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3 years ago

What is the power dissipated by a 50.0 ohm resistor?

Physics

1 answer:

Morgarella [4.7K]3 years ago

6 0

Answer:

The power, in watts, dissipated as heat in a resistor varies jointly with the resistance, in ohms, and the square of the current, in amperes. A 15-ohm resistor carrying a current of 1 ampere dissipates 15 watts. How much power is dissipated in a 5-ohm resistor carrying a current of 3 amperes?

Explanation:

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In transistor emitter current is equal to which current?

Paladinen [302]

In transistor,
Emitter current is equal to the sum of base current and collector current.
Thanks!

8 0

3 years ago

Two identical small metal spheres with q1 > 0 and |q1| > |q2| attract each other with a force of magnitude 72.1 mN when se

Brrunno [24]

1) $+2.19\mu C$

The electrostatic force between two charges is given by

$F=k\frac{q_1 q_2}{r^2}$ (1)

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of

$\frac{Q}{2}$

where Q is the total charge between the two spheres.

So we can actually rewrite the force as

$F=k\frac{(\frac{Q}{2})^2}{r^2}$

And since we know that

r = 1.41 m (distance between the spheres)

$F= 21.63 mN = 0.02163 N$

(the sign is positive since the charges repel each other)

We can solve the equation for Q:

$Q=2\sqrt{\frac{Fr^2}{k}}=2\sqrt{\frac{(0.02163)(1.41)^2}{8.98755\cdot 10^9}}}=4.37\cdot 10^{-6} C$

So, the final charge on the sphere on the right is

$\frac{Q}{2}=\frac{4.37\cdot 10^{-6} C}{2}=2.19\cdot 10^{-6}C=+2.19\mu C$

2) $q_1 = +6.70 \mu C$

Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that

$F = -72.1 mN = -0.0721 N$ (we put a negative sign since the force is attractive, which means that the charges have opposite signs)

r = 1.41 m is the separation between the charges

And also,

$q_2 = Q-q_1$

So we can rewrite eq.(1) as

$F=k \frac{q_1 (Q-q_1)}{r^2}$

Solving for q1,

$Fr^2=k (q_1 Q-q_1^2})\\kq_1^2 -kQ q_1 +Fr^2 = 0$

Since $Q=4.37\cdot 10^{-6} C$ , we can substituting all numbers into the equation:

$8.98755\cdot 10^9 q_1^2 -3.93\cdot 10^4 q_1 -0.141 = 0$

which gives two solutions:

$q_1 = 6.70\cdot 10^{-6} C\\q_2 = -2.34\cdot 10^{-6} C$

Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is

$q_1 = +6.70 \mu C$

8 0

4 years ago

A 3200-lb car is moving at 64 ft/s down a 30-degree grade when it runs out of fuel. Find its velocity after that if friction exe

Karolina [17]

Answer:

The velocity is 40 ft/sec.

Explanation:

Given that,

Force = 3200 lb

Angle = 30°

Speed = 64 ft/s

The resistive force with magnitude proportional to the square of the speed,

$F_{r}=kv^2$

Where, k = 1 lb s²/ft²

We need to calculate the velocity

Using balance equation

$F\sin\theta-F_{r}=m\dfrac{d^2v}{dt^2}$

Put the value into the formula

$3200\sin 30-kv^2=m\dfrac{d^2v}{dt^2}$

Put the value of k

$3200\times\dfrac{1}{2}-v^2=m\dfrac{d^2v}{dt^2}$

$1600-v^2=m\dfrac{d^2v}{dt^2}$

At terminal velocity $\dfrac{d^2v}{dt^2}=0$

So, $1600-v^2=0$

$v=\sqrt{1600}$

$v=40\ ft/sec$

Hence, The velocity is 40 ft/sec.

4 0

3 years ago

how far can a mother push a 20.0kg baby carriage, using a force of 62.0N at angle of 30.0°to the horizontal, if she can do 2920

enot [183]

Here is my solution using my app.

4 0

4 years ago

a person standing at the edge of a seaside cliff kicks a stone over the edge with a speed of 18 m/s. The cliff is 52 m above the

Alexxx [7]

You already have the speed, now you need the time.
I will use the formula for speed which is S=D/T.
S=Speed D=Distance T=Time.
So here we have, 18m/s = 52m/T
we do 18 divided by 52 which would be .3461.
.3461 seconds is how long it took the stone to reach the water.

7 0

4 years ago

Read 2 more answers