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Greeley [361]
3 years ago
14

A charge of 6.7 x 10 coulombs is located at a point where its potential energy is 5.6 x 10-12 joules. What is the electric poten

tial at that point?
Physics
1 answer:
Georgia [21]3 years ago
5 0

6.7 X 10 = 670

5.6 X 10 - 12 =44

714

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What two kinds of crust are involved in a subduction zone
Anarel [89]

Answer:

Oceanic crust and continental crust

Explanation:

A subduction zone is normally between oceanic crust which is made of basalt and continental crust which is made of granite. Oceanic crust is denser than continental crust. So when oceanic crust collides with continental crusts, it subsducts underneath the continental crust since it is denser.

5 0
3 years ago
Convert 7.93 lbs into grams. Hint: 1 kg = 2.2 lbs
Alekssandra [29.7K]

Answer:

7.93 lbs is equal to 3596.987 grams.

Explanation:

The weight in grams is equal to the pounds multiplied by 453.59237.

So... you would multiply 7.93 by 453.59237.

7.93 x 453.59237 = 3596.987

Hope that helped!

6 0
3 years ago
Read 2 more answers
Which is gerrymandering a problem for the house of representatives but not the senate?
Fudgin [204]

Senators are elected in statewide elections rather than in specific districts. <em>(D)</em>

3 0
3 years ago
A child is playing on a swing. As long as he does not swing too high the time it takes him to complete one full oscillation will
Aleks [24]

Answer:

We know that for a pendulum of length L, the period  (time for a complete swing) is defined as:

T = 2*pi*√(L/g)

where:

pi = 3.14

L = length of the pendulum

g = gravitational acceleration = 9.8 m/s^2

Now, we can think on the swing as a pendulum, where the child is the mass of the pendulum.

Then the period is independent of:

The mass of the child

The initial angle

Where the restriction of not swing to high is because this model works for small angles, and when the swing is to high the problem becomes more complex.

7 0
3 years ago
Proposed Exercise - Circular Movement
notka56 [123]

Answer:

ωB = 300 rad/s

ωC = 600 rad/s

Explanation:

The linear velocity of the belt is the same at pulley A as it is at pulley D.

vA = vD

ωA rA = ωD rD

ωD = (rA / rD) ωA

Pulley B has the same angular velocity as pulley D.

ωB = ωD

The linear velocity of the belt is the same at pulley B as it is at pulley C.

vB = vC

ωB rB = ωC rC

ωC = (rB / rC) ωB

Given:

ω₀A = 40 rad/s

αA = 20 rad/s²

t = 3 s

Find: ωA

ω = αt + ω₀

ωA = (20 rad/s²) (3 s) + 40 rad/s

ωA = 100 rad/s

ωD = (rA / rD) ωA = (75 mm / 25 mm) (100 rad/s) = 300 rad/s

ωB = ωD = 300 rad/s

ωC = (rB / rC) ωB = (100 mm / 50 mm) (300 rad/s) = 600 rad/s

5 0
3 years ago
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