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:
1.024 × 10⁸ m
Explanation:
The velocity v₀ of the orbit 8RE is v₀ = 8REω where ω = angular speed.
So, ω = v₀/8RE
For the orbit with radius R for it to maintain a circular orbit and velocity 2v₀, we have
2v₀ = Rω
substituting ω = v₀/8RE into the equation, we have
2v₀ = v₀R/8RE
dividing both sides by v₀, we have
2v₀/v₀ = R/8RE
2 = R/8RE
So, R = 2 × 8RE
R = 16RE
substituting RE = 6.4 × 10⁶ m
R = 16RE
= 16 × 6.4 × 10⁶ m
= 102.4 × 10⁶ m
= 1.024 × 10⁸ m
Answer:
The horizontal distance x between the two balloons is 54.15 m
Explanation:
The diagram described as obtained online is presented in the image attached to this solution.
Let the horizontal distance between the two balloons be x
Difference in height (vertical distance) between the two balloons = 61 - 48.2 = 12.8 m
Using trigonometric relations, it is evident that
Tan 13.3° = 12.8/x
x = 12.8/tan 13.3° = 12.8/0.2364 = 54.15 m
Answer:
1.77 x 10^-8 C
Explanation:
Let the surface charge density of each of the plate is σ.
A = 4 x 4 = 16 cm^2 = 16 x 10^-4 m^2
d = 2 mm
E = 2.5 x 10^6 N/C
ε0 = 8.85 × 10-12 C2/N ∙ m2
Electric filed between the plates (two oppositively charged)
E = σ / ε0
σ = ε0 x E
σ = 8.85 x 10^-12 x 2.5 x 10^6 = 22.125 x 10^-6 C/m^2
The surface charge density of each plate is ± σ / 2
So, the surface charge density on each = ± 22.125 x 10^-6 / 2
= ± 11.0625 x 10^-6 C/m^2
Charge on each plate = Surface charge density on each plate x area of each plate
Charge on each plate = ± 11.0625 x 10^-6 x 16 x 10^-4 = ± 1.77 x 10^-8 C
Answer:
the expression of current in the loop enclosed to the left of the capacitor plate is

Explanation:
As we know by Ampere's law that line integral of magnetic field around a closed loop is proportional to the current enclosed in the path
So we will have

so we have

so above is the expression of current in the loop enclosed to the left of the capacitor plate