Answer:
h = v₀² / 2g
, h = k/4g x²
Explanation:
In this exercise we can use the law of conservation of energy at two points, the lowest, before the shot and the highest point that the mouse reaches
Starting point. Lower compressed spring
Em₀ = K = ½ m v²
Final point. Highest on the path
= U = mg h
As or no friction the energy is conserved
Em₀ = Em_{f}
½ m v₀²² = m g h
h = v₀² / 2g
We can also use as initial energy the energy stored in the spring that will later be transferred to the mouse
½ k x² = 2 g h
h = k/4g x²
In order to answer this exercise you need to use the formulas
S = Vo*t + (1/2)*a*t^2
Vf = Vo + at
The data will be given as
Vf = final velocity = ?
Vo = initial velocity = 1.4 m/s
a = acceleration = 0.20 m/s^2
s = displacement = 100m
And now you do the following:
100 = 1.4t + (1/2)*0.2*t^2
t = 25.388s
and
Vf = 1.4 + 0.2(25.388)
Vf = 6.5 m/s
So the answer you are looking for is 6.5 m/s
Centripetal Force (Fcp) = ?
His arm length = Radius (R) = 0.75 m
Discus velocity = Linear Velocity (V) = 5 m/s
Discus mass (m) = 2 kg
Centripetal Acceleration (Acp) = V^2/R or W^2 x R
In this case i will use the V^2/R formula, because it uses the discus velocity (V).
Answer: Last option, 66 N.
Answer:
Explanation:
Given data
velocity v₀=20 cm/s at time t=3s
velocity vf=0 at time t=8 s
To find
Average Acceleration at time=3s to 8s
Solution
As we know that acceleration is first derivative of velocity with respect to time
