Answer:
(3) The period of the satellite is independent of its mass, an increase in the mass of the satellite will not affect its period around the Earth.
(4) he gravitational force between the Sun and Neptune is 6.75 x 10²⁰ N
Explanation:
(3) The period of a satellite is given as;
where;
T is the period of the satellite
M is mass of Earth
r is the radius of the orbit
Thus, the period of the satellite is independent of its mass, an increase in the mass of the satellite will not affect its period around the Earth.
(4)
Given;
mass of the ball, m₁ = 1.99 x 10⁴⁰ kg
mass of Neptune, m₂ = 1.03 x 10²⁶ kg
mass of Sun, m₃ = 1.99 x 10³⁰ kg
distance between the Sun and Neptune, r = 4.5 x 10¹² m
The gravitational force between the Sun and Neptune is calculated as;
Answer:
F = 196 N
Explanation:
For this exercise we will use Newton's second law, we define a reference system with the x axis in the direction of movement of the stones and the y axis vertically
Y axis
N- W = 0
N = mg
X axis
F -fr = ma
In this case, they ask us for the force to keep moving, so the stones go at constant speed, which implies that the acceleration is zero.
F- fr = 0
F = fr
the friction force has the equation
fr = μ N
fr = μ mg
we substitute
F = μ mg
let's calculate
F = 0.80 9.8 25
F = 196 N
Answer:
66.2 sec
Explanation:
C₁ = 1.0 F
C₂ = 1.0 F
ΔV = Potential difference across the capacitor = 6.0 V
C = parallel combination of capacitors
Parallel combination of capacitors is given as
C = C₁ + C₂
C = 1.0 + 1.0
C = 2.0 F
R = resistance = 33 Ω
Time constant is given as
T = RC
T = 33 x 2
T = 66 sec
V₀ = initial potential difference across the combination = 6.0 Volts
V = final potential difference = 2.2 volts
Using the equation
t = 66.2 sec
Since there is no friction between the ladder and the wall, there can be no vertical force component. That's the tricky part ;)
So to find the weight, divide the 100N <em>normal</em> force by earths gravitational acceleration, 9.8m/s^2
Then;
Draw an arrow at the base of the ladder pointing towards the wall with a value of 30N, to show the frictional force.
