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3 years ago

1. A 12 kg dog is running 5 m/s. How much kinetic energy does the dog have?

Physics

1 answer:

Neporo4naja [7]3 years ago

6 0

Answer:

Explanation:

1. KE = 1/2mv^2

= 1/2 (12)×5^2

= 6× 25

KE = 150j

2. KE = 1/2 mv^2

100j = 1/2m(5)^2

2(100j) = m×25

m = 200÷ 25

m = 8 kg

hope it helps

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Two stationary positive point charges, charge 1 of magnitude 3.25 nC and charge 2 of magnitude 2.00 nC , are separated by a dist

Liula [17]

Complete Question

Two stationary positive point charges, charge 1 of magnitude 3.25 nC and charge 2 of magnitude 2.00 nC , are separated by a distance of 58.0 cm . An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges.

Required:

What is the speed of the electron when it is 10.0 cm from the +3.25-nC charge?

Answer:

The velocity is $v = 80.82 \ m/s$

Explanation:

From the question we are told that

The magnitude of charge one is $q_1 = 3.25 nC = 3.25 *10^{-9} \ C$

The magnitude of charge two $q_2 = 2.00 \ nC = 2.00 *10^{-9} \ C$

The distance of separation is $d = 58.0 \ cm = 0.58 \ m$

Generally the electric potential of the electron at the midway point is mathematically represented as

$V = \frac{ q_1 }{\frac{d}{2} } + \frac{ q_2}{\frac{d}{2} }$

substituting values

$V = \frac{ 3.25 *10^{-9} }{\frac{ 0.58}{2} } + \frac{ 2 *10^{-9} }{\frac{ 0.58}{2} }$

$V = 1.8103 *10^{-8} \ V$

Now when the electron is 10 cm = 0.10 m from charge 1 , it is (0.58 - 0.10 = 0.48 m ) m from charge two

Now the electric potential at that point is mathematically represented as

$V_1 = \frac{q_1}{ 0.10} + \frac{q_2}{ 0.48}$

substituting values

$V_1 = \frac{3.25 *10^{-9}}{ 0.10} + \frac{2.0*10^{-9}}{ 0.48}$

$V_1 = 3.67*10^{-8} \ V$

Now the law of energy conservation ,

The kinetic energy of the electron = potential energy of the electron

i.e $\frac{1}{2} * m * v^2 = [V_1 - V]* q$

where q is the magnitude of the charge on the electron with value

$q = 1.60 *10^{-19} \ C$

While m is the mass of the electron with value $m = 9.11*10^{-31} \ kg$

$\frac{1}{2} * 9.11 *10^{-19} * v^2 = [ (3.67 - 1.8103) *10^{-8}]* 1.60 *10^{-19}$

$v = \sqrt{6532.4}$

$v = 80.82 \ m/s$

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A loudspeaker diaphragm is vibrating in simple harmonic motion with a frequency of 760 Hz and a maximum displacement of 0.85 mm.

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