A spherically symmetric charge distribution has a charge density given by ρ = a/r , where a is constant. Find the electric field
2 answers:
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The complete question is;
Block 1 sits on the floor with block 2 resting atop it. Block 3, which is stationary on a frictionless table, is attached to block 2 via a string that passes over a pulley depicted in the illustration below. Both the string and pulley have negligible mass.
Once block 1 is taken away without impacting block 2.
Derive an equation for the acceleration of block 3 considering arbitrary values for m3 and m2. Express your answer in terms of m3, m2, and relevant physical constants as needed.
Answer:
a = (m2)g/(m3 + m2)
Explanation:
Examining the attached illustration, by analyzing the free body diagram for block 3 and utilizing Newton's first law of motion, we reach the following formula;
T = (m3)a - - - (eq 1)
where;
T is the tension in the string
a is acceleration
m3 is the mass of block 3
Simultaneously, doing the same for Block 2, the free body diagram yields the equation; (m2)g - T = (m2)a
Rearranging for T results in;
T = (m2)g - (m2)a - - - (eq 2)
where;
g represents acceleration due to gravity
T is the tension in the string
a is acceleration
m2 is the mass of block 2
To deduce the acceleration, we will substitute (m3)a in place of T in eq 2.
Thus;
(m3)a = (m2)g - (m2)a
(m3)a + (m2)a = (m2)g
a(m3 + m2) = (m2)g
a = (m2)g/(m3 + m2)
Answer:
F = 0.535 N
Explanation:
We will apply energy concepts, considering both the peak and the bottom of the path.
Top
Em₀ = U = mg y
Bottom
= K = ½ m v²
Emo =
mg y = ½ m v²
v = √ (2gy)
y = L - L cos θ
v = √ (2g L (1 - cos θ))
Next, we will employ Newton's second law at the lowest point where the acceleration is centripetal.
F = ma
a = v² / r
For the turning radius, the cable length is r = L.
F = m 2g (1 - cos θ)
Now, let's find the result.
F = 2 1.25 9.8 (1 - cos 12)
F = 0.535 N