A grapefruit falls from a tree and hits the ground .75 seconds later. How far did the grapefruit drop? What was its speed?
1 answer:
1) The grapefruit is in free fall, so it moves by uniformly accelerated motion, with constant acceleration 
. Calling h its height at t=0, the height at time t is given by
We are told thatn when
the grapefruit hits the ground, so h(0.75 s)=0. If we substitute these data into the equation, we can find the initial height h of the grapefruit:
2) The speed of the grapefruit at time t is given by
where
is the initial speed of the grapefruit. Substituting t=0.75 s, we find the speed when the grapefruit hits the ground:
We are told thatn when
2) The speed of the grapefruit at time t is given by
where
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Answer:
The current in the circuit at a time interval of τ seconds after the switch has been closed is 0.123 A
Explanation:
The time constant for an R and C in series circuit is given by τ = RC.
R = 3000 ohms, C = 0.5 × 10⁻⁶ F = 5.0 × 10⁻⁷ F
τ = 3000 × 5 × 10⁻⁷ = 0.015 s
The voltage across a capacitor as it charges is given be
V(t) = Vs (1 - e⁻ᵏᵗ)
where k = 1/τ
At the point when t = τ, the expassion becomes
V(t = τ) = 1000 (1 - e⁻¹) = 0.632 × 1000 = 632 V
Current flows as a result of potential difference,.
Current in the circuit at this time t = τ is given by
I = (Vs - Vc)/R
Vs = source voltage = 1000 V
Vc = Voltage across the capacitor = 632 V
R = 3000 ohms
I = (1000 - 632)/3000 = 0.123 A
