The equation that relates distance, velocities, acceleration, and time is,
d = V₀t + 0.5gt²
where d is distance,
V₀ is the initial velocity,
t is time, and
g is the acceleration due to gravity (equal to 9.8 m/s²)
(1) Dropped rock,
(3 x 10² m ) = 0(t) + 0.5(9.8 m/s²)(t²)
The value of t from this equation is 24.73 s
(2) Thrown rock with V₀ = 26 m/s
(3 x 10² m) = (26)(t) + 0.5(9.8 m/s²)(t²)
The value of t from the equation is 5.61 s
The difference between the tim,
difference = 24.73 s - 5.61 s
difference = 19.12 s
<em>ANSWER: 19.12 s</em>
Answer:
1. 571.43m/s
2. 142.9m and 342.9m
Explanation:
1.Take the difference in time.
1.2-0.7=0.7 seconds
Take the distance between them and divide with differnce in time.
400÷0.7=571.43 seconds.
2.Take the time of the two men and divide by two.
0.5÷2= 0.25 secs
1.2÷2= 0.6 secs
multiply each with the velocity.
0.25×571.43=142.9m
0.6×571.43=342.9m
Given Information:
Resistance = R = 1.9 k
Ω
Capacitance = C = 30 uF
Initial charge = 30 uC
Final charge = 5 uC
Required Information:
Time taken to reduce the capacitor's charge to 5.0 μC = ?
Answer:
t = 0.101 seconds
Explanation:
The voltage across the capacitor is given by
V = V₀e^(–t/τ)
Where τ is time constant, V₀ is the initial voltage, V is the voltage after some time t
τ = RC
τ = 1900*30x10⁻⁶
τ = 0.057 sec
The initial voltage across the capacitor was
V₀ = Q/C
V₀ = 30/30
V₀ = 1 V
Voltage to reduce the charge to 5 uF
V = 5/30
V = 0.167 V
V = V₀e^(–t/τ)
0.167 = 1*e^(–t/0.057)
take ln on both sides
ln(0.167) = ln(e^(–t/0.057))
-1.789 = -t/0.057
t = 1.789*0.057
t = 0.101 seconds