When two oceanic plate boundaries meet, this crustal feature forms?

A planet with half Earth’s mass and half Earth’s radius is discovered. What would an astronaut who weighs 800 N on Earth weigh o

Citrus2011 [14]

1600 N
Hope this helped !

5 0

3 years ago

Earth is slightly closer to the sun in january than in july. How does the area swept out by earth’s orbit around the sun during

Anna007 [38]

This is because both areas are the same.

8 0

3 years ago

How much electrical potential energy is stored in a capacitor that has 8.0 × 10-10 C of charge on each plate and a potential dif

arsen [322]

THere is a standard relationship that gives this result where the capacity of the capacitor is used:
$E=1/2* \frac{Q^2}{C}$ .
We know though that Q/c=V and thus we can use the relationship:
E=Q*V/2 where we have just substituted in. If we also take into account that Q=VC, we can also get that E=V^2*C/2.
We are given the charge and the potential, so the best expression to use is the middle one.
Substituting, we get that E=1/2*8*10^(-10)*20=8*10^(-9).

The answer is B

8 0

3 years ago

An engine extracts 452.8kJ of heat from the burning of fuel each cycle, but rejects 266.7 kJ of heat (exhaust, friction,etc) dur

Svetlanka [38]

Answer: The thermal efficiency of the engine is 41.09 %.

Explanation:

Efficiency is the ratio of the useful work performed to the total energy expended or heat taken in.

Formula for thermal efficiency of engine is

$\eta=1- \frac{Q_2}{Q_1}\times 100$

$\eta$ = efficiency

${Q_2}$ = heat rejected = 266.7 kJ

${Q_1}$ = heat extracted = 452.8 kJ

Putting in the values we get:

$\eta=1- \frac{266.7 kJ}{452.8 kJ }\times 100$

$\eta=0.41\times 100$

$\eta =41.09\%$

The thermal efficiency of the engine is 41.09 %.

7 0

4 years ago

What is the electrical force between q1 and q2? Recall that k = 8.99 × 109 N•meters squared over Coulombs squared.. 4.3 × 10 N 3

Rudiy27

Answer:

Explanation:

Incomplete question but for understanding.

We want to find the electrical force between two charges, then you can use the coulombs law which states that the force of attraction or repulsion between two charges is directly proportional to the product of the two charges and inversely proportional to the square of their distance apart,

So,

F = kq1•q2 / r²

Where k is a constant and it is given as

K = 8.99 × 10^9 Nm²/C²

q1 and q2 are the charges and in this question it is not given, so the question is incomplete. Let assume that,

q1 = - 1.609 × 10^-19 C electron

q2 = 1.609 × 10^-19 C proton

Since unlike charges attract, then it is force of attraction

Also, r is the distance apart and it is not given, let assume the distance between the two charges is 2 × 10^-5m

Then,

F = kq1•q2 / r²

F = 8.99 × 10^9 × 1.609 × 10^-19 × 1.609 × 10^-19 / (2 × 10^-5)²

F = 5.82 × 10^-19 N

7 0

3 years ago

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