Answer:
28.5 m/s
18.22 m/s
Explanation:
h = 20 m, R = 20 m, theta = 53 degree
Let the speed of throwing is u and the speed with which it strikes the ground is v.
Horizontal distance, R = horizontal velocity x time
Let t be the time taken
20 = u Cos 53 x t
u t = 20/0.6 = 33.33 ..... (1)
Now use second equation of motion in vertical direction
h = u Sin 53 t - 1/2 g t^2
20 = 33.33 x 0.8 - 4.9 t^2 (ut = 33.33 from equation 1)
t = 1.17 s
Put in equation (1)
u = 33.33 / 1.17 = 28.5 m/s
Let v be the velocity just before striking the ground
vx = u Cos 53 = 28.5 x 0.6 = 17.15 m/s
vy = uSin 53 - 9.8 x 1.17
vy = 28.5 x 0.8 - 16.66
vy = 6.14 m/s
v^2 = vx^2 + vy^2 = 17.15^2 + 6.14^2
v = 18.22 m/s
Answer:
mechanical waves,
.
the quality of a sound governed by the rate of vibrations producing it; the degree of highness or lowness of a tone.
.
If the amplitude increases the volume increases and vice versa.
.
The type of medium affects a sound wave as sound travels with the help of the vibration in particles.
.
The higher the frequency, the shorter the wavelength.
Explanation:
Answer:
W = 47040 J
Explanation:
Given that,
The mass of a student, m = 60 kg
Height of the tower, h = 80 m
We need to find the work done in climbing the tower. The work done is given by :
W = mgh
So,
W = 60 × 9.8 × 80
W = 47040 J
So, the required work done is 47040 J.
It typically take longer for a heavier object to slow down therefor, a train will take more time. <span />
