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Korvikt [17]
3 years ago
11

What needs to be done to this circuit so that the light bulb lights up?

Physics
2 answers:
kramer3 years ago
6 0
A) close the switch
I took the test
Pachacha [2.7K]3 years ago
5 0
The answer is the first one a
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Neglecting air resistance which of the following statements is not true regarding the motion of a ball thrown vertically with an
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c or b

Explanation:

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3 years ago
Relative to the ground, what is the gravitational potential energy of 68 kg
Kisachek [45]

Answer:

Explanation:

p.E=mgh

P.E=68*10*443

p.E=301240j

8 0
2 years ago
You are loading a toy dart gun, which has two settings, the more powerful with the spring compressed twice as far as the lower s
Amiraneli [1.4K]

Answer:

option A

Explanation:

given,

compression for lower setting = x

work done to compress in lower setting = 5 J

compression in higher setting, x' = 2 x

work done in higher setting = ?

Work done in compression of spring at lower setting

W = \dfrac{1}{2}kx^2

5 = \dfrac{1}{2}kx^2............(1)

Work done in compression of spring at higher setting

W' = \dfrac{1}{2}kx'^2

W'= \dfrac{1}{2}k(2x)^2

W'= 4\times \dfrac{1}{2}kx^2

W'= 4\times W

from equation (1)

W'= 4\times 5

    W' = 20 J

Work take for the higher setting is equal to 20 J

Hence, the correct answer is option A

4 0
3 years ago
A 1/10th scale model of an airplane is tested in a wind tunnel. The reynolds number of the model is the same as that of the full
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Answer:

of the velocity of a full size plane in the air

7 0
4 years ago
Two particles, one with charge -5.45 × 10^-6 C and one with charge 4.39 × 10^-6 C, are 0.0209 meters apart. What is the magnitud
Levart [38]

Explanation:

Given that,

Charge 1, q_1=-5.45\times 10^{-6}\ C

Charge 2, q_2=4.39\times 10^{-6}\ C

Distance between charges, r = 0.0209 m

1. The electric force is given by :

F=k\dfrac{q_1q_2}{r^2}

F=9\times 10^9\times \dfrac{-5.45\times 10^{-6}\times 4.39\times 10^{-6}}{(0.0209)^2}

F = -492.95 N

2. Distance between two identical charges, r=0.0209\ m

Electric force is given by :

F=\dfrac{kq_3^2}{r^2}

q_3=\sqrt{\dfrac{Fr^2}{k}}

q_3=\sqrt{\dfrac{492.95\times (0.0209)^2}{9\times 10^9}}

q_3=4.89\times 10^{-6}\ C

Hence, this is the required solution.

3 0
3 years ago
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