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ss7ja [257]
3 years ago
6

In a 35 mm single lens reflex camera (SLR) the distance from the lens to the film is varied in order to focus on objects at vary

ing distances. Over what range must a lens of 45 mm focal length vary if the camera is to be able to focus on objects ranging in distance from infinity down to 1.4 m from the camera?
Physics
1 answer:
Evgesh-ka [11]3 years ago
8 0

Answer:

range of movement is 1.49 mm

Explanation:

given data

focal length = 45 mm

distance = 1.4 m

distance from the lens = 35 mm

distance from infinity down = 1.4 m =

to find out

range of movement

solution

we will apply here lens equation that is

1/f = 1/p + 1/q

here f = 45 and p =  infinity to 1400 mm

we find here image distance that is q

1/45 = 0 + 1/q             ......1

q = 45 mm

and

1/45 = 1/1400 + 1/q     ......2

q = 46.49

so range of movement

that is 46.49 - 45

range of movement is 1.49 mm

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W = 750 [J]

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What is the total amount of kinetic and potential energy in a system ?
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Its the sum of the potential energy and the kinetic energy

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Which of the following factors affects the pressure of an enclosed gas
PIT_PIT [208]

The answer is:

All the above

The explanation:

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because according to boyle's law when the temperature constant so the pressure and volume of a gas have an inverse relationship, when temperature is constant.

when:

PV = nRT

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3 0
3 years ago
Physics question 28 plz help me
Alexus [3.1K]

Answer:

a. I = 30 A

b. E = 1080000 J = 1080 KJ

c. ΔT = 12.86°C

d. Cost = $ 4.32  

Explanation:

a.

The current in the coil is given by Ohm's Law:

V = IR\\I = \frac{V}{R}

where,

I = current = ?

V = Voltage = 120 V

R = Resistance = 4 Ω

Therefore,

I = \frac{120\ V}{4 \Omega}\\

<u>I = 30 A</u>

<u></u>

b.

The energy can be calculated as:

E = VIt\\E = (120\ V)(30\ A)(5\ min)(\frac{60\ s}{1\ min})\\

<u>E = 1080000 J = 1080 KJ</u>

<u></u>

c.

For the increase in the temperature of water:

E = mC\Delta T\\

where,

m = mass of water = 20 kg

C = specific heat of water = 4.2 KJ/kg.°C

Therefore,

1080\ KJ = (20\ kg)(4.2\ KJ/kg.^oC)\Delta T

<u>ΔT = 12.86°C</u>

<u></u>

d.

First, we will calculate the total energy consumed:

E=(Power)(Time)\\E=VI(Time)\\E = (120\ V)(30\ A)(0.5\ h/d)(30\ d)\\E = 54000\ Wh\\E = 54 KWh

Now, for the cost:

Cost = (Unit\ Cost)(Energy)\\Cost = (\$ 0.08\KWh)(54\ KWh)

<u>Cost = $ 4.32</u>

7 0
3 years ago
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