Answer:
Tension in the string at this position: 3.1 N.
Explanation:
Convert the radius of the circle to meters:
.
What's the net force on the object?
The object is in a circular motion. As a result,
,
where
is the net force on the object,
is the mass of the object, 
is the velocity of the object, and
is the radius of the circular motion.
For this object,
.
The output unit of net force should be standard if the unit for mass, velocity, and radius are all standard. The net force shall always point towards the center. In this case the net force points downwards.
What are the forces on this object?
There are two forces on the object at this moment:
- Weight,
, which points downwards. 
. - Tension,
, which also points downwards. The size of the tension force needs to be found.
What's the size of the tension force?
Gravity and tension points in the same direction. The size of their resultant force is the sum of the two forces. In other words,
.
.
All three values in this question are given with two sig. fig. Round the value of to the same number of significant figures.
Answer:
Vectors have both magnitude and direction.
DE which is the differential equation represents the LRC series circuit where
L d²q/dt² + Rdq/dt +I/Cq = E(t) = 150V.
Initial condition is q(t) = 0 and i(0) =0.
To find the charge q(t) by using Laplace transformation by
Substituting known values for DE
L×d²q/dt² +20 ×dq/dt + 1/0.005× q = 150
d²q/dt² +20dq/dt + 200q =150
Mathematically it can be expressed as: I = FpΔt. Where F p is the average magnitude of the acting force and Δt = t 2 - t 1, the time lapse in the force acts.
The magnitude of the impulse applied to stop the cart is
I = FpΔt = (10N) * (5s) = 50 N.s
There are two stages to the flight: acceleration stage and deceleration stage.
m₁ = 200 kg, mass of the rocket
m₂ = 100 kg, mass of fuel
a₁ = 30.0 m/s², upward acceleration when burning fuel
Ignore air resistance and assume g = 9.8 m/s².
Acceleration stage:
The rocket starts from rest, therefore the initial vertical velocity is zero.
The distance traveled is given by
s₁ = (1/2)*(30.0 m/s²)*(30.0 s)² = 13500 m
Deceleration stage (due to gravity):
The initial velocity is u = (30.0 m/s²)*(30 s) = 900 m/s
The initial height is 13500 m
At maximum height, the vertical velocity is zero.
Let s₂ = the extra height traveled. Then
(900 m/s)² - 2*(9.8 m/s²)*(s₂ m) = 0
s₂ = 900²/19.6 = 41326.5 m
The maximum altitude is
s₁+s₂ = 54826.5 m
Answer: 54,826.5 m
