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:
es un motor de combustión interna con encendido por chispa.
Answer:
a

b
The value is 
Explanation:
From the question we are told that
The mass is
The spring constant is 
The instantaneous speed is 
The position consider is x = 0.750A meters from equilibrium point
Generally from the law of energy conservation we have that
The kinetic energy induced by the hammer = The energy stored in the spring
So

Here a is the amplitude of the subsequent oscillations
=> 
=> 
=> 
Generally from the law of energy conservation we have that
The kinetic energy by the hammer = The energy stored in the spring at the point considered + The kinetic energy at the considered point

=> 
=> 