If a star collapses to a tenth its size, gravitation at its surface increases by?
1 answer:
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First, you make a diagram of all the forces acting on the system. This is shown in the figure. We have to determine F1 and F4. Let's do a momentum balance. Momentum is conserved so the summation of all momentum is equal to zero. Momentum is force*distance.
To determine F1: (reference is F4, so F4=0)
∑Momentum = 0 = -F2 - F3 + F1
0 = (-4 kg)(9.81 m/s2)(0.25m)-(6kg)(9.81 m/s2)(0.5-0.3m)+F1(0.5-0.1m)
F1 = 53.96 N (left knife-edge)
To determine F4: (reference is F1, so F1=0)
∑Momentum = 0 = -F2 - F3 + F4
0 = (-4 kg)(9.81 m/s2)(0.25m)-(6kg)(9.81 m/s2)(0.5-0.2m)+F4(0.5-0.1m)
F4 = 68.67 N (right knife-edge)
To determine F1: (reference is F4, so F4=0)
∑Momentum = 0 = -F2 - F3 + F1
0 = (-4 kg)(9.81 m/s2)(0.25m)-(6kg)(9.81 m/s2)(0.5-0.3m)+F1(0.5-0.1m)
F1 = 53.96 N (left knife-edge)
To determine F4: (reference is F1, so F1=0)
∑Momentum = 0 = -F2 - F3 + F4
0 = (-4 kg)(9.81 m/s2)(0.25m)-(6kg)(9.81 m/s2)(0.5-0.2m)+F4(0.5-0.1m)
F4 = 68.67 N (right knife-edge)
Answer:
Option B
Explanation:
Magnification of Microscope is
Mo= Magnification of objective lens and
Me= magnification of the eyepiece.
Both magnifications( of objective and eyepiece) are inversely proportional to the focal length.
Magnification,
when the focal length is less magnification will be high and when the magnification is the low focal length of the microscope will be more.
Thus. Magnification will increase by decreasing the focal length.
The correct answer is Option B