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givi [52]
3 years ago
8

An image of the moon is focused onto a screen using a converging lens of focal length f= 34.3 cm. The diameter of the moon is 3.

48×10^6 m, and its mean distance from the earth is 3.85×10^8 m. What is the diameter of the moon's image?
Physics
1 answer:
Rashid [163]3 years ago
7 0

Answer:

The diameter of the moon's image is 0.31 cm.

Explanation:

Given that,

Focal length = 34.3 cm

Diameter of the moon d=3.48\times10^{6}\ m

Mean distance from the earth d=3.85\times10^{8}\ m

At that distance the object is assumes to be at infinity. hence the image will be formed at a distance equal to focal length

So, the image distance is 34.3 cm.

We need to calculate the  diameter of the moon's image

Using formula of magnification

m= \dfrac{image\ distance}{object\ distance}=\dfrac{height\ of\ image}{height\ of\ object}

\dfrac{0.343}{3.85\times10^{8}}=\dfrac{h'}{3.48\times10^{6}}

h'=\dfrac{0.343\times3.48\times10^{6}}{3.85\times10^{8}}

h'=0.0031\ m= 0.31\ cm

Hence, The diameter of the moon's image is 0.31 cm.

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Answer:

Explanation:

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Standard to install overload relay for an electric circuit is usually 20% ~ 25% more than the Running Current of the circuit.

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Step by Step Explanation:

30hp = 22371W (as 1hp = 745.7)

Assuming unity power factor (cosФ=1) and 208V phase to phase voltage:

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<em>{if 208V is phase to neutral voltage, then use following formula:</em>

<em> Rated Power (watt) = 3 . V.I. cosФ}</em>

\frac{22371}{\sqrt{3} * 208 * 1} = I

Rated Current = <u>62.169A</u>

So, required maximum rating for circuit breaker is:

20% to 25% of the rated current = 62.17*1.2 ~ 62.17*1.25

=74.6A ~ 77.7A

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a boy steps off the end of a high diving board and drops into the water below. it takes the boy 0.78 seconds to reach the water.
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Answer:

7.644 feet

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A block of mass 0.08 kg is pushed against a spring with spring constant k=31 N/m. The spring is compressed 0.15 meters from its
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Answer:

1.11 meters

Explanation:

As the spring is compressed, elastic potential energy is built up in the spring. The total elastic potential energy can be found using the following formula

Ep = 1/2 x k x s²          

where k = 31 N/m  (spring constant)

s = 0.15 m  (compression)

Ep = 3.4875 J

When the block of mass is released, the elastic potential energy (Ep) is converted to kinetic energy (Ek). From this we can find the initial velocity of the mass of block after release

Ek = 1/2 x m x u²    

   

where Ek = Ep = 3.4875J

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u = unknown (initial velocity)

u = 2.9526 m/s

Now that we know the initial velocity we need to find the deceleration of the mass of block due to friction. We will first find the force of friction from the following formula

F = ∪ x m x g          

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∪ = 0.4   (coefficient of friction)

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F = 0.31392 N

From this force we calculate the deceleration based on the following formula

F = m x a                  

where F = 0.31392   (frictional force)

m = 0.08 kg   (mass of block)

a = unknown  (acceleration)

a = -3.924 m/s²      -

*the negative sign is due to this value being deceleration

Now to find the total distance traveled we use the equation for motion

v² = u² + 2as            

where  v = 0 (final velocity)

u = 2.9526 m/s (initial velocity

a = -3.924 m/s² (deceleration due to friction)

s = unknown (distance traveled)

s = 1.11 meters

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Gelneren [198K]

Answer:

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Explanation:

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P = \frac{W}{\Delta t}=\frac{6000J}{6 s} = 1000 \frac{J}{s}

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