Explanation:
As per the problem,
When q > 0 then -q is a negative charge . Since, change in potential energy (
) increases.
or, 
> 0
or, 
Therefore, both positive and negative charge will move from 
to 
and as 
so both of them move through a negative potential difference.
Thus, we can conclude that the true statements are as follows.
- The positively charged object moves through a negative potential difference between A and B (that is, VB - VA < 0).
- The negatively charged object moves through a negative potential difference between A and B (that is, VB - VA < 0).
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
Answer:
(a), (c) and (e) s correct.
Explanation:
a. the power used by a circuit is the resistance times the current squared.
The power is given by P = I^2 R, so the statement is correct.
b. electric and magnetic fields are transporting the energy.
false
c. electrons are transporting the energy.
The energy is transferred by flow of electrons. It is correct.
d. the power used by a circuit is the voltage times the current squared.
The power is given by P = V I, the statement is wrong.
e. the power used by a circuit is the current times the voltage.
The power is given by P = V I, the statement is correct.
Answer:
3.64×10⁸ m
3.34×10⁻³ m/s²
Explanation:
Let's define some variables:
M₁ = mass of the Earth
r₁ = r = distance from the Earth's center
M₂ = mass of the moon
r₂ = d − r = distance from the moon's center
d = distance between the Earth and the moon
When the gravitational fields become equal:
GM₁m / r₁² = GM₂m / r₂²
M₁ / r₁² = M₂ / r₂²
M₁ / r² = M₂ / (d − r)²
M₁ / r² = M₂ / (d² − 2dr + r²)
M₁ (d² − 2dr + r²) = M₂ r²
M₁d² − 2dM₁ r + M₁ r² = M₂ r²
M₁d² − 2dM₁ r + (M₁ − M₂) r² = 0
d² − 2d r + (1 − M₂/M₁) r² = 0
Solving with quadratic formula:
r = [ 2d ± √(4d² − 4 (1 − M₂/M₁) d²) ] / 2 (1 − M₂/M₁)
r = [ 2d ± 2d√(1 − (1 − M₂/M₁)) ] / 2 (1 − M₂/M₁)
r = [ 2d ± 2d√(1 − 1 + M₂/M₁) ] / 2 (1 − M₂/M₁)
r = [ 2d ± 2d√(M₂/M₁) ] / 2 (1 − M₂/M₁)
When we plug in the values, we get:
r = 3.64×10⁸ m
If the moon wasn't there, the acceleration due to Earth's gravity would be:
g = GM / r²
g = (6.672×10⁻¹¹ N m²/kg²) (5.98×10²⁴ kg) / (3.64×10⁸ m)²
g = 3.34×10⁻³ m/s²
D. The direction of the force is toward the center of the object's circular path.
