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natita [175]
3 years ago
9

A Capacitor is a circuit component that stores energy and can be charged when current flows through it. A current of 3A flows th

rough a capacitor that has an initial charge of 2μC (micro Coulombs). After two microseconds, how much is the magnitude of the net electric charge (in μC) of the capacitor?
Physics
1 answer:
ddd [48]3 years ago
4 0

Answer:

8\mu C

Explanation:

t = Time taken = 2\mu s

i = Current = 3 A

q(0) = Initial charge = 2\mu C

Charge is given by

q=\int_0^t idt+q(0)\\\Rightarrow q=\int_0^{2\mu s} 3dt+2\mu C\\\Rightarrow q=3(2\mu s-0)+2\mu C\\\Rightarrow q=6\mu C+2\mu C\\\Rightarrow q=8\mu C

The magnitude of the net electric charge of the capacitor is 8\mu C

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8 0
3 years ago
Un vagón de carga de 5000-kg se mueve a 2m/s e impacta a un vagón de 10000-kg que se encuentra en reposo. Después de la colisión
Mariulka [41]

Answer:

0.67 m/s

Explanation:

Mass of car 1, m₁ = 5000 kg

Mass of car 2, m₂ = 10,000 kg

Initial speed of car 1, u₁ = 2 m/s

Final speed of car 2, u₂ = 0 (at rest)

We need to find the final velocity of both cars when inelastic collision occurs. The momentum will remain conserved in case of inelastic collision. Using the conservation of momentum. Let V is the final speed.

m_1u_1+m_2u_2=(m_1+m_2)V\\\\\V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\V=\dfrac{m_1u_1}{(m_1+m_2)}\\\\V=\dfrac{5000\times 2}{(5000+10000)}\\\\V=0.67\ m/s

So, after the inelastic collision, they will move with a speed of 0.67 m/s.

6 0
3 years ago
A generator converts blank energy into blank energy
jok3333 [9.3K]

Answer: Mechanical Energy and Electrical Energy

Explanation: A generator converts mechanical energy into electrical energy, while a motor does the opposite - it converts electrical energy into mechanical energy.

6 0
3 years ago
A mass of 0.4 kg hangs motionless from a vertical spring whose length is 0.95 m and whose unstretched length is 0.65 m. Next the
nexus9112 [7]

Answer:

Explanation:

spring constant of spring = mg / x

= .4 x 9.8 / ( .95 - .65 )

=13.07 N / m

energy stored in spring = 1/2 k x²

= .5 x 13.07 x ( 1.2 - .65 )²

= 1.976 J

Let it goes x m beyond its equilibrium position

Total energy at initial point

= 1.976 + 1/2 m v²

= 1.976 + .5 x .4 x 1.6²

= 2.488 J

energy at final point

= mgh + 1/2 k x²

.4 x 9.8 x  ( .55 + x ) + .5 x 13.07 x² = 2.488

6.535 x² + 2.156 + 3.92 x = 2.488

6.535 x² + 3.92 x - .332 = 0

x = .075 m

7.5 cm

4 0
3 years ago
Two asteroids collide and stick together. The first asteroid has a mass of 15\times 10^3\,\mathrm{kg}15×10 3 kg and is initially
statuscvo [17]

Answer:

Final speed is 900.06 m/s at 0.2215^{\circ}  

Solution:

As per the question:

Mass of the first asteroid, m = 15\times 10^{3}\kg

Mass of the second asteroid, m' = 20\times 10^{3}\kg

Initial velocity of the first asteroid, v = 770 m/s

Initial velocity of the second asteroid, v' = 1020 m/s

Angle between the two initial velocities, \theta = 20^{\circ}

Now,

Since, the velocities and hence momentum are vector quantities, then by the triangle law of vector addition of 2 vectors A and B, the resultant is given by:

\vec{R} = \sqrt{A^{2} + 2ABcos\theta + B^{2}}

Thus applying vector addition and momentum conservation, the final velocity is given by:

(m + m')v_{final} = \sqrt{(mv)^{2} + 2(mv)(m'v')cos20^{\circ} + (m'v')^{2}}                               (1)

Now,

(m +m')v_{final} = (35\times 10^{3})v_{final}

(mv)^{2} = (15\times 10^{3}\times 770)^{2} = 1.334\times 10^{14}

(m'v')^{2} = (20\times 10^{3}\times 1020)^{2} = 4.16\times 10^{14}

2(mv)(m'v')cos20^{\circ} = 2(15\times 10^{3}\times 770)(20\times 10^{3}\times 1020)cos20^{\circ} = 4.43\times 10^{14}

Now, substituting the suitable values in eqn (1), we get:

v_{final} = 900.06\ m/s

Now, the direction for the two vectors is given by:

\theta = sin^{- 1} \frac{m'v'sin20^{\circ}}{(m + m')v_{final}}

\theta = sin^{- 1} \frac{20\times 10^{3}\times 1020sin20^{\circ}}{(35\times 10^{3})\times 900.06} = 0.2215^{\circ}

5 0
3 years ago
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