Answer:
Mass of the pull is 77 kg
Explanation:
Here we have for
Since the rope moves along with pulley, we have
For the first block we have
T₁ - m₁g = -m₁a = -m₁g/4
T₁ = 3/4(m₁g) = 323.4 N
Similarly, as the acceleration of the second block is the same as the first block but in opposite direction, we have
T₂ - m₂g = m₂a = m₂g/4
T₂ = 5/4(m₂g) = 134.75 N
T₂r - T₁r = I·∝ = 0.5·M·r²(-α/r)
∴
Mass of the pull = 77 kg.
Answer:
Explanation:
Given:
<u>Now as we know that the value of gravity of any heavenly body is at height h is given as:</u>
∴The gravitational field strength will become one-fourth of what it is at the surface of the moon.
Answer:
E = k Q / [d(d+L)]
Explanation:
As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field
E = k ∫ dq/ r² r^
"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element and "r^" is a unit ventor from the load element to the point.
Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant
λ = Q / L
If we derive from the length we have
λ = dq/dx ⇒ dq = L dx
We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge
dE = k dq / x²2
dE = k λ dx / x²
Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider
E = k
We take out the constant magnitudes and perform the integral
E = k λ (-1/x)
Evaluating
E = k λ [ 1/d - 1/ (d+L)]
Using λ = Q/L
E = k Q/L [ 1/d - 1/ (d+L)]
let's use a bit of arithmetic to simplify the expression
[ 1/d - 1/ (d+L)] = L /[d(d+L)]
The final result is
E = k Q / [d(d+L)]
Answer:
0.087 m
Explanation:
Length of the rod, L = 1.5 m
Let the mass of the rod is m and d is the distance between the pivot point and the centre of mass.
time period, T = 3 s
the formula for the time period of the pendulum is given by
.... (1)
where, I is the moment of inertia of the rod about the pivot point and g is the acceleration due to gravity.
Moment of inertia of the rod about the centre of mass, Ic = mL²/12
By using the parallel axis theorem, the moment of inertia of the rod about the pivot is
I = Ic + md²
Substituting the values in equation (1)
12d² -26.84 d + 2.25 = 0
d = 2.15 m , 0.087 m
d cannot be more than L/2, so the value of d is 0.087 m.
Thus, the distance between the pivot and the centre of mass of the rod is 0.087 m.