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ohaa [14]
3 years ago
14

A battery charges a parallel-plate capacitor fully and then is removed. The plates are immediately pulled apart. (With the batte

ry disconnected, the amount of charge on the plates remains constant.) What happens to the potential difference between the plates as they are being separated?
Physics
1 answer:
Assoli18 [71]3 years ago
4 0

Answer:

<em>There will be an increase in potential difference.</em>

Explanation:

As we know that the potential difference depends upon the capacitance.

ΔV = Q/C

When battery is disconnected the charge remains constant on the plates but the capacitance decreases. As the capacitance has an inverse relation with the potential difference, there will be an increase in it.

In addition to that the potential difference can also be defined as the product of field and distance between the plates. As the charge is constant so the field is constant. Upon increasing the separation between the plates the potential difference will also increased.

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A bicycle has a momentum of 50.00 kg·m/s and a velocity of 5.0 m/s. What is the bicycle’s mass?
-BARSIC- [3]

Answer:

10 kilograms

Explanation:

The formula for mass

mass (kg)= momentum ( p) divided by velocity (v)

therefore this translates to 50 divided by 5

which gives the final answer as <u>10kgs</u>

7 0
2 years ago
A bag of cement weighing 325 N hangs in equilibrium from three wires. Two of the wires make angles of theta1=60.0 degrees and th
Murljashka [212]
The sketch of the system is: two strings, 1 and 2, are attached to the ceiling and to a third string, 3.The third string holds the bag of cement. 

The free body diagram of the weight with the string 3, drives to the tension T3 = weihgt => T3 = 325 N

The other free body diagram is around the joint of the three strings.

In this case, you can do the horizontal forces equilibrium equation as:

T1* cos(60) - T2*cos(40) = 0

And the vertical forces equilibrium equation:

Ti sin(60) + T2 sin(40) = T3 = 325 N

Then you have two equations with two unknown variables, T1 and T2

0.5 T1 - 0.766 T2 = 0

 0.866 T1 + 0.643T2 = 325

When you solve it you get, T1 = 252.8 N and T2 = 165 N

Answer: T1 = 252.8 N, T2 = 165N, and T3 = 325N



 
7 0
3 years ago
If a sample of a radioactive isotope has a half-life of 1 year, how much of the original sample will be left at the end of the s
Tasya [4]

Answer:

1/4 of the original

Explanation:

That would be TWO half lives:

1/2  * 1/2   = 1/4   <======= 1/4 would be left

4 0
10 months ago
Whats this symbol?<br> ∑
olga2289 [7]
That is the Sigma Symbol. It’s the addition of a sequence of numbers; the result is the sum of the total. if numbers are added sequentially from left to right, any intermediate result is a partial sum, prefix sum, or running total of the summation
4 0
3 years ago
It is known that the kinetics of recrystallization for some alloy obeys the Avrami equation, andthat the value of n in the expon
trapecia [35]

Answer:8.76\times 10^{-3} min^{-1}

Explanation:

Given

n=5

0.3 fraction recrystallize after 100 min

According to Avrami equation

y=1-e^{-kt^n}

where y=fraction Transformed

k=constant

t=time

0.3=1-e^{-k(100)^5}

e^{-k(100)^5} =0.7

Taking log both sides

-k\cdot (10^{10}=\ln 0.7

k=3.566\times 10^{-11}

At this Point we want to compute t_{0.5}\ i.e.\ y=0.5

0.5=1-e^{-kt^n}

0.5=e^{-kt^n}

0.5=e^{-3.566\times 10^{-11}\cdot (t)^5}

taking log both sides

\ln 0.5=-3.566\times 10^{-11}\cdot (t)^5

t^5=1.943\times 10^{10}

t=114.2 min

Rate of Re crystallization at this temperature

t^{-1}=8.76\times 10^{-3} min^{-1}

3 0
3 years ago
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