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Salsk061 [2.6K]
4 years ago
7

The distance between the lenses in a compound microscope is 18 cm. The focal length of the objective is 1.5 cm. If the microscop

e is to provide an angular magnification of -46 when used by a person with a normal near point (25 cm from the eye), what must be the focal length of the eyepiece
Physics
1 answer:
Blababa [14]4 years ago
6 0

Answer:

The focal length of eye piece is 6.52 cm.

Explanation:

Given that,

Angular Magnification of the microscope M = -46

the distance between the lens in microscope L= 16 cm

The focal length of objective f₀ = 1.5 cm

Normal near point N = 25 cm

Have to find focal length of eye piece f ₙ =?

The angular magnification is given by

M ≈ - (L-fₙ)N/f₀fₙ

Rearranging for fₙ

fₙ =L(1 - Mf₀/N)⁺¹

   =18/2.76

fₙ =  6.52 cm

The focal length of eye piece is 6.52 cm.

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Lake Erie contains roughly 4.00 ✕ 1011 m3 of water. (a) How much energy is required to raise the temperature of that volume of w
denis-greek [22]

Answer:

Part a)

Q = 6.36 \times 10^{18} J

Part b)

t = 144.11 years

Explanation:

Part a)

Energy required to raise the temperature of water from 17.8 degree C to 21.6 degree C is given by formula

Q = ms\Delta T

here we know that

Here the volume of the water is given as

V = 4.00 \times 10^{11} m^3

now the mass of water is

m = density \times volume

m = 4.00 \times 10^{14} kg

now the heat required is

Q = (4 \times 10^{14})(4186)(21.6 - 17.8)

Q = 6.36 \times 10^{18} J

Part b)

As we know that power is supplied at

P = 1400 MW

so here we know

P = \frac{Q}{t}

so here we have

t = \frac{Q}{P}

t = \frac{6.36 \times 10^{18}}{1400 \times 10^6}

t = 4.54 \times 10^9 s

t = 144.11 years

7 0
3 years ago
A cannon ball is fired with an initial velocity of 75 m/s at an angle of 58 degree above the ground. What maximum height will it
KiRa [710]

Answer:

h = 206.4 m

range = 515.9 m

Explanation:

from the question we are given the following:

initial velocity (u) = 75 m/s

angle  above surface = 58 degrees

acceleration due to gravity (g) = 9.8 m/s^{2}

find the maximum height (h) and the horizontal distance

maximum height (h) = \frac{u^{2}sinθ^{2}  }{2g}

h =  \frac{75^{2}sin58^{2}  }{2 x 9.8}

h = 206.4 m              

the horizontal distance here is the range

range = \frac{u^{2}sin2θ }{g}

range = \frac{75^{2}sin(2 x 58) }{9.8}

range = 515.9 m

7 0
3 years ago
A dog has a mass of69kg and an acceleration of 2m/s/s. What is the force of the dog?
Anna35 [415]

Answer:

<h2>138 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 69 × 2

We have the final answer as

<h3>138 N</h3>

Hope this helps you

7 0
3 years ago
If your speed slows down you will have what acceleration?
vitfil [10]
When speed slows down you have kinetic energy.
8 0
4 years ago
Read 2 more answers
If a projectile travels in the air for 8 seconds when does a projectile reach its highest point
emmainna [20.7K]

Given

The projectile is in air for a time of t=8 sec

To find

The time it takes to reach the highest point

Explanation

A projectile moves up to the highest point and then again moves down following a parabolic path.

So it will reach the highest point at a time half the time it requires to follow teh parabolic path.

The time taken to reach the highest point is 4 sec

Conclusion

The time taken is 4 sec.

5 0
2 years ago
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