The direction of an electric field is the direction (5 points)

1 answer:

5 0

The direction of an electric field is determined from the behavior of a positive test charge that is set free in the electric field.This charge moves along a distinct vector showing the direction of the electric field Therefore the answer is b. a positive charge will move in the field.

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An object of unknown mass is initially at rest and dropped from a height h. It reaches the ground with a velocity v1 . The same

Hatshy [7]

Answer:

$v_2=\sqrt{2}v_1$

Explanation:

The velocity v₁ can be calculated with the kinematic formula:

$v_1^{2} =v_0^{2} +2gh$

Since the object is initially at rest, v₁ becomes:

$v_1=\sqrt{2gh}$

Where g is the acceleration due to gravity. Now, the velocity v₂ can be calculated with the same formula, but now the initial velocity is v₁:

$v_2^{2}=v_1^{2} +2gh$

Substituting v₁ in this expression and solving for v₂, we get:

$v_2^{2}=(\sqrt{2gh} )^{2} +2gh=4gh\\\\\implies v_2=\sqrt{4gh}=2\sqrt{gh}$

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$\frac{v_2}{v_1}=\frac{2\sqrt{gh} }{\sqrt{2gh}}=\sqrt{2}\\ \\\implies v_2=\sqrt{2}v_1$

It means that v₂ is √2 times v₁.

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Complete Question

The complete question is shown on the first uploaded image

Answer:

the value of $y = 13.3$