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The potential difference across a and b is 15 v. determine the electrical charge on the 3 μf capacitor?

Slav-nsk [51]

The potential difference across a and b is 15 v. determine the electrical charge on the 3 μf capacitor will be 45 * $10^{-6}$ C

Capacitance, property of an electric conductor, or set of conductors, that is measured by the amount of separated electric charge that can be stored on it per unit change in electrical potential. Capacitance also implies an associated storage of electrical energy.

Charge (Q) stored in a capacitor is the product of its capacitance (C) and the voltage (V) applied to it. The capacitance of a capacitor should always be a constant, known value. So we can adjust voltage to increase or decrease the cap's charge. More voltage means more charge, less voltage... less charge.

charge = capacitance * voltage

Q = CV

= 3 * $10^{-6}$ * 15 v

= 45 * $10^{-6}$ C

To learn more about capacitance here

brainly.com/question/14746225

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8 0

2 years ago

A projectile is launched into the air with an initial speed of 40 m/s and a launch angle of 20° above the horizontal. The projec

miv72 [106K]

Explanation:

V=40m/s

Vy=V.sina=40.sin20=40 . 0.342=13.68m/s

Vx=V.cosa=40.cos20=40 . 0.766=30.64m/s

Projectile travels during 5 seconds and the ramge becomes:

x=V.t=30.64 . 5=153.2m

7 0

3 years ago

A girl of mass m1=60 kilograms springs from a trampoline with an initial upward velocity of v1=8.0 meters per second. At height

AleksandrR [38]

a) 5.0 m/s

This first part of the problem can be solved by using the conservation of energy. In fact, the mechanical energy of the girl just after she jumps is equal to her kinetic energy:

$E_i=\frac{1}{2}m_1v_1^2$

where m1 = 60 kg is the girl's mass and v1 = 8.0 m/s is her initial velocity.

When she reaches the height of h = 2.0 m, her mechanical energy is sum of kinetic energy and potential energy:

$E_f = \frac{1}{2}m_1 v_2 ^2 + m_1 gh$

where v2 is the new speed of the girl (before grabbing the box), and h = 2.0m. Equalizing the two equations (because the mechanical energy is conserved), we find

$\frac{1}{2}m_1 v_1^2 = \frac{1}{2}m_1 v_2 ^2 + m_1 gh\\v_1^2 = v_2^2 +2gh\\v_2 = \sqrt{v_1^2 -2gh}=\sqrt{(8.0 m/s)^2-(2)(9.8 m/s^2)(2.0 m)}=5.0 m/s$

b) 4.0 m/s

After the girl grab the box, the total momentum of the system must be conserved. This means that the initial momentum of the girl must be equal to the total momentum of the girl+box after the girl catches the box:

$p_i = p_f\\m_1 v_2 = (m_1 + m_2) v_3$

where m2 = 15 kg is the mass of the box. Solving the equation for v3, the combined velocity of the girl+box, we find

$v_3 = \frac{m_1 v_2}{m_1 + m_2}=\frac{(60 kg)(5.0 m/s)}{60 kg+15 kg}=4 m/s$

c) 2.8 m

We can use again the law of conservation of energy. The total mechanical energy of the girl after she catches the box is sum of kinetic energy and potential energy:

$E_i = \frac{1}{2}(m_1+m_2) v_3^2 + (m_1+m_2)gh=\frac{1}{2}(75 kg)(4 m/s)^2+(75 kg)(9.8 m/s^2)(2.0m)=2070 J$

While at the maximum height, the speed is zero, so all the mechanical energy is just potential energy:

$E_f = (m_1 +m_2)gh_{max}$

where h_max is the maximum height. Equalizing the two expressions (because the mechanical energy must be conserved) and solving for h_max, we find

$E_i = (m_1+m_2)gh_{max}\\h_{max}=\frac{E_i}{(m_1+m_2)g}=\frac{2070 J}{(75 kg)(9.8 m/s^2)}=2.8 m$

4 0

3 years ago

How much power is needed to lift the 200-N object to a height of 10 m in 4 s?

harkovskaia [24]

Answer:

500 watts

Explanation:

Recall that the definition of power is the amount of energy delivered per unit of time.

In our case, the energy delivered is potential energy which we can estimate as the product of the weight of the object times the distance it is lifted above ground:

200 N x 10 m = 2000 Nm

then the power is the quotient of this potential energy divided the time it took to lift the object to that position:

Power = 2000 / 4 Nm/s = 500 Nm/s = 500 watts

6 0

3 years ago

? Which statement is true of an object in equilibrium?

Degger [83]

The answer is C,<span> The sum of all forces acting on the object is zero. hope that helps!!</span>

7 0

3 years ago