The mass on the left has a downslope weight of
W1 = 3.5kg * 9.8m/s² * sin35º = 19.7 N
The mass on the right has a downslope weight of
W2 = 8kg * 9.8m/s² * sin35º = 45.0 N
The net is 25.3 N pulling downslope to the right.
(a) Therefore we need 25.3 N of friction force.
Ff = 25.3 N = µ(m1 + m2)gcosΘ = µ * 11.5kg * 9.8m/s² * cos35º
25.3N = µ * 92.3 N
µ = 0.274
(b) total mass is 11.5 kg, and the net force is 25.3 N, so
acceleration a = F / m = 25.3N / 11.5kg = 2.2 m/s²
tension T = 8kg * (9.8sin35 - 2.2)m/s² = 27 N
Check: T = 3.5kg * (9.8sin35 + 2.2)m/s² = 27 N √
hope this helps. :)
Answer:
<h2>For 24 seconds
force exerted is 5092 N towards opposite direction of motion of bus.</h2><h2>For 3.90 seconds
force exerted is 31333 N towards opposite direction of motion of bus.</h2>
Explanation:
We have equation of motion v = u + at
Initial velocity, u = 20 m/s
Final velocity, v = 0 m/s
Case 1:-
Time, t = 24 s
Substituting
v = u + at
0 = 20 + a x 24
a = -0.8333 m/s²
Force = Mass x Acceleration = 6110 x -0.8333 = -5092 N
Force exerted is 5092 N towards opposite direction of motion of bus.
Case 2:-
Time, t = 3.90 s
Substituting
v = u + at
0 = 20 + a x 3.90
a = -5.13 m/s²
Force = Mass x Acceleration = 6110 x -5.13 = -31333 N
Force exerted is 31333 N towards opposite direction of motion of bus.
Answer:
There is no displacement.
Explanation:
Because the runner is running laps and returning to the original place, there is no displacement as displacement is relative to the change in location from the original position.
Hope this helps. . .
ly UwU
Answer:
5.82x10^-3
Explanation:
If the decimal moves right I think it goes negative in the exponent because the starting number is small...or less than 1
I think
I'm assuming the question is supposed to be like this:
How does the Orbital velocity of a satellite depends on the mass of the satellite?
Answer
It is independent of mass of satellite.