Answer:
Δt ≈ 2.9137 ≈ 3 seconds per kilometer
Explanation:
The speed of light is approximately 300000 km /s, while the speed fo the sound in the air is 0.3432 km/s.
The light takes therefore this time to travel one kilometer
On the other hand the sound takes this time to travel one kilometer
t = 2.9137 s
Then the delay time is 2.9137 -
Δt ≈ 2.9137 ≈ 3 s
For problems that involve an object accelerating along an inclined
plane, the weight can be used to determine the x and y force component that
causes the motion by using the formula:
The force component along the x direction is equal to weight
of the object multiplies by sin of angle of inclination of the ramp.
Fx=wsin(a). while force component along the y direction is equal to weight of
the object multiplies by cos of angle of inclination of the ramp, Fy = wcos(a).
Answer:
(a) Angular speed of resultant combination will be rev/min
(b) Fraction of total energy loss will be 0.9
Explanation:
We have given angular speed of the first wheel
According to question moment of inertia of second wheel is 9 times the moment of inertia of first wheel
So
(a) Now according to law of conservation of angular momentum
(b) Change in rotational kinetic energy will be equal to .......eqn1
Putting these values in eqn 1
Change in kinetic energy will be 0.9
Hello!
Recall the period of an orbit is how long it takes the satellite to make a complete orbit around the earth. Essentially, this is the same as 'time' in the distance = speed * time equation. For an orbit, we can define these quantities:
← The circumference of the orbit
speed = orbital speed, we will solve for this later
time = period
Therefore:
Where 'r' is the orbital radius of the satellite.
First, let's solve for 'v' assuming a uniform orbit using the equation:
G = Gravitational Constant (6.67 × 10⁻¹¹ Nm²/kg²)
m = mass of the earth (5.98 × 10²⁴ kg)
r = radius of orbit (1.276 × 10⁷ m)
Plug in the givens:
Now, we can solve for the period:
Q = magnitude of charge at each of the three locations A, B and C = 2 x 10⁻⁶ C
r₁ = distance of charge at origin from charge at B = 50 - 0 = 50 cm = 0.50 m
r₂ = distance of charge at origin from charge at C = 100 - 0 = 100 cm = 1 m
F₁ = magnitude of force by charge at B on charge at origin
F₂ = magnitude of force by charge at C on charge at origin
Magnitude of force by charge at B on charge at origin
inserting the values
F₁ = 0.144 N
Magnitude of force by charge at C on charge at origin
inserting the values
F₂ = 0.036 N
Net force on the charge at the origin is given as
F = F₁ + F₂
F = 0.144 + 0.036
F = 0.18 N
from the diagram , direction of net force is towards left or negative x-direction.