Weight = (mass) x (acceleration of gravity) .
On Earth, acceleration of gravity is 9.8 m/s² (rounded) .
650 N = (mass) x (9.8 m/s²)
Divide each side by (9.8 m/s²): 650 N / 9.8 m/s² = mass
Mass = 66.3 kilograms (rounded)
<span>When the question says the ball lands a distance of 235 meters from the release point, we can assume this means the horizontal distance is 235 meters.
Let's calculate the time for the ball to fall 235 meters to the ground.
y = (1/2)gt^2
t^2 = 2y / g
t = sqrt{ 2y / g }
t = sqrt{ (2) (235 m) / (9.81 m/s^2) }
t = 6.9217 s
We can use the time t to find the horizontal speed.
v = d / t
v = 235 m / 6.9217 s
v = 33.95 m/s
Since the horizontal speed is the speed of the plane, the speed of the plane is 33.95 m/s</span>
Answer:
The horizontal distance covered by the firework will be 
Explanation:
Let acceleration due to gravity on the planet be g, initial velocity of the firework be u and angle made with the horizontal be ∅.
writing equation of motion in vertical direction:


and 
therefore 
writing equation of motion in horizontal direction:


therefore the equation becomes 
therefore horizontal distance traveled =
Answer:
5.23km/s
Explanation:
Given
Radius of Earth = 6.37 * 10^6 m
Altitude of Satellite = 8200km = 8200 * 10³m = 8.2 * 10^6 m
Gravity Acceleration on Satellite Altitude = 1.87965m/s²
For a satellite to remain in circular orbit, then it means the acceleration of gravity must be exact as the centripetal acceleration.
Centripetal Acceleration = V²/R
So, Acceleration of Gravity (A)= Centripetal Acceleration = V²/R
Make V the subject of formula
A = V²/R
V² = AR
V = √AR
Where R = (radius of earth) + (altitude of satellite)
R = 6.37 * 10^6 + 8.2 * 10^6
R = 14.57 * 10^6m
A = 1.87965m/s²
V = √(1.87965 * 14.57x10^6)
V = √27386500.5
V = 5233.211299001789
V = 5233.2113 m/s ------- Approximated
V = 5.23km/s
22. reduction
25. Le Chatelier's principle