$

Therefore, the charge is

$Q=EA\epsilon_o
$

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$

Therefore, required charge is 58.44 C

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3 years ago

An ideally efficient heat pump delivers 1000 J of heat to room air at 300 K. If it extracted heat from 260 K outdoor air, how mu

choli [55]

Answer:

Wnet, in, = 133.33J

Explanation:

Given that

Pump heat QH = 1000J

Warm temperature TH= 300K

Cold temperature TL= 260K

Since the heat pump is completely reversible, the combination of coefficient of performance expression is given as,

From first law of thermodynamics,

COP(HP, rev) = 1/(1-TL/TH)

COP(HP, rev) = 1/(1-260/300)

COP(HP, rev) = 1/(1-0.867)

COP(HP, rev) = 1/0.133

COP(HP, rev) = 7.5

The power required to drive the the heat pump is given as

Wnet, in= QH/COP(HP, rev)

Wnet, in = 1000/7.5

Wnet, in = 133.333J. QED

So the 133.33J was the amount heat that was originally work consumed in the transfer.

Extra....

According to the first law, the rate at which heat is removed from the low temperature reservoir is given as

QL=QH-Wnet, in

QL=1000-133.333

QL=866.67J

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4 years ago