The eyepiece of a compound telescope has a focal length of 12 cm and the objective lens has a focal length of 16 m. What is the
1 answer:
Answer:
The magnification is 133X
Explanation:
For a refractor telescope using convex lenses, the overall magnification os calculated as by dividing the focal length of the objective by that of the eyepiece.
So:
Care must be taken with the units, in this case they are in meters and centimeters, they should be converted into the same unit before calculating:
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Answer:
You are asked to design a cylindrical steel rod 50.0 cm long, with a circular cross section, that will conduct 170.0 J/s from a furnace at 350.0 ∘C to a container of boiling water under 1 atmosphere.
Explanation:
Given Values:
L = 50 cm = 0.5 m
H = 170 j/s
To find the diameter of the rod, we have to find the area of the rod using the following formula.
Here Tc = 100.0° C
k = 50.2
H = k × A ×
Solving for A
A =
A =
A = = 6.77 ×
m²
Now Area of cylinder is :
A = d²
solving for d:
d =
d = 9.28 cm
The crate moves at constant velocity, this means that its acceleration is zero, so the net force acting on the crate is zero (Newton's second law).
There are only two forces acting on the crate: the force F applied by the worker and the frictional force, acting in the opposite direction:
, where 
is the coefficient of friction and 
is the mass of the crate. Since the net force should be equal to zero, the two forces must have same magnitude, so we have:
And so, this is the force that the worker must apply to the crate.
There are only two forces acting on the crate: the force F applied by the worker and the frictional force, acting in the opposite direction:
And so, this is the force that the worker must apply to the crate.
Answer:
a) t= 4.81 s,and t= 23.51.
b)d=6.88 m
c)v=4.8 m/s
Explanation:
Acceleration of train ,a= 0.6 m/s²
u = 0 m/s
Your speed ,V= 8.5 m/s
Lets take after t time he you catch the train
Distance travel by train in t time
----------1
d= V ( t- 4)
d= 8.5 ( t- 4) --------2
By equating equation 1 and 2
0.3 t² = 8.5 ( t- 4)
0.3 t² -8.5 t + 34 = 0
t= 4.81 s and t= 23.51
It means that you will catch train after t= 4.81 s,and t= 23.51.
t=23.51 sec means that you will catch the train after 23.51 sec also because acceleration of train is low.
Distance travel
d= 8.5 ( t- 4)
t= 4.81 s
d= 8.5 ( 4.81- 4) m
d=6.88 m
Lets speed = v
0.3 t² = v ( t- 4)
0.3 t² - v t + 4 v = 0
To have one solution
D= 0 Should be zero.
v²- 4 x 0.3 x 4 v
v = 4 x 0.3 x 4
v=4.8 m/s