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:
P = 75 W
Explanation:
given,
Distance, L = 8 m
Force,F = 150 N
Time, t = 16 s
Work by the climber
Work done = Force x displacement
W = F. L
W = 150 x 8
W = 1200 J
We know,
P = 75 W
Hence, Power climber is using to climb is equal to 75 W.
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It <span>Ions were once atoms with the same number of electrons and protons. Since they have opposite charges atoms are neutral. When they become ions the lose or gain electrons and become unbalanced. ... These different charges are attracted to each other via electric forces.</span>
