Answer:
it will take 36.12 ms to reduce the capacitor's charge to 10 μC
Explanation:
Qi= C×V
then:
Vi = Q/C = 30μ/20μ = 1.5 volts
and:
Vf = Q/C = 10μ/20μ = 0.5 volts
then:
v = v₀e^(–t/τ)
v₀ is the initial voltage on the cap
v is the voltage after time t
R is resistance in ohms,
C is capacitance in farads
t is time in seconds
RC = τ = time constant
τ = 20µ x 1.5k = 30 ms
v = v₀e^(t/τ)
0.5 = 1.5e^(t/30ms)
e^(t/30ms) = 10/3
t/30ms = 1.20397
t = (30ms)(1.20397) = 36.12 ms
Therefore, it will take 36.12 ms to reduce the capacitor's charge to 10 μC.
'In transverse waves, the particles of the medium move perpendicular to the direction of the flow of energy' is true for transverse waves only.
'In longitudinal waves, the particles of the medium move parallel to the direction of the flow of energy' is true for longitudinal waves only.
'Many wave motions in nature are a combination of longitudinal and transverse motion' is true for both longitudinal and transverse waves.
<u>Explanation:</u>
Longitudinal waves are those where the direction of propagation of particles are parallel to the medium' particles. While transverse waves propagate perpendicular to the medium' particles.
As wave motions are assumed to be of standing waves which comprises of particles moving parallel as well as perpendicular to the medium, most of the wave motions are composed of longitudinal and transverse motion.
So the option stating the medium' particle moves perpendicular to the direction of the energy flow is true for transverse waves. Similarly, the option stating the medium' particle moves parallel to the direction of flow of energy is true for longitudinal waves only.
And the option stating that wave motions comprises of combination of longitudinal and transverse motion is true for both of them.
Answer:
10 seconds
Explanation:
x = x₀ + v₀ t + ½ at²
250 = 0 + (0) t + ½ (5) t²
250 = 2.5 t²
t² = 100
t = 10
It takes 10 seconds to land from a height of 250 ft.
