Answer:
3.88m/s
Explanation:
Using the law of conservation of momentum
m1u1+m2u2 = (m1+m2)v
m1 and m2 are the masses
u1 and 2 are the initial velocities
v is the final velocity
Given
m1 = 64kg
u1 = 4.2m/s
m2 = 25kg
u2 = 3.2m/s
Required
Final velocity v
Substitute the given values into the formula
64(4.2)+25(3.2) = (65+25)v
268.8+80 = 90v
348.8 = 90v
v = 348.8/90
v = 3.88m/s
Hence the velocity of the kayak after the swimmer jumps off is 3.88m/s
Answer:
3.7 km/h
Explanation:
Let's call v the proper speed of the boat and v' the speed of the water in the river.
When the boat travels in the direction of the current, the speed of the boat is:
v + v'
And it covers 50 km in 3 h, so we can write
(1)
When the boat travels in the opposite direction, the speed of the boat is
v - v'
And it covers 50 km in 5.4 h, so
(2)
So we have a system of two equations: by solving them simultaneously, we find the value of v and v':

Subtracting the second equation from the first one we get:

So, the speed of the water is 3.7 km/h.
F has direct relation with a
then doubling F cause acc. to get double i:e 6×2=12
Answer:
Muscle contraction thus results from an interaction between the actin and myosin filaments that generates their movement relative to one another. The molecular basis for this interaction is the binding of myosin to actin filaments, allowing myosin to function as a motor that drives filament sliding.