Answer:
<em>At t=T/2 the angular speed equals to its average angular speed </em>
Explanation:
<u>Angular Motion</u>
Let w be the angular speed of a rotating object, its angular acceleration, and T the time the acceleration is acting upon the object. The basic formula for the angular motion is
We are told the initial speed is zero, so
The average angular speed from t=0 to t=T can be found by
This value is reached at a certain time we need to compute, knowing that
Or equivalently
Simplifying we have
At t=T/2 the angular speed equals to its average angular speed
Answer:
0.8712 m/s²
Explanation:
We are given;
Velocity of first car; v1 = 33 m/s
Distance; d = 2.5 km = 2500 m
Acceleration of first car; a1 = 0 m/s² (constant acceleration)
Velocity of second car; v2 = 0 m/s (since the second car starts from rest)
From Newton's equation of motion, we know that;
d = ut + ½at²
Thus,for first car, we have;
d = v1•t + ½(a1)t²
Plugging in the relevant values, we have;
d = 33t + 0
d = 33t
For second car, we have;
d = v2•t + ½(a2)•t²
Plugging in the relevant values, we have;
d = 0 + ½(a2)t²
d = ½(a2)t²
Since they meet at the next exit, then;
33t = ½(a2)t²
simplifying to get;
33 = ½(a2)t
Now, we also know that;
t = distance/speed = d/v1 = 2500/33
Thus;
33 = ½ × (a2) × (2500/33)
Rearranging, we have;
a2 = (33 × 33 × 2)/2500
a2 = 0.8712 m/s²
Answer:
1
Explanation:
Let's call the magnitude of the charge of 10000 electrons .
Because A loses q, it has a net charge (electrons are negative; losing them creates positive charges). Conversely, B has gained q, so it has a charge of . The distance between them is 1 m.
By Coulomb's law, the force between them is
When A loses an additional 10000 electrons, the charge on it becomes while that on B is . The distance is now 2 m.
The force is now