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Aliun [14]
2 years ago
12

What is the height of a building that an object is dropped from if it has a mass of 3 kg and hits the ground with a velocity of

50 m/s?
Physics
1 answer:
Marizza181 [45]2 years ago
8 0

Answer:

125 m

Explanation:

m = 3kg

v = 50m/s

u = 0m/s

a = +g = 10m/s²

s = H = ?

using the formula,

v² = u² + 2as

50² = 0² + 2(10)(H)

2500 = 20H

H = 2500/20

H = 125m

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Explanation:

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A uniform electric field is directed parallel to the +y axis. If a positive test charge begins at the origin and moves upward al
lukranit [14]

Answer: option 1 : the electric potential will decrease with an increase in y

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V = kq/y

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Where V = electric potential,

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We have that

V = k/y

We see the potential(V) is inversely proportional to distance (y).

This implies that an increase in distance results to a decreasing potential and a decrease in distance results to an increase in potential.

This fact makes option 1 the correct answer

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Indentify the 3 types of heat tranfer
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What is the net force acting on the buggy. ?The net force is pointing to the ?
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A spherical capacitor contains a charge of 3.00 nC when connected to a potential difference of 230 V. If its plates are separate
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Answer:

Part(a): the capacitance is 0.013 nF.

Part(b): the radius of the inner sphere is 3.1 cm.

Part(c): the electric field just outside the surface of inner sphere is \bf{2.81 \times 10^{4}~n~C^{-1}}.

Explanation:

We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and '\epsilon_{0}' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as

C = \dfrac{4 \pi \epsilon_{0}}{(\dfrac{1}{a} - \dfrac{1}{b})}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)

Part(a):

Given, charge contained by the capacitor Q = 3.00 nC and potential to which it is subjected to is V = 230V.

So the capacitance (C) of the shell is

C &=& \dfrac{Q}{V} = \dfrac{3 \times 10^{-90}~C}{230~V} = 1.3 \times 10^{-11}~F = 0.013~nF

Part(b):

Given the inner radius of the outer shell b = 4.3 cm = 0.043 m. Therefore, from equation (1), rearranging the terms,

&& \dfrac{1}{a} = \dfrac{1}{b} + \dfrac{1}{C/4 \pi \epsilon_{0}} = \dfrac{1}{0.043} + \dfrac{1}{1.3 \times 10^{-11} \times 9 \times 10^{9}} = 31.79\\&or,& a = \dfrac{1}{31.79}~m = 0.031~m = 3.1~cm

Part(c):

If we apply Gauss' law of electrostatics, then

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3 0
3 years ago
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