Answer:
Q = 4 Q₀
Explanation:
This is an exercise on capacitors, where the capacitance is
C = 
if we apply the given conditions
C = \epsilon_{o} \ \frac{2A}{0.5d}
C = 4 \epsilon_{o} \ \frac{A}{d}
let's call the capacitance Co with the initial values
C₀ = \epsilon_{o} \ \frac{A}{d}
C = 4 C₀
The charge on each plate of a capacitor is
Q = C ΔV
If the potential difference is maintained, the new charge is
Q = 4 C₀ ΔV
let's call
Q₀ = C₀ ΔV
we substitute
Q = 4 Q₀
A particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. Find the velocity and the speed when t = 4. f(t) = 12t² + 35 t + 1
Answer:
Velocity = 131 m/s
Speed = 131 m/s
Explanation:
Equation of motion, s = f(t) = 12t² + 35 t + 1
To get velocity of the particle, let us find the first derivative of s
v (t) = ds/dt = 24t + 35
At t = 4
v(4) = 24(4) + 35
v(4) = 131 m/s
Speed is the magnitude of velocity. Since the velocity is already positive, speed is also 131 m/s
Answer:
Explanation:
<u>Given:</u>
Force = f = 60 N
Mass = m = 12 kg
<u>Required:</u>
Acceleration = a = ?
<u>Formula:</u>
F = ma
<u>Solution:</u>
Rearranging formula
a = F / m
a = 60 / 12
a = 5 ms⁻²
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Hope this helped!
<h3>~AH1807</h3><h3>Peace!</h3>
Answer:
Explanation:
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