The strength of the gravitational field is given by:
where
G is the gravitational constant
M is the Earth's mass
r is the distance measured from the centre of the planet.
In our problem, we are located at 300 km above the surface. Since the Earth radius is R=6370 km, the distance from the Earth's center is:
And now we can use the previous equation to calculate the field strength at that altitude:
And we can see this value is a bit less than the gravitational strength at the surface, which is
.
Answer:
m = B²qR² / 2 V
Explanation:
If v be the velocity after acceleration under potential difference of V
kinetic energy = loss of electric potential energy
1/2 m v² = Vq ,
v² = 2 Vq / m ----------------------- ( 1 )
In magnetic field , charged particle comes in circular motion in which magnetic force provides centripetal force
magnetic force = centripetal force
Bqv = mv² / R
v = BqR / m
v² = B²q²R² / m² ------------------------- (2)
from (1) and (2)
B²q²R² / m² = 2 Vq / m
m = B²q²R² / 2 Vq
m = B²qR² / 2 V
Answer:
Force A=-−2,697.75 N
Force B=13, 488.75 N
Explanation:
Taking moments at point A, the sum of clockwise and anticlockwise moments equal to zero.
25 mg-20Fb=0
25*1100g=20Fb
Fb=25*1100g/20=1375g
Taking g as 9.81 then Fb=1375*9.81=13,488.75 N
The sum of upward and downward forces are same hence Fa=1100g-1375g=-275g
-275*9.81=−2,697.75. Therefore, force A pulls downwards
Note that the centre of gravity is taken to be half the whole length hence half of 50 is 25 m because center of gravity is always at the middle
Answer:
- tension: 19.3 N
- acceleration: 3.36 m/s^2
Explanation:
<u>Given</u>
mass A = 2.0 kg
mass B = 3.0 kg
θ = 40°
<u>Find</u>
The tension in the string
The acceleration of the masses
<u>Solution</u>
Mass A is being pulled down the inclined plane by a force due to gravity of ...
F = mg·sin(θ) = (2 kg)(9.8 m/s^2)(0.642788) = 12.5986 N
Mass B is being pulled downward by gravity with a force of ...
F = mg = (3 kg)(9.8 m/s^2) = 29.4 N
The tension in the string, T, is such that the net force on each mass results in the same acceleration:
F/m = a = F/m
(T -12.59806 N)/(2 kg) = (29.4 N -T) N/(3 kg)
T = (2(29.4) +3(12.5986))/5 = 19.3192 N
__
Then the acceleration of B is ...
a = F/m = (29.4 -19.3192) N/(3 kg) = 3.36027 m/s^2
The string tension is about 19.3 N; the acceleration of the masses is about 3.36 m/s^2.
