Answer:
PLEASE BRAINLESS ME INEED
# CARRY ON LEARNING
Acceleration units are in m/s^2 , so you take 25.5 m/s divided by 5.75s. Acceleration is the rate of change of velocity :)
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On connecting a 9V battery to a Capacitor consisting of two circular plates of radius 0.066 m separated by an air gap of 2. 0 mm. The charge on the positive plate is 544.7 ×10⁻¹² C.
Capacitance of the capacitor is determined by the area of the plate of Capacitor and distance between the plates of capacitor.
Let the area of the Capacitor be A , radius of circular plates be r and distance between the plates of capacitor be d.
Given, Voltage, V = 9V
Radius, r = 0.066m
Distance, d = 2mm = 0.002m
Area of the Capacitor, A = πr²
A = π(0.066)²
A = 0.013m²
Capacitance, C = ε₀A / d
C = 8.85×10⁻¹²×0.013/0.002
C = 60.5 ×10⁻¹² F
C = 60.5 pF
We know that Q = CV where Q is the charge on capacitor.
Q = 60.5 ×10⁻¹² × 9
Q= 544.7 ×10⁻¹² C
Since, both plates of a capacitor acquire equal and opposite charge.
Hence the charge on the positive plate of the capacitor is 544.7 ×10⁻¹² C.
Learn more about Capacitance here, brainly.com/question/14746225
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Im pretty sure the answer would be A
Answer:
The velocity of the second car when it passes the first car is 40 m/s
Explanation:
The position and velocity of the cars is given by the following equations:
x = x0 + v0 · t + 1/2 · a · t²
v = v0 + a · t
Where:
x = position at time t
x0 = initial position
v0 = initial velocity
t = time
a = acceleration
v = velocity at time t
If the velocity is constant, then a = 0 and x = x0 + v · t
When the second car passes the first car, the position of both cars is the same:
x first car = x second car
x0 + v · t = x0 + v0 · t + 1/2 · a · t² (x0 = 0 and v0 = 0)
v · t = 1/2 · a · t²
2 · v /a = t
2 · 20 m/s / 2.0 m/s² = t
t = 20 s
Using the equation of velocity, we can calculate the velocity of the second car at t = 20 s
v = v0 + a · t
v = 0 m/s + 2.0 m/s² · 20 s = 40 m/s
The velocity of the second car when it passes the first car is 40 m/s