Answer:
37.5 N Hard
Explanation:
Hook's law: The force applied to an elastic material is directly proportional to the extension provided the elastic limit of the material is not exceeded.
Using the expression for hook's law,
F = ke.............. Equation 1
F = Force of the athlete, k = force constant of the spring, e = extension/compression of the spring.
Given: k = 750 N/m, e = 5.0 cm = 0.05 m
Substitute into equation 1
F = 750(0.05)
F = 37.5 N
Hence the athlete is pushing 37.5 N hard
Answer:
26325 m\s
Explanation:
Data:
v = ?
f = 117 Hz
w = 225
Formula:
v = fw
Solution:
v = ( 117)(225)
v = 26325 m\s <em>A</em><em>n</em><em>s</em><em> </em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>
By looking at the potential energies before and after the reaction, we can tell that the reaction is exothermic (final < initial) or endodermic (final > initial).
Also, the amount of activation energy gives an idea of the external energy required to initiate the reaction (for example, by heating the reactants).
Furthermore, by the same principle, we can also deduce the activation energy for the reverse reaction.
If a catalyst is available, the diagram will show a reduced activation energy, compared to a reaction without catalyst. However, it will also show that the catalyst does not alter the initial and final energies of the reaction.
Answer:
Ff = 839.05 N
Explanation:
We can use the equation:
Ff = μ*N
where <em>N</em> can be obtained as follows:
∑ Fc = m*ac ⇒ N - F = m*ac = m*ω²*R ⇒ N = F + m*ω²*R
then if
F = 32 N
m = 133 Kg
R = 0.635 m
ω = 95 rev /min = (95 rev / min)(2π rad / 1 rev)(1 min / 60 s) = 9.9484 rad /s
we get
N = 32 N + (133 Kg)*(9.9484 rad /s)²*(0.635 m) = 8390.53 N
Finally
Ff = μ*N = 0.10*(8390.53 N) = 839.05 N
