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Below is the solution:
W done by Normal = 0. (make the incline flat, Normal force goes directly up: no work done)
<span>W done by gravity = w*displacement = (11kg*9.8) * 7.5sin(35) = -463J </span>
<span>W done by friction is the opposite of the work done by weight because the object is not moving. Therefore W done by friction = 463J</span>
Answer:
A: They produce a real image.
Explanation:
The images formed on the retina of the eye for a normal visibility must always be real.
Only a real image can be physically projected on any physical object whereas the virtual images are visible due to reflections.
- The nearsightedness is corrected with the help of a concave lens since it is the condition of the eye lens remaining thick and curved to converge the rays entering the eyes after a shorter distance which results in their image formation even before the retinal surface so to initially diverge them a bit so that they converge on the retinal surface and form the image there we use concave lens. Vice-versa of the above justification in the case of farsightedness.
Answer:
2.0 m
Explanation:
Energy is conserved.
Initial KE = Final PE + Work done by friction
½ mv² = mgh + Fd
½ mv² = mgh + mgμd
½ v² = gh + gμd
½ v² − gh = gμd
d = (½ v² − gh) / (gμ)
d = (½ (7.5 m/s)² − (10 m/s²) (2.1 m)) / ((10 m/s²) (0.35))
d = 2.0 m
1. 
The Schwarzschild radius of an object of mass M is given by:
(1)
where
G is the gravitational constant
M is the mass of the object
c is the speed of light
The black hole in the problem has a mass of

where
is the solar mass. Substituting,

and substituting into eq.(1), we find the Schwarzschild radius of this black hole:

2) 242.8 solar radii
We are asked to find the radius of the black hole in units of the solar radius.
The solar radius is

Therefore, the Schwarzschild radius of the black hole in solar radius units is
