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olganol [36]
2 years ago
12

To clearly see an image, the wavelength used must be at most 1/4 of the size of the object that is to be imaged. what frequency

is needed to image a fetus at 8 weeks of gestation that is 1.6 cm long?
Physics
1 answer:
LuckyWell [14K]2 years ago
5 0
From the statement, the wavelength should be 1.6 cm * 1/4 = 0.4 cm. Then, we use the wave equation formula. The frequency is equal to the speed of light divided by the wavelength. First, let's use the SI units.

0.4 cm = 0.004 m

The speed of light is equal to 3 x 10^8 m/s

Frequency = 3 x 10^8 m/s ÷ 0.004 m
Frequency = 7.49 x 10^10 s^-1
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3 years ago
What is the minimum value of the friction coefficient between the boxes that will keep them from slipping when the 100 N force i
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Answer:

The friction coefficient's minimum value will be "0.173".

Explanation:

The given query seems to be incomplete. Below is the attached file of the complete question.

According to the question,

(a)

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⇒ F_{net}=ma

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(b)

For m₃,

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2 years ago
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Answer:

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2 years ago
A vessel at rest at the origin of an xy coordinate system explodes into three pieces. Just after the explosion, one piece, of ma
ahrayia [7]

Incomplete question as we have not told to find what.So the complete question is here

A vessel at rest at the origin of an xy coordinate system explodes into three pieces. Just after the explosion, one piece, of mass m, moves with velocity (-60 m/s)i and a second piece, also of mass m, moves with velocity (-60 m/s)j. The third piece has mass 3m.Just after the explosion, what are the (a) magnitude and (b) direction of the velocity of the third piece?

Answer:

V_{3}=(20i+20j)m/s

Explanation:

Given data

The vessel at rest

Piece one,of mass m,moves with velocity=(-60 m/s)i

Piece two,of mass m,moves with velocity=(-60 m/s)j

Piece three,of mass 3m

As the linear momentum is conserved in this system,Because the system is closed and no external  force acting on it

So momentum is given as

p_{initial}=p_{final}

As the vessel at rest so the initial momentum is zero

So

m_{1}V_{1}+m_{2}V_{2}+m_{3}V_{3}=0\\m_{3}V_{3}=-m_{1}V_{1}-m_{2}V_{2}\\V_{3}=\frac{-m_{1}V_{1}-m_{2}V_{2}}{m_{3}} \\V_{3}=\frac{-m_{1}(-60m/s)i-m_{2}(-60m/s)j}{3m}\\V_{3}=(20i+20j)m/s

 

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