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professor190 [17]
2 years ago
11

A 6- F capacitor is charged to 90 V and is then connected across a 700- resistor. What is the initial charge on the capacitor

Physics
1 answer:
Lorico [155]2 years ago
7 0

Answer:

540C.

Explanation:

A capacitor of capacitance C when charged to a voltage of V will have a charge Q given as follows;

Q = CV          ----------(i)

From the question, the initial charge on the capacitor is the charge on it before it was connected to the resistor. In other words, the initial charge on the capacitor will have a maximum value which can be calculated using equation (i) above.

Where;

C = 6F

V = 90V

Substitute these values into equation (i) as follows;

Q = 6 x 90

Q = 540 C

Therefore, the initial charge on the capacitor is 540C.

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I think in parallel circuits.
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an 11 kg tool box is floating stationary in an orbiting spacecraft when a 79 kg astronaut, initially at rest, gives the tool box
snow_tiger [21]

The astronaut final velocity is 0.06 m/s to the right

Explanation:

In absence of external forces, the total momentum of the box-astronaut system is conserved.

At the beginnig, the total momentum of the two is zero, since they are at rest:

p_i = 0

While the final momentum, after the astronaut throws the box, is:

p_f=mv+MV

where

m = 11 kg is the mass of the box

M = 79 kg is the mass of the astronaut

v = -0.45 m/s (to the left) is the velocity of the box (we take left as negative direction)

V is the final velocity of the astronaut

The total momentum is conserved, so

p_i = p_f\\0=mv+MV

And solving , we find V:

V=-\frac{mv}{M}=-\frac{(11)(-0.45)}{79}=0.06 m/s

And the positive sign indicates that the direction is to the right.

Learn more about momentum:

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6 0
3 years ago
PLEASE HELP FOR BRAINLIEST ANSWER!
Alex787 [66]
I believe the answer is the fourth one, hope this helps
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3 years ago
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Where is a divergent boundary most likely to be found?
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Explanation:

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2 years ago
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The half-life of Iodine-131 is 8.0252 days. If 14.2 grams of I-131 is released in Japan and takes 31.8 days to travel across the
MakcuM [25]

Answer:

Explanation:

Half-life problems are modeled as exponential equations.  The half-life formula is P=P_o\left (\dfrac{1}{2} \right)^{\frac{t}{k}} where P_o is the initial amount, k is the length of the half-life, t is the amount of time that has elapsed since the initial measurement was taken, and P is the amount that remains at time t.

P=14.2\left (\dfrac{1}{2} \right)^{\frac{t}{8.0252}}

<u>Deriving the half-life formula</u>

If one forgets the half-life formula, one can derive an equivalent equation by recalling the basic an exponential equation, y=a b^{t}, where t is still the amount of time, and y is the amount remaining at time t.  The constants a and b can be solved for as follows:

Knowing that amount initially is 14.2g, we let this be time zero:

y=a b^{t}

(14.2)=ab^{(0)}

14.2=a *1

14.2=a

So, a=14.2, which represents out initial amount of the substance, and our equation becomes: y=14.2 b^{t}

Knowing that the "half-life" is 8.0252 days (note that the unit here is "days", so times for all future uses of this equation must be in "days"), we know that the amount remaining after that time will be one-half of what we started with:

\left(\frac{1}{2} *14.2 \right)=14.2 b^{(8.0252)}

\dfrac{7.1}{14.2}=\dfrac{14.2 b^{8.0252}}{14.2}

0.5=b^{8.0252}

\sqrt[8.0252]{\frac{1}{2}}=\sqrt[8.0252]{b^{8.0252}}

\sqrt[8.0252]{\frac{1}{2}}=b

Recalling exponent properties, one could find that  \left ( \frac{1}{2} \right )^{\frac{1}{8.0252}}=b, which will give the equation identical to the half-life formula.  However, recalling this trivia about exponent properties is not necessary to solve this problem.  One can just evaluate the radical in a calculator:

b=0.9172535661...

Using this decimal approximation has advantages (don't have to remember the half-life formula & don't have to remember as many exponent properties), but one minor disadvantage (need to keep more decimal places to reduce rounding error).

So, our general equation derived from the basic exponential function is:

y=14.2* (0.9172535661)^t  or y=14.2*(0.5)^{\frac{t}{8.0252}} where y represents the amount remaining at time t.

<u>Solving for the amount remaining</u>

With the equation set up, substitute the amount of time it takes to cross the Pacific to solve for the amount remaining:

y=14.2* (0.9172535661)^{(31.8)}          y=14.2*(0.5)^{\frac{(31.8)}{8.0252}}

y=14.2* 0.0641450581                    y=14.2*(0.5)^{3.962518068}

y=0.9108598257                              y=14.2* 0.0641450581

                                                        y=0.9108598257

Since both the initial amount of Iodine, and the amount of time were given to 3 significant figures, the amount remaining after 31.8days is 0.911g.

8 0
1 year ago
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