Answer:
(A) We are using them faster than they are replenished by nature
Answer:
(a) 5.7 s
(b) 39 m/s
Explanation:
(a) u = 18 m/s
At the maximum height, the final velocity of ball is zero. lte teh time taken by the ball to go from 50 m height to maximum height is t.
use first equation of motion.
v = u + g t
0 = 18 - 10 x t
t = 1.8 s
Let the maximum height attained by the ball when it thrown from 50 m height is h'.
Use third equation of motion
v^2 = u^2 + 2 g h'
0 = 18^2 - 2 x 10 x h'
h' = 16.2 m
Total height from the ground H = h + h' = 50 + 16.2 = 76.2 m
Let t' be the time taken by the ball to hit the ground as it falls from maximum height.
use third equation of motion
H = ut + 1/2 x g t'^2
76.2 = 0 + 1/2 x 10 x t'^2
t' = 3.9 s
Total time taken by the ball to hit the ground = T = t + t' = 1.8 + 3.9 = 5.7 s
(b) Let v be the velocity with which the ball strikes the ground.
v^2 = u^2 + 2 g H
v^2 = 0 + 2 x 10 x 76.2
v = 39 m/s
Answer:
The mass of object is calculated as 5.36 kg
Explanation:
The known terms to find the mass are:
acceleration of object (a) = 22.35 
Force exerted (F) = 120N
mass of an object (m) = ?
From Newton's second law of motion;
F = ma
or, 120 = m × 22.35
or, m= 
kg
∴ m = 5.36 kg
We can solve for the resultant x and y components by using
the sine and cosine functions.
resultant x = 2.5 cos 35 + 5.2 cos 22 = 6.87 km
resultant y = 2.5 sin 35 + 5.2 sin 22 = 3.38 km
The resultant displacement is calculated using hypotenuse
equation:
displacement = sqrt (6.87^2 + 3.38^2)
displacement = 7.66 km
The resultant angle is:
θ = tan^-1 (3.38 / 6.87)
θ = 26.20°
Therefore the magnitude and direction is:
7.66 km, 26.20° to the ground
Answer:
he tail of the arrow moves a distance of 0.5 m as the arrow is shot. yare yare daze
Explanation:
