Weight of the carriage 
Normal force 
Frictional force 
Acceleration 
Explanation:
We have to look into the FBD of the carriage.
Horizontal forces and Vertical forces separately.
To calculate Weight we know that both the mass of the baby and the carriage will be added.
- So Weight(W)

To calculate normal force we have to look upon the vertical component of forces, as Normal force is acting vertically.We have weight which is a downward force along with
, force of
acting vertically downward.Both are downward and Normal is upward so Normal force 
- Normal force (N)

- Frictional force (f)

To calculate acceleration we will use Newtons second law.
That is Force is product of mass and acceleration.
We can see in the diagram that
and
component of forces.
So Fnet = Fy(Horizontal) - f(friction) 
- Acceleration (a) =

So we have the weight of the carriage, normal force,frictional force and acceleration.
To help you I need to assume a wavelength and then calculate the momentum.
The momentum equation for photons is:
p = h / λ , this is the division of Plank's constant by the wavelength.
Assuming λ = 656 nm = 656 * 10 ^ - 9 m, which is the wavelength calcuated in a previous problem, you get:
p = (6.63 * 10 ^-34 ) / (656 * 10 ^ -9) kg * m/s
p = 1.01067 * 10^ - 27 kg*m/s which must be rounded to three significant figures.
With that, p = 1.01 * 10 ^ -27 kg*m/s
The answers are rounded to only 2 significan figures, so our number rounded to 2 significan figures is 1.0 * 10 ^ - 27 kg*m/s
So, assuming the wavelength λ = 656 nm, the answer is the first option: 1.0*10^-27 kg*m/s.
By law of refraction we know that image position and object positions are related to each other by following relation

here we know that



now by above formula


so apparent depth of the bottom is seen by the observer as h = 3.39 cm
Answer:
0.056 psi more pressure is exerted by filled coat rack than an empty coat rack.
Explanation:
First we find the pressure exerted by the rack without coat. So, for that purpose, we use formula:
P₁ = F/A
where,
P₁ = Pressure exerted by empty rack = ?
F = Force exerted by empty rack = Weight of Empty Rack = 40 lb
A = Base Area = 452.4 in²
Therefore,
P₁ = 40 lb/452.4 in²
P₁ = 0.088 psi
Now, we calculate the pressure exerted by the rack along with the coat.
P₂ = F/A
where,
P₂ = Pressure exerted by rack filled with coats= ?
F = Force exerted by filled rack = Weight of Filled Rack = 65 lb
A = Base Area = 452.4 in²
Therefore,
P₂ = 65 lb/452.4 in²
P₂ = 0.144 psi
Now, the difference between both pressures is:
ΔP = P₂ - P₁
ΔP = 0.144 psi - 0.088 psi
<u>ΔP = 0.056 psi</u>
Answer:
, 
Explanation:
The jet is flying at constant velocity: this means that its acceleration is zero, so the net force acting on the jet is also zero.
Therefore, we can write:

where
is the thrust force generated by each engine of the jet
is the drag force
Solving for Fd,

The velocity of the jet is

So, the rate at which the drag force does work (which is the power) is

and substituting

we find

Converting into horsepower,
