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mash [69]
3 years ago
7

An 7.5 × binocular has 3.7-cm-focal-length eyepieces. What is the focal length of the objective lenses? Express your answer to t

wo significant figures and include the appropriate units.
Physics
1 answer:
elixir [45]3 years ago
5 0

To solve this problem we will apply the concept of magnification, which is given as the relationship between the focal length of the eyepieces and the focal length of the objective. This relationship can be expressed mathematically as,

\mu = \frac{f_0}{f_e}

Here,

\mu = Magnification

f_e = Focal length eyepieces

f_0 = Focal length of the Objective

Rearranging to find the focal length of the objective

f_0 = \mu f_e

Replacing with our values

f_0 = 7.5* 3.7cm

f_0 = 27.75cm

Therefore the focal length of th eobjective lenses is 27.75cm

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You cover the following displacements every day going to school: d1=50 m, E and d2=95 m, N. You do this for 12 minutes. a) What
puteri [66]

The speed will be 0.2 m/s and the velocity will be 0 m/s.

Speed = Total Distance / Total time

We have given total distance as ( 50 + 95 ) metres and total time as 12 minutes or we can say 720 seconds.

Speed = 145/ 720 m/s

Speed = 0.2 m/s

Velocity = Total Displacement / Total time

As the initial and final is the home, hence the net displacement is 0 in that case.

In this case also the total time we have given is 12 minutes or we can say 720 seconds.

Velocity = 0 / 720 m/s

Velocity = 0 m/s

So to conclude with we can say that the speed is 0.2 m/s and the velocity is 0 m/s.

Learn more about velocity here:

brainly.com/question/80295?source=archive

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5 0
2 years ago
A
densk [106]
The elevator is going up
5 0
2 years ago
How much work is done to increase the speed of a 1.0 kg toy car by 5.0 m/s?
soldi70 [24.7K]

Answer:

The correct option is (b).

Explanation:

We need to find the work done to increase the speed of a 1 kg toy car by 5 m/s.

We know that, the work done is equal to the kinetic energy of an object i.e.

W=\Delta K\\\\W=\dfrac{1}{2}mv^2\\\\W=\dfrac{1}{2}\times 1\times 5^2\\W=12.5\ J

So, 12.5 J of work is done to increase the speed of a 1.0 kg toy car by 5.0 m/s.

6 0
3 years ago
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Greeley [361]

Answer:

this is impossible for me

Explanation:

7 0
3 years ago
A 87.0 kg astronaut is working on the engines of a spaceship that is drifting through space with a constant velocity. The astron
Ket [755]

Answer:

259.62521 seconds

Explanation:

m_1 = Mass of astronaut = 87 kg

m_2 = Mass of wrench = 0.57 kg

v_1 = Velocity of astronaut

v_2 = Velocity of wrench = 22.4 m/s

Here, the linear momentum is conserved

m_1v_1=m_2v_2\\\Rightarrow v_1=\frac{m_2v_2}{m_1}\\\Rightarrow v_1=\frac{0.57\times 22.4}{87}\\\Rightarrow v_1=0.14675\ m/s

Time = Distance / Speed

Time=\frac{38.1}{0.14675}=259.62521\ s

The time taken to reach the ship is 259.62521 seconds

4 0
3 years ago
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