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andreev551 [17]
3 years ago
9

Terrence connects several lightbulbs in a parallel circuit. Electricity flows through the circuit, and all the bulbs light up. 

  What will happen if Terrence removes lightbulb number 2
Physics
1 answer:
Ganezh [65]3 years ago
7 0
Light bulb number two would go out and the others would stay lit because it is parallel and not a series circuit that if you remove one from it they all go out. 
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What is the gravitational force between mars and Phobos
alina1380 [7]

Answer:

F=5.16\times 10^{15}\ N

Explanation:

We have,

Mass of Mars is, m_M=6.42\times 10^{23}\ kg

Mass of its moon Phobos, m_P=1.06\times 10^{16}\ kg

Distance between Mars and Phobos, d = 9378 km

It is required to find the gravitational force between Mars and Phobos. The force between two masses is given by

F=G\dfrac{m_Mm_P}{d^2}

Plugging all values, we get :

F=6.67\times 10^{-11}\times \dfrac{6.42\times 10^{23}\times 1.06\times 10^{16}}{(9378\times 10^3)^2}\\\\F=5.16\times 10^{15}\ N

So, the gravitational force is 5.16\times 10^{15}\ N.

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ludmilkaskok [199]

Answer:

1) ironing a shirt 2) writing on surfaces 3) working of an eraser

5 0
3 years ago
A 10.0-g bullet is fired into a 200-g block of wood at rest on a horizontal surface. after impact, the block slides 8.00 m befor
miss Akunina [59]
<span>Step 1 -- determine the acceleration of the 200-g block after bullet hits it a = (coeff of friction) * g g = acceleration due to gravity = 9.8 m/sec^2 (constant) a = 0.400*9.8 a = 3.92 m/sec^2 Step 2 -- determine the speed of the block after the bullet hits it Vf^2 - Vb^2 = 2(a)(s) where Vf = final velocity = 0 (since it will stop) Vb = velocity of block after bullet hits it a = -3.92 m/sec^2 s = stopping distance = 8 m (given) Substituting values, 0 - Vb^2 = 2(-3.92)(8) Vb^2 = 62.72 Vb = 7.92 m/sec. M1V1 + M2V2 = (M1 + M2)Vb where M1 = mass of the bullet = 10 g (given) = 0.010 kg. V1 = velocity of bullet before impact M2 = mass of block = 200 g (given) = 0.2 kg. V2 = initial velocity of block = 0 Vb = 7.92 m/sec Substituting values, 0.010(V1) + 0.2(0) = (0.010 + 0.2)(7.92) Solving for V1, V1 = 166.32 m/sec. Therefore the answer is (B) 166 m/s!</span>
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3 years ago
A 96.0 kg hoop rolls along a horizontal floor so that its center of mass has a speed of 0.240 m/s. How much work must be done on
Ede4ka [16]

Answer:

The work done by the hoop is equal to 5.529 Joules.

Explanation:

Given that,

Mass of the hoop, m = 96 kg

The speed of the center of mass, v = 0.24 m/s

To find,

The work done by the hoop.

Solution,

The initial energy of the hoop is given by the sum of linear kinetic energy and the rotational kinetic energy. So,

K_i=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2

I is the moment of inertia, I=mr^2

Since, \omega=\dfrac{v}{r}

K_i=mv^2

K_i=96\times (0.24)^2=5.529\ J

Finally it stops, so the final energy of the hoop will be, K_f=0

The work done by the hoop is equal to the change in kinetic energy as :

W=K_f-K_i

W = -5.529 Joules

So, the work done by the hoop is equal to 5.529 Joules. Therefore, this is the required solution.

4 0
3 years ago
Connecting many devices in a single socket does not affect the flow of current in a
eimsori [14]

<u>A</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u><u>:</u><u>-</u><em> </em><em>F</em><em>a</em><em>l</em><em>s</em><em>e</em>

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<em>False because it can leads to overloading and further to short circut.</em>

<h2><em><u>H</u></em><em><u>o</u></em><em><u>p</u></em><em><u>e</u></em><em><u> </u></em><em><u>I</u></em><em><u>t</u></em><em><u> </u></em><em><u>W</u></em><em><u>i</u></em><em><u>l</u></em><em><u>l</u></em><em><u> </u></em><em><u>H</u></em><em><u>e</u></em><em><u>l</u></em><em><u>p</u></em><em><u> </u></em><em><u>Y</u></em><em><u>o</u></em><em><u>u</u></em><em><u>!</u></em></h2>
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3 years ago
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