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
Answer:
Work done by the machine (W) = 500 × 1.5 = 750 J
Work supplied to the machine (W) = 100 × 10 = 1000 J
Here, work supplied to the machine is input work = 1000 J
In short, the key value added of CDR data over census or survey approaches is the potential to access current and comprehensive evidence on population size, density, and dynamics, information that is fundamentally necessary for managing any humanitarian emergency or disease-related disaster but which is often
As per the given Figure attached here we know that both charges q1 and q2 will apply same force on charge q3 and hence the resultant force due to both charges will be along Y axis vertically upwards
So here we know that
now from the above equation
so both of the charges will apply 0.288 N force on q3 charge along the line joining them
now the net force due to vector sum is given by
here we know that angle is
now we have
so net force on q3 is 0.46 N vertically upwards along +Y axis
Answer:
143 °
Explanation:
a ) If d be the distance between slits , λ be wavelength of light used and at angle θ nth dark fringe is formed then
d sinθ = ( 2n+1) λ/2
for first dark fringe
d sinθ = λ/2
d /λ = 1/ 2 sinθ
1 / 2 sin15
= 1.93
b )
For intensity of fringe at angle θ, the relation is
I = I₀ cos²θ
I / I₀ = cos²θ/2
Given I / I₀ =0. 1
0.1 = cos²θ/2
θ/2 = 71.5
θ = 143 °
