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Dmitriy789 [7]
3 years ago
5

A boy shoves his stuffed toy zebra, which has mass m, down a frictionless chute, starting at a height D above the bottom of the

chute and with an initial speed of v. The toy animal emerges horizontally from the bottom of the chute and continues sliding along a horizontal surface with coefficient of kinetic friction μ. At what distance d from the bottom of the chute does the toy zebra come to rest? Express your answer in terms of the given variables and g, the acceleration due to gravity.
Physics
1 answer:
suter [353]3 years ago
6 0

Answer:

d=\frac{0.5v^2+gD}{\mu\times g}

Explanation:

mass of the boy = m kg

staring height of the boy = D meters

initial velocity = v

coefficient of kinetic friction = μ

total energy of the toy is

E_total = K.E. +P.E.

= 0.5mv^2 +mgD

= m( 0.5v^2+ gD)

The work done by friction force

W= μ×F_normal ×distance

W= μ×m×g×d

hence, the distance d covered by the zebra toy

and since, the total energy is conserved

W= E_total

μ×m×g×d= m( 0.5v^2+ gD)

d=\frac{0.5v^2+gD}{\mu\times g}

hence, the distance d from the bottom of the chute does the toy algebra comes to rest.

d=\frac{0.5v^2+gD}{\mu\times g}

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A parallel plate air capacitor has a capacitance of 10 to the power -9. What potential difference is required for a charge of 5×
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The potential difference across the capacitor is 5 × 10∧4 volts and the energy stored in it is 1. 25 Joules

<h3>What is the energy in a capacitor?</h3>

The energy stored in a capacitor is an electrostatic potential energy.

It is related to the charge(Q) and voltage (V) between the capacitor plates.

It is represented as 'U'.

<h3>How to determine the potential difference</h3>

Formula:

Potential difference, V is the ratio of the charge to the capacitance of a capacitor.

It is calculated using:

V = Q ÷ C

Where Q = charge 5 × 10∧-5C and C = capacitance 10∧-9

Substitute the values into the equation

Potential difference, V = 5 × 10∧-5 ÷  10∧-9 = 5 × 10∧4 volts

<h3>How to determine the energy stored</h3>

Formula:

Energy, U = 1 ÷ 2 (QV)

Where Q= charge and V = potential difference across the capacitor

Energy, U = 1 ÷ 2 ( 5 × 10∧-5 × 5 × 10∧4)

= 0.5 × 25 × 10∧-1

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Therefore, the potential difference across the capacitor is 5 × 10∧4 volts and the energy stored in it is 1. 25 Joules

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