Answer:
94.13 ft/s
Explanation:
<u>Given:</u>
= time interval in which the rock hits the opponent = 10 s - 5 s = 5 s
= distance to be moved by the rock long the horizontal = 98 yards
= displacement to be moved by the rock during the time of flight along the vertical = 0 yard
<u>Assume:</u>
= magnitude of initial velocity of the rock
= angle of the initial velocity with the horizontal.
For the motion of the rock along the vertical during the time of flight, the rock has a constant acceleration in the vertically downward direction.
Now the rock has zero acceleration along the horizontal. This means it has a constant velocity along the horizontal during the time of flight.
On dividing equation (1) by (2), we have
Now, putting this value in equation (2), we have
Hence, the initial velocity of the rock must a magnitude of 94.13 ft/s to hit the opponent exactly at 98 yards.
Theoretically, 35 x 18 = 630
The one that will change the velocity of a periodic wave is :
B. Changing the medium of the wave
Waves is always determined by the properties of the medium, which means that changing the medium will change the velocity of the wave
hope this helps
Answer:
0.00016 kg
Explanation:
Given:
Power = P = 1.2 × 10⁹ Watts
Power = work done / Time
efficiency = 0.30
Input power = 1.2 × 10⁹ / 0.30 = 4 × 10⁹ W
Energy = 4 × 10⁹ x 60 x 60 = 1.44 x 10¹³ joules
E = m c² , where c is the speed of light and m is the mass.
⇒ mass = m = E / c² = (1.44 x 10¹³) / (3 × 10⁸ )²
= 0.00016 kg
