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daser333 [38]
3 years ago
7

A ball is dropped from the ceiling and free falls to the floor. Which one of the

Physics
1 answer:
Alex73 [517]3 years ago
5 0

Answer:

A

Explanation:

Gravity

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Eye strain and self confidence issues.
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You place a box weighing 200.2 N on a inclined plane that makes a 37.1 angle with the horizontal. Compute the component of the g
salantis [7]

Answer:

141.56 N.

Explanation:

Data given:

Weight of the box= 200.2 N

Angle with the horizontal= 37.1°

Solution;

Gravitational force on the box, F_g= weight of the box

                                                                           = 200.2 N

Component of gravitational force along plane = F_g * sin( ∅ )

                                                                                   = W * (sin∅)

                                                                                   = (200.1) * sin (37.1°)

                                                                                   = 141.56 N

8 0
3 years ago
(b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as meas
aleksley [76]

Complete Question: A charge q1 = 2.2 uC is at a distance d= 1.63m from a second charge q2= -5.67 uC. (b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as measured from q1.)

Answer:

d= 0.46 m

Explanation:

The electric potential is defined as the work needed, per unit charge, to bring a positive test charge from infinity to the point of interest.

For a point charge, the electric potential, at a distance r from it, according to Coulomb´s Law and the definition of potential, can be expressed as follows:

V = \frac{k*q}{r}

We have two charges, q₁ and q₂, and we need to find a point between them, where the electric potential due to them, be zero.

If we call x to the distance from q₁, the distance from q₂, will be the distance between both charges, minus x.

So, we can find the value of x, adding the potentials due to q₁ and q₂, in such a way that both add to zero:

V = \frac{k*q1}{x} +\frac{k*q2}{(1.63m-x)} = 0

⇒k*q1* (1.63m - x) = -k*q2*x:

Replacing by the values of q1, q2, and k, and solving for x, we get:

⇒ x = (2.22 μC* 1.63 m) / 7.89 μC = 0.46 m from q1.

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When looking at liquids in a container, that have layered on one another;
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