(a) The maximum potential difference across the resistor is 339.41 V.
(b) The maximum current through the resistor is 0.23 A.
(c) The rms current through the resistor is 0.16 A.
(d) The average power dissipated by the resistor is 38.4 W.
<h3>Maximum potential difference</h3>
Vrms = 0.7071V₀
where;
V₀ = Vrms/0.7071
V₀ = 240/0.7071
V₀ = 339.41 V
<h3> rms current through the resistor </h3>
I(rms) = V(rms)/R
I(rms) = (240)/(1,540)
I(rms) = 0.16 A
<h3>maximum current through the resistor </h3>
I₀ = I(rms)/0.7071
I₀ = (0.16)/0.7071
I₀ = 0.23 A
<h3> Average power dissipated by the resistor</h3>
P = I(rms) x V(rms)
P = 0.16 x 240
P = 38.4 W
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Answer:
0.687 m/s
Explanation:
Initial energy = final energy
1/2 mu² = mgh + 1/2 mv²
1/2 u² = gh + 1/2 v²
Given u = 2.00 m/s, g = 9.8 m/s², and h = 0.180 m:
1/2 (2.00 m/s)² = (9.8 m/s²) (0.180 m) + 1/2 v²
v = 0.687 m/s
Answer: The initial force is reduced a factor 1/4 when the separation between charge is doubled
Explanation: As it well known the electric force between two charges is given by:
Finitial=k*q1*q2/d^2 where d is the distance between charges and k is a constant
if the distance is doubled this means 2*dinitial thus the new force is equal to F initial* 1/4
Answer:
Hello your question is incomplete below is the complete question
Calculate Earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the sun, Take the eccentricity of Earth's orbit to be 1/60 and its Semimajor axis to be 93,000,000
answer : V = 1.624* 10^-5 m/s
Explanation:
First we have to calculate the value of a
a = 93 * 10^6 mile/m * 1609.344 m
= 149.668 * 10^8 m
next we will express the distance between the earth and the sun
--------- (1)
a = 149.668 * 10^8
E (eccentricity ) = ( 1/60 )^2
= 90°
input the given values into equation 1 above
r = 149.626 * 10^9 m
next calculate the Earths velocity of approach towards the sun using this equation
------ (2)
Note :
Rc = 149.626 * 10^9 m
equation 2 becomes
(
therefore : V = 1.624* 10^-5 m/s
(a) The distance of the image formed by the concave mirror is 19.1 cm.
(b) The image formed is diminished and real.
<h3>
Image distance </h3>
The distance of the image formed by the concave mirror is calculated as follows;
1/f = 1/v + 1/u
1/v = 1/f - 1/u
1/v = 1/15 - 1/70
1/v = 0.05238
v = 1/0.05238
v = 19.1 cm
The image distance is smaller than object distance, thus the image formed is diminished and real.
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