Answer:
Vf = 9.622 m/s
Explanation:
Distance covered S=0.80m, Time t=0.087s and
g=9.81 m/s²
Required: Final velocity Vf when it hit the floor
Sol:
We have first find Vi so
Vi=Vit + 0.5 a t2
Vi =8.768m/s
Now Vf= Vi + at
Vf = 9.622 m/s
Answer:
The speed v of the particle at t=5.00 seconds = 43 m/s
Explanation:
Given :
mass m = 5.00 kg
force f(t) = 6.00t2−4.00t+3.00 N
time t between t=0.00 seconds and t=5.00 seconds
From mathematical expression of Newton's second law;
Force = mass (m) x acceleration (a)
F = ma
...... (1)
acceleration (a)
......(2)
substituting (2) into (1)
Hence, F 



Integrating both sides

The force is acting on the particle between t=0.00 seconds and t=5.00 seconds;
......(3)
Substituting the mass (m) =5.00 kg of the particle, equation of the varying force f(t)=6.00t2−4.00t+3.00 and calculating speed at t = 5.00seconds into (3):







v = 43 meters per second
The speed v of the particle at t=5.00 seconds = 43 m/s
Answer:
Explanation:
Givens
vi = 0
a = 9.81
d = 4.50 m
vf = ?
Formula
vf^2 = vi^2 + 2 * a * d
Solution
Substitute the knowns into the formula
vf^2 =0 + 2 * 9.81 * 4.50
vf^2 = 88.29 Take the square root of both sides.
sqrt(vf^2) = sqrt(88.29)
vf = 9.40 m/s
The answer would be Power