Answer: 60m/s
Explanation:
From the diagram:
Θ = 30°
Vertical resolution (y-axis) :
Voy = VoSinΘ
g in the upward direction = negative (-) = - g
Vfinal = 0
Distance (H) traveled along y =
Time taken to reach maximum height :
From v = u + at
0 = usinΘ - gt
gt = usinΘ
t = usinΘ / g
Horizontal resolution:
S = ut + 1/2at^2
Substituting t = usinΘ / g ; Voy = usinΘ
S = (usinΘ × usinΘ / g) - 1/2 g × (usinΘ /g)^2
S = (u^2sin^2Θ / g) - (u^2sin^2Θ / 2g)
S = (u^2sin^2Θ) / 2g
Now if S = maximum height = 45m
Then,
45 = [Vo^2sin^2(30°)] / 2(10)
45 =[ Vo^2 * (0.5)^2] / 20
45 =( Vo^2 * 0.25) / 20
20 * 45 = Vo^2 * 0.25
900 / 0.25 = Vo^2
3600 = Vo^2
Vo = sqrt(3600)
Vo = 60m/s
Answer:
Yes
Explanation:
An object can be moving (have kinetic energy) and be elevated above the ground at the same time (and also have potential energy).
a = ( V2 - V1)/( t2 - t1)
3.2 = ( 23.5m/s - 15.2m/s)/(t - 0)
3.2m/s = 8.3/t
t(3.2) = 8.3
t = 8.3/3.2
t = 2.59 seconds
Answer:
2
Explanation:
We know that in the Fraunhofer single-slit pattern,
maxima is given by
Given values
θ=2.12°
slit width a= 0.110 mm.
wavelength λ= 582 nm
Now plugging values to calculate N we get
Solving the above equation we get
we N= 2.313≅ 2
B dropping a ball
C tentative and testable
