Answer:
When the ball goes to first base it will be 4.23 m high.
Explanation:
Horizontal velocity = 30 cos17.3 = 28.64 m/s
Horizontal displacement = 40.5 m
Time
Time to reach the goal posts 40.5 m away = 1.41 seconds
Vertical velocity = 30 sin17.3 = 8.92 m/s
Time to reach the goal posts 40.5 m away = 1.41 seconds
Acceleration = -9.81m/s²
Substituting in s = ut + 0.5at²
s = 8.92 x 1.41 - 0.5 x 9.81 x 1.41²= 2.83 m
Height of throw = 1.4 m
Height traveled by ball = 2.83 m
Total height = 2.83 + 1.4 = 4.23 m
When the ball goes to first base it will be 4.23 m high.
Answer:
c. 1600J
Explanation:
The loss in potential energy of the boy is given by:

where
m = 40 kg is the mass of the boy
g = 9.8 m/s^2 is the acceleration of gravity
is the total change in the height of the boy (4 metres + 2 cm due to the compression of the spring)
Substituting, we find

Answer:
i/f = i/o + i/i f = focal, o = object, i = image
1 / i = 1 / f - 1 / o = (o - f) / o f
i = o * f / ( o - f) image distance
i = 12.5 * 22 / (12.5 - 22) = -28.9 cm
Image is real
Image is 28.9 cm to left of lens
M = - i / o = = 28.9 / 12.5 = 2.3 magnification (convex lens)
-30 I think sorry if I’m wrong