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ipn
3 months 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:
Keith_Richards [3.2K]3 months ago
7 0

Answer:

The telescope captures light at a rate 1600 times greater than the eyesight of humans!

Explanation:

The capability to gather light with an optical device correlates directly to the size of its aperture.

Thus, when assessing the light-gathering potential of two entities, it boils down to the ratio of their aperture sizes.

Assuming the telescope has a diameter of D = 1 m

And the diameter of a human eye is d = 2.5 cm = 0.025 m

The comparison of light-gathering powers between the telescope and the human eye is given by D² ÷ d²

= (D²/d²) = (1²/0.025²) = 1600 times.

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I hope this information is helpful!

</pindeed>
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An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate
inna [3103]

Answer:

the time it takes after impact for the puck is 2.18 seconds

Explanation:

initially given information

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thickness = 0.1 mm = 1 ×10^{-4} m

dynamic viscosity = 1.75 ×10^{-5} Ns/m²

temperature of air = 15°C

to determine

time needed for the puck to reduce its speed by 10%

solution

we note that velocity changes from 0 to v

assuming initial velocity = v

therefore final velocity = 0.9v

implying a change in velocity is du = v

and clearance dy = h

shear stress acting on the surface is expressed as

= µ \frac{du}{dy}

therefore

= µ \frac{v}{h}............1

substituting the values

= 1.75 ×10^{-5} × \frac{v}{10^{-4}}

= 0.175 v

the area between the air and puck is given by

Area = \frac{\pi }{4} d^{2}

area = \frac{\pi }{4} 0.1^{2}

area = 7.85 × \frac{v}{10^{-3}} m²

thus, the force on the puck can be represented as

Force = × area

force = 0.175 v × 7.85 × 10^{-3}

force = 1.374 × 10^{-3} v

now applying Newton's second law

force = mass × acceleration

- force = mass \frac{dv}{dt}

- 1.374 × 10^{-3} v = 0.03 \frac{0.9v - v }{t}

solving for t = \frac{0.1 v * 0.03}{1.37*10^{-3} v}

the time needed after impact for the puck is 2.18 seconds

3 0
3 months ago
The mean free path of a helium atom in helium gas at standard temperature and pressure is 0.2 um.What is the radius of the heliu
Yuliya22 [3333]

Answer: 0.10233nm

Explanation:

The mean free path of an atom can be calculated using the following equation:

\lambda=\frac{RT}{\sqrt{2} \pi d^{2}N_{A}P}    (1)

Where:

\lambda=0.2\mu m=0.2(10)^{-6}m

R=8.3145J/mol.K is referred to as the Universal gas constant

T=0\°C=273.115K represents the absolute standard temperature

d denotes the diameter of helium atoms

N_{A}=6.0221(10)^{23}/mol symbolizes Avogadro's number

P=1atm=101.3kPa=101.3(10)^{3}Pa=101.3(10)^{3}J/m^{3} indicates absolute standard pressure

<pFrom this, we can solve for d using (1), aiming to determine the radius r of the helium atom:

d=\sqrt{\frac{RT}{\sqrt{2}\pi\lambda N_{A}P}}    (2)

d=\sqrt{\frac{(8.3145J/mol.K)(273.115K)}{\sqrt{2}\pi(0.2(10)^{-6}m)(6.0221(10)^{23}/mol)(101.3(10)^{3}J/m^{3})}}    (3)

d=2.0467(10)^{-10}m    (4)

If the radius equals half of that diameter:

r=\frac{d}{2}  (5)

Eventually:

r=\frac{2.0467(10)^{-10}m}{2}  (6)

r=1.0233(10)^{-10}m  (7)

Nonetheless, we were tasked with finding this radius in nanometers. Knowing 1nm=(10)^{-9}m:

r=1.0233(10)^{-10}m.\frac{1nm}{(10)^{-9}m}=0.10233nm  (8)

Ultimately:

r=0.10233nm Represents the radius of the helium atom in nanometers.

5 0
4 months ago
If an electronic circuit experiences a loss of 3 decibels with an input power of 6 watts, what would its output power be, to the
Yuliya22 [3333]

Answer:

The output power of the circuit is 3 Watts.

Given:

a loss in decibels = 3 dB

Input power = 6 Watts

To find:

What is the output power?

Formula used:

Output power = Input power × loss in ratio

Solution:

3 dB loss corresponds to a ratio of 0.5

Output power can be calculated as follows:

Output power = Input power × loss in ratio

Output power = 6 × 0.5

Output power = 3 Watts

Therefore, the output power of the circuit is 3 Watts.

4 0
3 months ago
Read 2 more answers
Laser beam with wavelength 632.8 nm is aimed perpendicularly at opaque screen with two identical slits on it, positioned horizon
Softa [3030]

Response:

1 slit width = 0.158 mm, slit separation = 0.633 mm, distance between diffraction maxima = 12.7 mm

Explanation:

This involves defining several terms:

λ, the wavelength of the beam

D, the distance from the plane to the slit

x, the distance between minima in the diffraction pattern (in single slit setups)

w, the fringe width for double slit setups

1. In a double slit experiment, the fringe width is also recognized as the distance between Maxima.

Thus, w=λD/d, leading to d=λD/w when rearranged.

So, d= (632.8 x 10^-9 x 1 x 10^3)/1

which equals d= 0.633mm.

For single slit diffraction, minima are defined by a*sinΘ= m*x

where a is the slit width and m is an integer.

For small angles, it simplifies to Θ= (x/D) = (λ/a), allowing us to solve for a:

a = λD/a

a = (632.8x10^-9 x 1)/4

yielding a= 0.158mm.

2. A grating with 20 slits/mm gives d= 1/20mm= 0.05x10^-3.

To find y (the distance between maxima), apply y= λL/d:

Finally, y= (632.8 x 10^-9 x 1)/0.05x10^-3, which results in y= 12.7mm.

6 0
3 months ago
Select the areas that would receive snowfall because of the lake effect.
Keith_Richards [3271]

Result:

- Grand Marais

- Two Harbors

- Duluth

Explanation:

The locations likely to receive snowfall due to lake effect include Grand Marais, Two Harbors, and Duluth. These areas are situated directly along the shores of Lake Superior, one of the world’s largest lakes. Its substantial water volume significantly influences the local climate, generating a lot of humidity in the air and considerable evaporation, both of which lead to cloud formation and, when temperatures drop adequately, result in significant snowfall rather than rain.

8 0
3 months ago
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