Answer:
Magnitude of the net force acting on the kayak = 39.61 N
Explanation:
Considering motion of kayak:-
Initial velocity, u = 0 m/s
Distance , s = 0.40 m
Final velocity, v = 0.65 m/s
We have equation of motion v² = u² + 2as
Substituting
v² = u² + 2as
0.65² = 0² + 2 x a x 0.4
a = 0.53 m/s²
We have force, F = ma
Mass, m = 75 kg
F = ma = 75 x 0.53 = 39.61 N
Magnitude of the net force acting on the kayak = 39.61 N
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Medium stars become dwarf stars, large stars become either neutron (pulsar) stars or black holes
a. 0.5 T
- The amplitude A of a simple harmonic motion is the maximum displacement of the system with respect to the equilibrium position
- The period T is the time the system takes to complete one oscillation
During a full time period T, the mass on the spring oscillates back and forth, returning to its original position. This means that the total distance covered by the mass during a period T is 4 times the amplitude (4A), because the amplitude is just half the distance between the maximum and the minimum position, and during a time period the mass goes from the maximum to the minimum, and then back to the maximum.
So, the time t that the mass takes to move through a distance of 2 A can be found by using the proportion
![1 T : 4 A = t : 2 A](https://tex.z-dn.net/?f=1%20T%20%3A%204%20A%20%3D%20t%20%3A%202%20A)
and solving for t we find
![t=\frac{(1T)(2 A)}{4A}=0.5 T](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B%281T%29%282%20A%29%7D%7B4A%7D%3D0.5%20T)
b. 1.25T
Now we want to know the time t that the mass takes to move through a total distance of 5 A. SInce we know that
- the mass takes a time of 1 T to cover a distance of 4A
we can set the following proportion:
![1 T : 4 A = t : 5 A](https://tex.z-dn.net/?f=1%20T%20%3A%204%20A%20%3D%20t%20%3A%205%20A)
And by solving for t, we find
![t=\frac{(1T)(5 A)}{4A}=\frac{5}{4} T=1.25 T](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B%281T%29%285%20A%29%7D%7B4A%7D%3D%5Cfrac%7B5%7D%7B4%7D%20T%3D1.25%20T)