Gary rides his bike down the street in 30 s. His speed was a constant 5 m/s. What is the value of his acceleration during this t
1 answer:
You might be interested in
(a) +2.60 m/s
The motion of the fish dropped by the seagul is a projectile motion, which consists of two independent motions:
- a horizontal uniform motion, at constant speed
- a vertical motion, at constant acceleration (acceleration of gravity, , downward)
In this part we are only interested in the horizontal motion. As we said the horizontal component of the fish's velocity does not change, therefore its value when the fish reaches the ocean is equal to its initial value, which is the speed at which the seagull was flying (because it was flying horizontally):
(b) -17.2 m/s
The vertical component of the fish's velocity instead follows the equation:
where
is the initial vertical velocity, which is zero
is the acceleration of gravity
t is the time
Since the fish reaches the ocean at t = 1.75 s, we can substitute this time into the formula to find the final vertical velocity:
where the negative sign indicates the direction (downward).
(c)
The horizontal component of the fish's velocity would increase
The vertical component of the fish's velocity would stay the same.
As we said from part (a) and (b):
- The horizontal component of the fish's velocity is constant during the motion and it is equal to the initial velocity of the seagull -> so if the seagull's initial speed increases, the horizontal velocity of the fish will increase too
- The vertical component of the fish's velocity does not depend on the original speed of the seagull, therefore it is not affected.
Answer:
a.) τ = 2.85 s b.) Q = 3.19 * 10^-5 C c.) t = 1.691 s
Explanation:
So we are told that it is a RC circuit. We are told = 12.0 V, R = 1.07 MΩ and C = 2.66 µF.
a.) The time constant for RC circuit, τ = RC. Substituting our known values we get:
τ = RC where R = (1.07 * 10 ^ 6)Ω and C = (2.66 * 10 ^ -6) F
τ = (1.07 * 10 ^ 6)Ω * (2.66 * 10 ^ -6) F = 2.8462 s ≈ 2.85 s
τ = 2.85 s
b.) The relationship between capacitance, potential, charge is given:
The capacitor is fully charge when t approaches infinity, therefore:
When t approaches infinity, the term e becomes very small (e^-∞ = 0), therefore we can simplify the equation and plug in our values
Q = 3.19 * 10^-5 C
c.) Using the same equation as before, we can substitute Q in and solve for Q:
t = 1.691 s
Hope this helps! I'm not sure what the units you want, so convert to the desired units.