It will be 3 wavelengths because 1 cycle = 1 wavelength.
Answer:
E = 16.464 J
Explanation:
Given that,
Mass of tetherball, m = 0.8 kg
It is hit by a child and rises 2.1 m above the ground, h = 21. m
We need to find the maximum gravitational potential energy of the ball. The formula for the gravitational potential energy is given by :
E = mgh
g is acceleration due to gravity
E = 0.8 kg × 9.8 m/s² × 2.1 m
= 16.464 J
So, the maximum potential energy of the ball is 16.464 J.
Answer:
H = 1/2 g t^2 where t is time to fall a height H
H = 1/8 g T^2 where T is total time in air (2 t = T)
R = V T cos θ horizontal range
3/4 g T^2 = V T cos θ 6 H = R given in problem
cos θ = 3 g T / (4 V) (I)
Now t = V sin θ / g time for projectile to fall from max height
T = 2 V sin θ / g
T / V = 2 sin θ / g
cos θ = 3 g / 4 (T / V) from (I)
cos θ = 3 g / 4 * 2 sin V / g = 6 / 4 sin θ
tan θ = 2/3
θ = 33.7 deg
As a check- let V = 100 m/s
Vx = 100 cos 33.7 = 83,2
Vy = 100 sin 33,7 = 55.5
T = 2 * 55.5 / 9.8 = 11.3 sec
H = 1/2 * 9.8 * (11.3 / 2)^2 = 156
R = 83.2 * 11.3 = 932
R / H = 932 / 156 = 5.97 6 within rounding
Answer
Integral EdA = Q/εo =C*Vc(t)/εo = 3.5e-12*21/εo = 4.74 V∙m <----- A)
Vc(t) = 21(1-e^-t/RC) because an uncharged capacitor is modeled as a short.
ic(t) = (21/120)e^-t/RC -----> ic(0) = 21/120 = 0.175A <----- B)
Q(0.5ns) = CVc(0.5ns) = 2e-12*21*(1-e^-t/RC) = 30.7pC
30.7pC/εo = 3.47 V∙m <----- C)
ic(0.5ns) = 29.7ma <----- D)
In physical science, there are two types of quantity: scalar and vector. While scalar quantities only include the magnitude, vector quantities include both the magnitude and the direction. Displacement is an example of vector quantities. Therefore, it includes magnitude and direction.
