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.
Answer:
-0.0789 m
Explanation:
Recall that the y-component comes associated with the sin(18.4) through the following trigonometric relationship:
y = 0.250 sin(-18.4) ≈ -0.0789 m
Notice it is negative since it is below the x-axis.
Answer:
a) 2.063*10^-4
b) 1.75*10^-4
Explanation:
Given that: d= 1.628 mm = 1.628 x 10-3 I= 12 mA = 12.0 x 10-8 A The Cross-sectional area of the wire is:
a) <em>The Potential difference across a 2.00 in length of a 14-gauge copper </em>
<em> wire: </em>
L= 2.00 m
From Table Copper Resistivity 
= 1.72 x 10-8 S1 • m The Resistance of the Copper wire is:
=0.0165Ω
The Potential difference across the copper wire is:
V=IR
=2.063*10^-4
b) The Potential difference if the wire were made of Silver: From Table: Silver Resistivity p= 1.47 x 10-8 S1 • m
The Resistance of the Silver wire is:
=0.014Ω
The Potential difference across the Silver wire is:
V=IR
=1.75*10^-4
I,think potential energy is mgh so 65*100*9,81
