A) Starting from rest, we have the entire cycle determined by 0.8s.
If we assume a constant movement, half of that time is when it reaches the highest point, that is, in 0.4s.
The distance as a function of speed and acceleration is given by,
At the initial point the speed is zero and the acceleration is equivalent to gravity.
B) When returning to the ground, the final speed is zero. Therefore, the equation that relates velocity to acceleration is given by,
I think the answer is no. humans consume more than the earth can sustain.
Ω₀ = the initial angular velocity (from rest)
t = 0.9 s, time for a revolution
θ = 2π rad, the angular distance traveled
Let
α = the angular acceleration
ω = the final angular velocity
The angular rotation obeys the equation
(1/2)*(α rad/s²)*(0.9 s)² = (2π rad)
α = 15.514 rad/s²
The final angular velocity is
ω = (15.514 rad/s²)*(0.9 s) = 13.963 rad/s
If the thrower's arm is r meters long, the tangential velocity of release will be
v = 13.963r m/s
Answer: 13.963 rad/s
Answer:
SI unit of k (spring constant) = N/m
Explanation:
We have expression for force in a spring extended by x m given by
F = kx
Where k is the spring constant value.
Taking units on both sides
Unit of F = Unit of k x Unit of x
N = Unit of k x m
Unit of k = N/m
SI unit of k (spring constant) = N/m
Answer:
2.605m
Explanation:
Using the formula for calculating Range (distance travelled in horizontal direction)
Range R = U√2H/g
U is the speed = 4.8m/s
H is the maximum height = ?
g is the acc due to gravity = 9.8m/s²
R = 3.5m
Substitute into the formula and get H
3.5 = 4.8√2H/9.8
3.5/4.8 = √2H/9.8
0.7292 = √2H/9.8
square both sides
0.7292² = 2H/9.8
2H = 0.7292² * 9.8
2H = 5.21
H = 5.21/2
H = 2.605m
Hence the height of the ball from the ground is 2.605m
