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3 years ago

What is the magnitude of the electric force between two point charges with Q1 = -1.5 C and Q2 = 0.8 C at a distance of 1 km?

Physics

1 answer:

Ilia_Sergeevich [38]3 years ago

8 0

Answer:

F = -10800 N

Explanation:

Given that,

Charge 1, $q_1=-1.5\ C$

Charge 2, $q_2=0.8\ C$

Distance between the charges, $d=1\ km=10^3\ m$

We need to find the electric force acting between two point charges. Mathematically, it is given by :

$F=k\dfrac{q_1q_2}{d^2}$

$F=-9\times 10^9\times \dfrac{1.5\times 0.8}{(10^3)^2}$

F = -10800 N

So, the magnitude of electric force between two point charges is 10800 N. Hence, this is the required solution.

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Answer:

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3 years ago

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- A fridge has a maximum static friction force

Dennis_Churaev [7]

Answer:

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Explanation:

The maximum static friction force is

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3 years ago

List the 3 types of collisions that occur when your vehicle hits an object.

koban [17]

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3 years ago

If V has a magnitude of 14 units and the same direction qs a vector 3i+6j+2k find v

melomori [17]

Answer:

v = 6i + 12j + 4k

Explanation:

Find the magnitude of the direction vector.

√(3² + 6² + 2²) = 7

Normalize the direction vector.

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Multiply by the magnitude of v.

v = 14 (3/7 i + 6/7 j + 2/7 k)

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3 years ago

A grandfather clock has a pendulum that consists of a thin brass disk of radius r = 13.62 cm and mass 1.199 kg that is attached

Feliz [49]

Answer:

Explanation:

Expression for time period of a pendulum is as follows

T = $2\pi\sqrt{\frac{l}{g} }$

l is length of pendulum from centre of bob and g is acceleration due to gravity

Given

Time period T = 1.583

g = 9.846

Substituting the values

1.583 = $2\pi\sqrt{\frac{l}{9.846} }$

l = $\frac{(1.583)^2\times9.846}{4\times(\frac{22}{7})^2 }$

l = .6244 m

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= 48.82 cm

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2 years ago