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3 years ago

Define project torque

Physics

1 answer:

pogonyaev3 years ago

8 0

Answer:

ExplanProject Torque was the free North American version of Level-R. It is a multiplayer online racing game (MMORG) with partially chargeable content, or micro-transactions. It features gameplay elements such as tuning and customization. The game is titled as Level R in Europe with a slightly different game interface and menus. Europe, US, Russia, Indonesia, Thailand, China, and Japan versions, are developed by Invictus Games and are subject to Invictus copyrightation:

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If 2.0 x 10^-4 C charge passes a point in 5.0 x 10^-5 s, what is the rate of current flow?

vampirchik [111]

Current is defined as the rate of charge flowing a point every second. Having a current of 1 Ampere signifies 1 Coulomb is flowing in a circuit every second. It is measured by the use of an ammeter which is positioned in series to the component to be measured. The current in the problem is calculated as follows:

I = 2.0 x 10^-4 C / 5.0 x 10^-5 s
<span>I = 4 A</span>

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3 years ago

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larisa86 [58]

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3 years ago

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What type of device forms images through refraction?

Tcecarenko [31]

Answer:

A lens

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2 years ago

A skier starts from rest down a slope 500.0 M long, the skier accelerates at a constant rate of 2.00 m/s/s, what's the velocity

nevsk [136]

We can use the kinematic equation
$(v_f)^2 = (v_i)^2 + 2*a*d$
where Vf is what we are looking for
Vi is 0 since we start from rest
a is acceleration
and d is the distance

we get
(Vf)^2 = (0)^2 + 2*(2)*(500)
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8 0

3 years ago

Chapter 21, Problem 009 Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.12

PilotLPTM [1.2K]

Answer:

a) -1.325 μC

b) 4.17μC

Explanation:

First, you need to know that charge is conserved. So, the adition of the charges, as there is no lost in charge, should always be the same. Also, after the wire is removed, both spheres will have the same charge, trying to find equilibrium. In summary:

$q_1 + q_2 = constant\\q_1_f = q_2_f |Then\\q_1_f + q_2_f = 2q_1_f = q_1_o+q_2_o\\q_1_f = q_2_f = \frac{q_1_o+q_2_o}{2}$

We know both q1f and q2f must be positive, because the negative charge at the beginning was the the smaller.

The electrostatic force is equal to:

$F_e = k\frac{q_1q_2}{r^2}$

K is the Coulomb constant, equal to 9*10^9 Nm^2/C^2

Now, we are told that the electrostatic force after the wire is equal to 0.0443 N:

$F_e_f = k \frac{q_1_fq_2_f}{r^2} = k\frac{\frac{q_1_o+q_2_o}{2}\frac{q_1_o+q_2_o}{2}}{r^2} = k\frac{(q_1_o+q_2_o)^2}{4r^2} \\(q_1_o+q_2_o) = \sqrt{\frac{F_e_f*4r^2}{k}} = \sqrt{\frac{0.0443N *4(0.641m)^2}{9*10^9Nm^2/C^2} } = 2.844 *10^{-6}C \\ q_1_o = 2.844*10^{-6}C - q_2_o$

Originally, the force is negative because it was an attraction force, therefore, its direction was opposite to the direction of the repulsive force after the wire:

$F_e_o = k\frac{q_1_oq_2_o}{r^2}\\ q_1_oq_2_o = \frac{F_e_o*r^2}{k} = \frac{-0.121N(0.641m)^2}{9*10^9Nm^2/C^2} = -5.524*10^{-12}$

$(2.844*10^{-6}C - q_2_o)q_2_o = -5.524*10^{-12}\\0 = q_2_o^2 - 2.844*10^{-6}q_2_o - 5.524*10^{-12}$

Solving the quadratic equation:

$q_2_o = 4.17*10^{-6}C | -1.325 * 10^{-6}C$

for this values q_1 wil be:

$q_1_o = -1.325 *10^{-6}C | 4.17*10^{-6}C$

So as you can see, the negative charge will always be -1.325 μC and the positive 4.17μC

5 0

3 years ago