Answer:
50 V
Explanation:
The formula electric potential is given as,
V = kq/r............. Equation 1
q = Vr/k ................. Equation 2
Where q = charge at that point, V = Electric potential, k = coulombs constant, r = distant.
Given: V = 100 V, r = 2.0 m, k = 9.0×10⁹ Nm/C².
Substitute into equation 1
q = (100×2)/(9.0×10⁹ )
q = (200/9)(10⁹)
q = 22.22×10⁻⁹
q = 2.22×10⁻⁸ C.
The potential at point 4.0 m
Given: r = 4.0 m, q = 2.22×10⁻⁸ C, k = 9.0×10⁹ Nm²/C²
Substitute into equation 2
V = 9.0×10⁹(2.22×10⁻⁸)/4
V = 49.95 V
V ≈ 50 V
Hence the potential = 50 V
Answer:
35.55 ohm
Explanation:
C = 80 micro farad
F = 56 hertz
Reactance
Xc = 1 / (2 × pi × F × C)
Xc = 1 / ( 2 × 3.14 × 56 × 80 × 10^-6)
Xc = 35.55 ohm
Answer:
4.80 m
Explanation:
We are given the mass of the high jumper, its initial velocity, and the acceleration of gravity. We are trying to find the vertical displacement of the high jumper.
Let's set the upwards direction to be positive and the downwards direction to be negative.
List out the relevant known variables.
- v₀ = 9.7 m/s
- a = -9.8 m/s²
- Δx = ?
We still need one more variable in order to use the constant acceleration equations. Since we are trying to find the max height of the jumper, we can use the fact that at the top of its trajectory, its final velocity will be 0 m/s.
4. v = 0 m/s
Using these four variables, let's find the constant acceleration equation that contains these variables:
Substitute the known values into the equation and solve for Δx.
- (0)² = (9.7)² + 2(-9.8)Δx
- 0 = 94.09 + (-19.6)Δx
- -94.09 = -19.6Δx
- Δx = 4.80
The high jumper can jump to a max height of 4.80 m.
As Kinetic Energy=half of Mass into Velocity square so as velocity is doubled the K.E of the object is doubled..
