Answer:
the horizontal distance covered by the cannonball before it hits the ground is 327.5 m
Explanation:
Given;
height of the cliff, h = 210 m
initial horizontal velocity of the cannonball, Ux = 50 m/s
initial vertical velocity of the cannonball, Uy = 0
The time for the cannonball to reach the ground is calculated as;
The horizontal distance covered by the cannonball before it hits the ground is calculated as;
Therefore, the horizontal distance covered by the cannonball before it hits the ground is 327.5 m
Answer:
The starting velocity for ball 1 is 1.00 meter/second. Its ending velocity is 0.25 meter/second.
The change in velocity for ball 1 is 0.25 – 1.00 = -0.75 meter/seconds
Answer:
Explanation:
given
T = 3months = 7.9 × 10⁶s
orbital speed = 88 × 10³m/s
V= 2πr÷T
∴ r = (V×T) ÷ 2π
r = (88km × 7.9 × 10⁶s) ÷ 2π
r = 1.10 × 10⁸km
using kepler's 3rd law
mass of both stars = (seperation diatance)³/(orbital speed)²
M₁ + M₂ = (2r)³/(
year)²
= (1.06 × 10²⁵)/(6.2×10¹³)
1.71×10¹²kg
since M₁ = M₂ =1.71×10¹²kg ÷ 2
M₁ = M₂ = 8.55×10¹¹kg
Acceleration = (change in speed) / (time for the change)
Acceleration = (4 m/s) / (8 seconds)
Acceleration = 0.5 m/s²
Force = (mass) x (acceleration)
Force = (85 kg) x (0.5 m/s²)
Force = 42.5 Newtons
