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3241004551 [841]
3 years ago
13

Compare the light gathering power of a 1 meter diameter telescope to that of the human eye ,which has a diameter of roughly 2.5

centimeters. How many times more light can the telescope gather than the human eye?
Physics
1 answer:
lesantik [10]3 years ago
7 0

Answer:

The telescope can gather light 1600 times more than the human eyes can!

Explanation:

The light gathering ability of an optical element is directly proportional to its area of opening.

So, in comparing the light gathering abilities for two objects, it is just the ratio of their area of opening.

Let the diameter of the telescope be D = 1 m

And the diameter of the human eyes be d = 2.5 cm = 0.025 m

Light gathering ability of the telescope compared to the eyes = D² ÷ d²

= (D²/d²) = (1²/0.025²) = 1600 times.

The telescope can gather light 1600 times more than the human eyes can!

Hope this Helps!!!

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Part complete during a collision with a wall, the velocity of a 0.200-kg ball changes from 20.0 m/s toward the wall to 12.0 m/s
Inga [223]

The average force applied to the ball= 106.7 N

Explanation:

Force is given by

f= ΔP/t

ΔP= change in momentum= m Vf- m Vi

m= mass =0.2 kg

Vf= final velocity= 12 m/s

Vi=initial velocity= -20 m/s ( negative because it is going towards the wall which is treated as negative axis)

t= time= 60 ms= 0.06 s

now ΔP= 0.2 [ 12-(-20)]

ΔP=0.2 (32)=6.4 kg m/s

now force F= ΔP/t

F= 6.4/0.06

F=106.7 N

8 0
3 years ago
In a device like the one shown below, the cylinder is allowed to fall a distance of 300 m. As a result, the temperature of the w
Delvig [45]

Answer:

Increase in the temperature of water would be 0.9 degree C

Explanation:

As we know by energy conservation

Change in the gravitational potential energy of the cylinder = increase in the thermal energy of the water

Here we know that the gravitational potential energy of the cylinder is given as

U = mgh

here we have

h = 300 m

now we can say

Mc\Delta T = (m \times 9.8 \times 300)

now if the cylinder falls from height h = 100 m

then we have

Mc\Delta T' = (m \times 9.8 \times 100)

now from above two equations

\frac{\Delta T'}{\Delta T} = \frac{100}{300}

\Delta T' = 2.7 \times \frac{1}{3} = 0.9 Degree C

8 0
3 years ago
What is meant by the phrase "a consistent method of measurement"?
julia-pushkina [17]
A measurement that will always give the same answer.
5 0
3 years ago
A ball is whirled on the end of a string in a horizontal circle of radius R at speed v. By which one of the following means can
Vera_Pavlovna [14]

Explanation:

When an object moves in a circular path, it will have circular acceleration. Its magnitude of acceleration is given by :  

a=\omega^2R

Since, \omega=\dfrac{2\pi }{T}R

a=(\dfrac{2\pi}{T})^2R

T is the time period

R is the radius of the circular path

To increase the centripetal acceleration bu a factor of 1.5 or 3/2, radius of circle must be increase by a factor of 6 and T is increased by a factor of 2 such that,

R'=6R and T'=2T

So,

a'=(\dfrac{2\pi}{T'})^2R'

a'=(\dfrac{2\pi}{(2T)})^2(6R)

a'=\dfrac{6}{4}(\dfrac{2\pi}{T})^2R

a'=\dfrac{3}{2}(\dfrac{2\pi}{T})^2R

Hence, this is the required solution.

4 0
3 years ago
Find a numerical value for rhoearth, the average density of the earth in kilograms per cubic meter. Use 6378km for the radius of
Radda [10]

According to the information provided to define an average density, it is necessary to use the concepts related to mass calculation based on gravitational constants and radius, as well as the calculation of the volume of a sphere.

By definition we know that the mass of a body in this case of the earth is given as a function of

M = \frac{gr^2}{G}

Where,

g= gravitational acceleration

G = Universal gravitational constant

r = radius (earth at this case)

All of this values we have,

g = 9.8m/s^2\\G  = 6.67*10^{-11} m^3/kg*s^2\\r = 6378*10^3 m

Replacing at this equation we have that

M = \frac{gr^2}{G} \\M = \frac{(9.8)(6378*10^3)^2}{6.67*10^{-11}} \\M = 5.972*10^{24}kg

The Volume of a Sphere is equal to

V = \frac{4}{3}\pi r^3\\V = \frac{4}{3} \pi (6378*10^3)^3\\V = 1.08*10^{21}m^3

Therefore using the relation between mass, volume and density we have that

\rho = \frac{m}{V}\\\rho = \frac{5.972*10^{24}}{1.08*10^{21}}\\\rho = 5.52*10^3kg/m^3

6 0
3 years ago
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