Answer:
P = 33.6 [N]
Explanation:
To solve this problem we must use Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.
∑F = m*a
where:
F = forces [N]
m = mass = 14 [kg]
a = acceleration = 6 [m/s²]
![F = 14*6\\F = 84 [N]](https://tex.z-dn.net/?f=F%20%3D%2014%2A6%5C%5CF%20%3D%2084%20%5BN%5D)
In the second part of this problem we must find the work done, where the work in physics is known as the product of force by distance, it is important to make it clear that force must be applied in the direction of movement.
where:
W = work [J]
F = force = 84 [N]
d = displaciment = 40 [m]
![W = 84*40\\W = 3360 [J]](https://tex.z-dn.net/?f=W%20%3D%2084%2A40%5C%5CW%20%3D%203360%20%5BJ%5D)
Finally, the power can be calculated by the relationship between the work performed in a given time interval.
where:
P = power [W]
W = work = 3360 [J]
t = time = 100 [s]
Now replacing:
![P=3360/100\\P=33.6[W]](https://tex.z-dn.net/?f=P%3D3360%2F100%5C%5CP%3D33.6%5BW%5D)
The power is given in watts
Answer:
1.65
Explanation:
The equation of the forces along the horizontal direction is:
(1)
where
F = 65 N is the force applied with the push
is the frictional force
m = 4 kg is the mass
is the acceleration
The force of friction can be written as 
(2), where
is the coefficient of kinetic friction
R is the normal force exerted by the floor
The equation of forces along the vertical direction is
(3)
since the bookcase is in equilibrium. Substituting (2) and (3) into (1), we find
And solving for ,
Answer:
f ’= 97.0 Hz
Explanation:
This is an exercise of the doppler effect use the frequency change due to the relative movement of the fort and the observer
in this case the source is the police cases that go to vs = 160 km / h
and the observer is vo = 120 km / h
the relationship of the doppler effect is
f ’= f₀ (v + v₀ / v- 
)
let's reduce the magnitude to the SI system
v_{s} = 160 km / h (1000 m / 1km) (1h / 3600s) = 44.44 m / s
v₀ = 120 km / h (1000m / 1km) (1h / 3600s) = 33.33 m / s
we substitute in the equation of the Doppler effect
f ‘= 100 (330+ 33.33 / 330-44.44)
f ’= 97.0 Hz
