The solution for this problem is:
For 1st minimum, let m be equal to 1.
d = slit width
D = screen distance.
Θ = arcsin (m * lambda/ (d))
= 0.13934 rad, 7.9836 deg
y = D*tan (Θ)
y = 6.50 * tan (7.9836)
= 0.91161 m is the distance from the central maximum to the first-order minimum
D. 51 N. The minimum applied force that will cause the television slide is 51 N.
In order to solve this problem we have to use the force of static friction equation Fs = μs*n, where μs is the coefficient of static friction, and n is the normal force m*g.
With μs = 0.35, and n = 15kg*9.8m/s² = 147 N
Fs = (0.35)(147 N)
Fs = 51.45 N
Fs ≅ 51 N
Answer:
The necessary separation between the two parallel plates is 0.104 mm
Explanation:
Given;
length of each side of the square plate, L = 6.5 cm = 0.065 m
charge on each plate, Q = 12.5 nC
potential difference across the plates, V = 34.8 V
Potential difference across parallel plates is given as;
Where;
d is the separation or distance between the two parallel plates;
Therefore, the necessary separation between the two parallel plates is 0.104 mm
Answer:
We know that potential energy of a body;
= mass(m)× gravitational acceleration(g) × height(h)
Lets find out the mass of the body
P.E. = mgh
=> 6500J = mass × 9.8m/s^2 × 12m
=>6500J = mass × ( 9.8 × 12 ) × ( m/s^2 × m)
=> 6500 Nm = m × 117.6 × m^2 / s^2
=> 6500/117.6 Ns^2/m = mass [°.° Ns^2/m = kg]
=> 55.272 Kg = mass
Therefore the mass of the body = 55.272 kg ~ <em>6</em><em>0</em><em> </em><em>k</em><em>g</em><em> </em>(Ans)
Hope it helps you
