If it takes 1 minute for 45c of charge to pass a point in an electric circuit. What is the current through the circuit

Is this charging by induction or conduction?

sineoko [7]

Conduction i believe

3 0

3 years ago

You are at the edge of a diving board that is 9 meters above the water. If you weigh 500 Newtons, what is your potential energy?

Semenov [28]

Answer:

4500 J

Explanation:

First, let's define some equations and derivations.

Our potential energy formula is:

$\displaystyle U = mgh$

Where m is mass (in kg), g is the gravitational constant (in m/s²), and h is height (in m).

We also know that mg is equal to the weight of an object (in N), from Newton's 2nd Law of Motion: F = ma (Force is equal to [constant] mass times acceleration).

Therefore, we can simply substitute force into the equation:

$\displaystyle U = Fh$

Where F is the force (in N) and h is still height (in m).

Now we can calculate the amount of potential energy in our system, measured in joules.

Substitute in the given variables, F = 500 N and h = 9 m:

$\displaystyle U = (500 \ N)(9 \ m)$

Using simple Pre-Algebra rules, we find that:

$\displaystyle U = 4500 \ J$

This tells us that the we have 4500 joules of potential energy when I am 9 meters above the water on the edge of the diving board.

6 0

3 years ago

Read 2 more answers

Two particles, one with charge -5.45 × 10^-6 C and one with charge 4.39 × 10^-6 C, are 0.0209 meters apart. What is the magnitud

Levart [38]

Explanation:

Given that,

Charge 1, $q_1=-5.45\times 10^{-6}\ C$

Charge 2, $q_2=4.39\times 10^{-6}\ C$

Distance between charges, r = 0.0209 m

1. The electric force is given by :

$F=k\dfrac{q_1q_2}{r^2}$

$F=9\times 10^9\times \dfrac{-5.45\times 10^{-6}\times 4.39\times 10^{-6}}{(0.0209)^2}$

F = -492.95 N

2. Distance between two identical charges, $r=0.0209\ m$

Electric force is given by :

$F=\dfrac{kq_3^2}{r^2}$

$q_3=\sqrt{\dfrac{Fr^2}{k}}$

$q_3=\sqrt{\dfrac{492.95\times (0.0209)^2}{9\times 10^9}}$

$q_3=4.89\times 10^{-6}\ C$

Hence, this is the required solution.

3 0

3 years ago

B. Assuming the acceleration is still -9.81 m/s2, what is the instantaneous velocity of the

goblinko [34]

We have that the instantaneous velocity of the shuttlecock when it hits the ground is

$V_{int}=\sqrt{U^2+19.6H}$

From the question we are told

Assuming the acceleration is still -9.81 m/s2, what is the instantaneous velocity of the

shuttlecock when it hits the ground? Show your work below.

Generally the equation for acceleration is mathematically given as

$a=v \frac{dv}{dx}\\\\\Therefore\\\\\2ah=v^2-u^2$

Where

acceleration is still -9.81 m/s2,

Hence,

$V^2-U^2=2(-9.81)*-H$

Therefore

$V_{int}=\sqrt{U^2+19.6H}$

For more information on this visit

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7 0

3 years ago

Please help me with this question, especially the angle part. I have solved the initial velocity already. I don't know what to d

Helga [31]

I think you almost got it.

At the top, the velocity only has horizontal component, so v=12 m/s is v_x, which is v*cos(theta), because v_x is constant, so the same when it was launched or now.

With the value of the initial speed (28 m/s, which is the total speed), you can set

v_x = v * cos( theta ) ---> 12 = 28*cos(theta) --> cos(theta)=12/28=3/7

or theta = 64.62 deg, it is D. Think about it. I hope you see it.

5 0

3 years ago