Answer:
r = 20.22 m
Explanation:
Given that,
Charge,
Electric field,
We need to find the distance. We know that, the electric field a distance r is as follows :
So, the required distance is 20.22 m.
Answer:
Hello, the tripping of a 230-kilovolt transmission line.
Explanation:
the tripping of a 230-kilovolt transmission line near Ontario, Canada, at 5:16 p.m., which caused several other heavily loaded lines also to fail. Hopefully this helps you find what your looking for!.
Answer:
The speed of the car at the end of the 2nd second = 8.0 m/s
Explanation:
The equations of motion will be used to solve this problem.
A car starts from rest,
u = initial velocity of the car = 0 m/s
Accelerates at a constant rate in a straight line,
a = constant acceleration of the car = ?
In the first second the car moves a distance of 2.0 meters,
t = 1.0 s
x = distance covered = 2.0 m
x = ut + (1/2)at²
2 = 0 + (1/2)(a)(1²)
a = 4.0 m/s²
How fast will the car be moving at the end of the second second
Now,
a = 4.0 m/s²
u = initial velocity of the car at 0 seconds = 0 m/s
v = final velocity of the car at the end of the 2nd second = ?
t = 2.0 s
v = u + at
v = 0 + (4×2)
v = 8.0 m/s
Answer:
(a) 98 N
(b) 158 N
(c) 38 N
Explanation:
<h2>
Part (a)</h2>
When the acceleration is 0 m/s², the net force on the mass is 0 N. Therefore, the tension force is equal to the weight force due to Newton's Second Law:
- ∑F_y = T - w = ma_y
- ∑F_y = T - w = m(0 m/s²)
- ∑F_y = T - w = 0
- ∑F_y = T = w
Since the tension in the cable and the weight of the mass are equal to each other, we can solve for the weight force of the mass by using:
- w = mg
- w = (10 kg)(9.8 m/s²)
- w = 98 N
Since T = w, we can say that T = 98 N.
<h2>Part (b)</h2>
Let's set the upwards direction to be positive and the downwards direction to be negative. We can use Newton's Second Law to solve for the tension in the cable if the acceleration is 6 m/s² upward:
- ∑F_y = T - w = ma_y
- ∑F_y = T - mg = m(6 m/s²)
- ∑F_y = T - mg = 6m
Plug the known values into the equation and solve for T.
- T - mg = 6m
- T - (10 kg)(9.8 m/s²) = 6(10 kg)
- T - 98 = 60
- T = 158 N
The tension in the cable if the acceleration is +6 m/s² is 158 N.
<h2>
Part (c)</h2>
The process is the same, but this time acceleration is -6 m/s².
- ∑F_y = T - w = ma_y
- ∑F_y = T - mg = m(-6 m/s²)
- ∑F_y = T - mg = -6m
Plug known values into the equation and solve for T.
- T - mg = -6m
- T - (10 kg)(9.8 m/s²) = -6(10 kg)
- T - 98 = -60
- T = 38 N
The tension in the cable if the acceleration is -6 m/s² is 38 N.