what is the force of gravity attraction between an object with a mass of 0.5 kg and another object has a mass of 0.33 kg and a d
1 answer:
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Answer:
2400 cm
Explanation:
The milky way has diameter of 100000 light years.
The local group is the collection of galaxies which includes the milky way.
10 million light years is the diameter of the local group.
Hence, the diameter of the scale model of the local group would be 2400 cm
Complete question is;
Block 1 is resting on the floor with block 2 at rest on top of it. Block 3, at rest on a smooth table with negligible friction, is attached to block 2 by a string that passes over a pulley, as shown in the attachment below. The string and pulley have negligible mass.
Block 1 is removed without disturbing block 2.
Derive an equation for the acceleration of block 3 for any arbitrary values of m3 and m2. Express your answer in terms of m3, m2, and physical constants as appropriate.
Answer:
a = (m2)g/(m3 + m2)
Explanation:
Looking at the attached image, if we consider the free body diagram for block 3, by using Newton's first law of motion, we will arrive at the formula;
T = (m3)a - - - (eq 1)
where;
T is the tension in the string
a is acceleration
m3 is mass of block 3
Meanwhile doing the same with Block 2, the free body diagram would give us the formula; (m2)g - T = (m2)a
Making T the subject gives us;
T = (m2)g - (m2)a - - - (eq 2)
where;
g is acceleration due to gravity
T is the tension in the string
a is acceleration
m2 is mass of block 2
To solve for the acceleration, we will just substitute (m3)a for 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)
