Answer:
the expected distance is 4.32 m
Explanation:
given data
half life time = 1.8 × s
speed = 0.8 c = 0.8 × 3 ×
to find out
expected distance over
solution
we know c is speed of light in air is 3 × m/s
we calculate expected distance by given formula that is
expected distance = half life time × speed .........1
put here all these value
expected distance = half life time × speed
expected distance = 1.8 × × 0.8 × 3 ×
expected distance = 4.32
so the expected distance is 4.32 m
Answer:
Explanation:
It is a question relating to charging of capacitor . For charging of capacitor , the formula is as follows.
Q = CV ( 1 - )
λ = 1/CR , C is capacitance and R is resistance.
= 1/(500 x 10⁻⁶ x 20 x 10³ )
= .1
λ t = .1 x 20
λ t = 2
CV = 500 X 10⁻⁶ X 5
= 2500 X 10⁻⁶ C
Q = 2500 x 10⁻⁶ ( 1 - )
= 2500 x 10⁻⁶ x .86566
= 2161.66 μ C .
voltage = Charge / capacitor
2161.66 μ C / 500μ F
= 4.32 V
Answer:
find the diagram in the attachment.
Explanation:
Let vi = 12 m/s be the intial velocy when the ball is thrown, Δy be the displacement of the ball to a point where it starts returning down, g = 9.8 m/s^2 be the balls acceleration due to gravity.
considering the motion when the ball thrown straight up, we know that the ball will come to a stop and return downwards, so:
(vf)^2 = (vi)^2 + 2×g×Δy
vf = 0 m/s, at the highest point in the upward motion, then:
0 = (vi)^2 + 2×g×Δy
-(vi)^2 = 2×g×Δy
Δy = [-(vi)^2]/2×g
Δy = [-(-12)^2]/(2×9.8)
Δy = - 7.35 m
then from the highest point in the straight up motion, the ball will go back down and attain the speed of 12 m/s at the same level as it was first thrown
