voltage across 2.0μf capacitor is 5.32v
Given:
C1=2.0μf
C2=4.0μf
since two capacitors are in series there equivalent capacitance will be
[tex] \frac{1}{c} = \frac{1}{c1} + \frac{1}{c2} [/tex]
=1.33μf
As the capacitance of a capacitor is equal to the ratio of the stored charge to the potential difference across its plates, giving: C = Q/V, thus V = Q/C as Q is constant across all series connected capacitors, therefore the individual voltage drops across each capacitor is determined by its its capacitance value.
Q=CV
given,V=8v
charge on 2.0μf capacitor is
=5.32v
learn more about series capacitance from here: brainly.com/question/28166078
#SPJ4
I believe it would be D a change in direction of motion
Answer:
0.301 m
Explanation:
Torque = Force × Radius
τ = Fr
40.0 Nm = 133 N × r
r = 0.301 m
The mechanic must apply the force 0.301 m from the nut.
Answer:
12.5 m/s
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 0 m/s
Height (h) = 8 m
Final velocity (v) at 8 m above the lowest point =?
NOTE: Acceleration due to gravity (g) = 9.8 m/s²
The velocity of the roller coaster at 8 m above the lowest point can be obtained as follow:
v² = u² + 2gh
v² = 0² + (2 × 9.8 × 8)
v² = 0 + 156.8
v² = 156.8
Take the square root of both side
v = √156.8
v = 12.5 m/s
Therefore, the velocity of the roller coaster at 8 m above the lowest point is 12.5 m/s.
