Modify how could you charge the electric circuit shown below to allow lightbulb a to stay lit even if lightbulb b is removed fro
1 answer:
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Answer:
a) v = √(v₀² + 2g h), b) Δt = 2 v₀ / g
Explanation:
For this exercise we will use the mathematical expressions, where the directional towards at is considered positive.
The velocity of each ball is
ball 1. thrown upwards vo is positive
v² = v₀² - 2 g (y-y₀)
in this case the height y is zero and the height i = h
v = √(v₀² + 2g h)
ball 2 thrown down, in this case vo is negative
v = √(v₀² + 2g h)
The times to get to the ground
ball 1
v = v₀ - g t₁
t₁ =
ball 2
v = -v₀ - g t₂
t₂ = - \frac{v_{o} + v }{ g}
From the previous part, we saw that the speeds of the two balls are the same when reaching the ground, so the time difference is
Δt = t₂ -t₁
Δt =
Δt = 2 v₀ / g
Speed is the secular unit, which means it only measures "how fast it goes."
Velocity is a victory unit, which means it measures "both how fast it goes AND the direction."
Acceleration is the change of velocity over time.
Velocity is a victory unit, which means it measures "both how fast it goes AND the direction."
Acceleration is the change of velocity over time.