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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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What should Miguel do first and why? What type of healthcare professional will respond to the call?
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Answer:

  1. Miguel should hand the phone over to a medical personnel.
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From chapter 2 of the book: Medical Assisting: Administrative Skills, Miguel Perez is an administrative assistant. The duties of an administrative assistant in an healthcare professional setting range from performing medical clerical services like keeping patient's files organized, scheduling appointments and answering calls.

So, when a patient has medical concerns and calls the doctor's office, Miguel should answer the phone, know the patient's concerns, put the patient on hold and hand the phone over to a medical personnel - either a nurse or a physician. This is because Miguel is not a trained healthcare professional and cannot offer medical advice or assistance.

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a golfer is on the edge of a 12.5m bluff overlooking the 18th hole which is located 67.1m from the base of the bluff. she launch
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Answer: 26.359 m/s

Explanation:

This problem is related to parabolic motion and can be solved by the following equations:

x=V_{o}cos \theta t (1)

y=y_{o}+V_{o} sin \theta t - \frac{1}{2}gt^{2} (2)

V=V_{o}-gt (3)

Where:

x=67.1 m is the horizontal distance traveled by the golf ball

V_{o} is the golf ball's initial velocity

\theta=0\° is the angle (it was  a horizontal shot)

t is the time

y=0 m is the final height of the ball

y_{o}=12.5 m is the initial height of the ball

g=9.8 m/s^{2} is the acceleration due gravity

V is the final velocity of the ball

Let's begin by finding t  from (2):

t=\sqrt{\frac{2 y_{o}}{g}} (4)

t=\sqrt{\frac{2 (12.5 m)}{9.8 m/s^{2}}} (5)

t=1.597 s (6)

Substituting (6) in (1):

67.1 m=V_{o} cos(0\°) 1.597 s (7)

Finding V_{o}:

V_{o}=42.01 m/s (8)

Substituting V_{o} in (3):

V=42.01 m/s-(9.8 m/s^{2})(1.597 s) (9)

Finally:

V=26.359 m/s

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