Here is what we know:
a = ?, s = ?, u = 18.1m/s, v = zero/rest, t = 7.0s, m = 2110kg
(a = acceleration, s= displacement, u = initial velocity, v = final velocity, t = time and m is mass)
Now we choose a kinematic formula. Since we know v, u and t, we will use the formula: v = u+at and rearrange it so that we can find a.
a = v-u/t
a = 0-18.1/7.0 = -2.5857...
therefore, a = -2.6m/s
We have our acceleration, now let’s find the net force. To find the force we use one of Newtons laws of motion.
We will use Newtons second law since it describes what happens when one or more forces act upon an object.
F = ma
F = (2110kg)(-2.6m/s)
F = -5486 kg
Therefore the net force F = -5486 N
The correct answer for 5 is Electrons
Given that the space station is free of gravitational force, it is required that it spins an certain speed to acquire centripetal acceleration.
In this case, you want that the centripetal acceleration, Ac, equals g (gravitational acceleration on the earth), becasue this will cause a centripetal force equal to the weight on earth.
The formula for centripetal acceleration is Ac = [angular velocity]^2*R
where R = [1/2]50.0m = 25.0 m
Ac = 9.81 m/s^2
=> [angular velocity]^2 = Ac/R = 9.81m/s^2v/ 25.0m = 0.3924 (rad/s)^2
[angular velocity] = √(0.3924) rad/s = 0.63 rad/s
Answer: 0.63 rad/s
Answer:
Explanation:
We are given that
Current,I=2.85 A
Length of segment=l=2.35 m
d=0.275 m
We have to find the magnitude of magnetic field due to this segment of wire at a point P which is at midpoint of the straight wire.
Magnetic field,B=
Where
Using the formula