Answer:
Gravitational field strength is the force experienced by a unit mass. Gravitational force is the amount of force acting on a body. It is the product of field strength times the mass under consideration. Gravitational pull is just a more colloquial name for gravitational force.
Explanation:
hope it helps u
In a way it’s true because you can get a ticket for getting caught littering
Density = (mass) / (volume)
4,000 kg/m³ = (mass) / (0.09 m³)
(4,000 kg/m³) x (0.09 m³) = mass
mass = 360 kg
force of gravity = (mass) x (acceleration of gravity) = (360 kg) x (9.8 m/s²) = (360 x 9.8) kg-m<span>/s² </span><span>= </span>3,528 newtons .
Answer:
Explanation:
Given an LC circuit
Frequency of oscillation
f = 299 kHz = 299,000 Hz
AT t = 0 , the plate A has maximum positive charge
A. At t > 0, the plate again positive charge, the required time is
t =
t = 1 / f
t = 1 / 299,000
t = 0.00000334448 seconds
t = 3.34 × 10^-6 seconds
t = 3.34 μs
it will be maximum after integral cycle t' = 3.34•n μs
Where n = 1,2,3,4....
B. After every odd multiples of n, other plate will be maximum positive charge, at time equals
t" = ½(2n—1)•t
t'' = ½(2n—1) 3.34 μs
t" = (2n —1) 1.67 μs
where n = 1,2,3...
C. After every half of t,inductor have maximum magnetic field at time
t'' = ½ × t'
t''' = ½(2n—1) 1.67μs
t"' = (2n —1) 0.836 μs
where n = 1,2,3...
Answer:
0.20
Explanation:
The box is moving at constant velocity, which means that its acceleration is zero; so, the net force acting on the box is zero as well.
There are two forces acting in the horizontal direction:
- The pushing force: F = 99 N, forward
- The frictional force: 
, backward, with
coefficient of kinetic friction
m = 50 kg mass of the box
g = 9.8 m/s^2 gravitational acceleration
The net force must be zero, so we have
which we can solve to find the coefficient of kinetic friction:
