Answer:
A. -2.16 * 10^(-5) N
B. 9 * 10^(-7) N
Explanation:
Parameters given:
Distance between their centres, r = 0.3 m
Charge in first sphere, Q1 = 12 * 10^(-9) C
Charge in second sphere, Q2 = -18 * 10^(-9) C
A. Electrostatic force exerted on one sphere by the other is:
F = (k * Q1 * Q2) / r²
F = (9 * 10^9 * 12 * 10^(-9) * -18 * 10^(-9)) / 0.3²
F = -2.16 * 10^(-5) N
B. When they are brought in contact by a wire and are then in equilibrium, it means they have the same final charge. That means if we add the charges of both spheres and divided by two, we'll have the final charge of each sphere:
Q1 + Q2 = 12 * 10^(-9) + (-18 * 10^(-9))
= - 6 * 10^(-9) C
Dividing by two, we have that each sphere has a charge of -3 * 10^(-9) C
Hence the electrostatic force between them is:
F = [9 * 10^9 * (-3 * 10^(-9)) * (-3 * 10^(-9)] / 0.3²
F = 9 * 10^(-7) N
C Weight is the gravitational pull on an object
Answer:
(a) 7.72×10⁵ J
(b) 4000 J
(c) 1.82×10⁻¹⁶ J
Explanation:
Kinetic Energy: This can be defined energy of a body due to its motion. The expression for kinetic energy is given as,
Ek = 1/2mv²................... Equation 1
Where Ek = Kinetic energy, m = mass, v = velocity
(a)
For a moving automobile,
Ek = 1/2mv².
Given: m = 2.0×10³ kg, v = 100 km/h = 100(1000/3600) m/s = 27.78 m/s
Substitute into equation 1
Ek = 1/2(2.0×10³)(27.78²)
Ek = 7.72×10⁵ J
(b)
For a sprinting runner,
Given: m = 80 kg, v = 10 m/s
Substitute into equation 1 above,
Ek = 1/2(80)(10²)
Ek = 40(100)
Ek = 4000 J
(c)
For a moving electron,
Given: m = 9.10×10⁻³¹ kg, v = 2.0×10⁷ m/s
Substitute into equation 1 above,
Ek = 1/2(9.10×10⁻³¹)(2.0×10⁷)²
Ek = 1.82×10⁻¹⁶ J
Fair enough, but you'll have to tell us the volume of the bar first.
