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24.041m
Answer:
The power decreases by 36%
Explanation:
Given:
At 20° C
Power, P₀ = 300 W
Potential difference, V = 150 volts
Now, power is given as
P = V²/R
where, R is the resistance
on substituting the values, we get
300 = 150²/R₀
or
R₀ = 75 Ω
Now, the variation of resistance with temperature is given as
R = R₀[1 + α(T - T₀)]
where, α is the temperature coefficient of resistivity = 0.0003125 (°C⁻¹)
now, at
T₀ = 20° C
R₀ = 75 Ω
for
T = 1820° C
we have
R = R₀[1 + α(T - T₀)]
substituting the values
we get
R = 75×[1 + 0.0003125 × (1820 - 20)]
or
R = 117.18 Ω
Now using the formula for power
We have,
P = V²/R
or
P = 150²/117.18 = 192 W
Therefore, the percentage change will be
= 
on substituting the values , we get
= 
= -36%
here, negative sign depicts the decrease in power
The net displacement at a point on the string where the pulses cross is 0.2 m.
The term "displacement" refers to a shift in an object's position. It has a magnitude and a direction, making it a vector quantity. An arrow pointing from the starting point to the finishing point serves as its symbol.
A string that is connected to a post at one end is used to transmit a sequence of pulses, each measuring 0.1 meters in amplitude.
At the post, the pulses are reflected and return along the string without losing any of their amplitude.
Now, let's say the ends are free.
There is no inversion on reflection if the end is free. The amplitude at their intersection is 2A.
Now, since A = 0.1 m
Then, 2A = 2(0.1) = 0.2 m
As a result, the net displacement at the string's intersection of two pulses is 0.2 m.
The correct option is (c).
Learn more about amplitude here:
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Potential equals kenecric at the bottom so potential would also increas
Answer:
A. -2.16 * 10^(-5) N
B. 9 * 10^(-7) N
Explanation:
Parameters given:
Distance between their centres, r = 0.3 m
Charge in first sphere, Q1 = 12 * 10^(-9) C
Charge in second sphere, Q2 = -18 * 10^(-9) C
A. Electrostatic force exerted on one sphere by the other is:
F = (k * Q1 * Q2) / r²
F = (9 * 10^9 * 12 * 10^(-9) * -18 * 10^(-9)) / 0.3²
F = -2.16 * 10^(-5) N
B. When they are brought in contact by a wire and are then in equilibrium, it means they have the same final charge. That means if we add the charges of both spheres and divided by two, we'll have the final charge of each sphere:
Q1 + Q2 = 12 * 10^(-9) + (-18 * 10^(-9))
= - 6 * 10^(-9) C
Dividing by two, we have that each sphere has a charge of -3 * 10^(-9) C
Hence the electrostatic force between them is:
F = [9 * 10^9 * (-3 * 10^(-9)) * (-3 * 10^(-9)] / 0.3²
F = 9 * 10^(-7) N
