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dalvyx [7]
3 years ago
9

An object 16.8 cm tall is placed in front of a converging lens. A real image, 46 cm tall, is formed on the other side of the len

s. What is the magnification of the lens?
Physics
1 answer:
worty [1.4K]3 years ago
7 0

Answer:

2.74

Explanation:

Magnification = image distance/object distance

Mag = v/u

Given

v = 46cm

u = 16.8

Magnification = 46/16.8

Magnification = 2.74

Hence the magnification is 2.74

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One of your summer lunar space camp activities is to launch a 1130 kg1130 kg rocket from the surface of the Moon. You are a seri
maxonik [38]

Answer:

∆U = 2.296×10^10Joules

Explanation:

Gravitational potential energy is defined as the energy possessed by an object under the influence of gravity due to its virtue of position.

Potential energy U = Fr where;

F is the force of attraction between the masses of the moon and the rocket.

r is the radius or height of the object.

From Newton's law of universal gravitation, F = GMm/r²

Potential energy U = (-GMm/r²)×r

Potential energy U = -GMm/r

The force is negative because the objects act upward.

M is the mass of the rocket

m is the mass of the moon

Gravitational potential energy possessed by the rocket

U1 = -GMm/r1

r1 is the altitude covered by the rocket

Gravitational potential energy possessed by the Moon

U2 = -GMm/(r2+r1)

r2 is the radius of the moon

Change in gravitational potential energy ∆U = U2-U1

∆U = -GMm/(r2+r1)-(-GMm/r1)

∆U = -GMm/(r2+r1) + GMm/r1

∆U = -GMm{1/(r2+r1)-1/r1}

Given

G = 6.67×10^-11m³/kgs²

M = 1130kg

m = 7.36×10²²kg

r1 = 215km = 215,000m

r2 = 1740km = 1,740,000m

∆U = -6.67×10^-11× 7.36×10²² × 1130{1/(215,000+1,740,000)-1/215000}

∆U= -55.47×10¹⁴{1/1955000-1/215000}

∆U = -55.47×10¹⁴{5.12×10^-7 - 4.65×10^-6}

∆U = -284×10^7 + 257.94×10^8

∆U = 22,954,000,000Joules

∆U = 2.296×10^10Joules

8 0
3 years ago
Which of the following is a key assumption of the scientific method
lesantik [10]

Answer:

Though you have not gave the choices, I do believe it is “testing”

Explanation:

3 0
3 years ago
In which region of the ear does resonance allow the brain to interpret sound answer
lakkis [162]
I'm not too sure what your asking but here are two answers that may help.
The ear drum amplifies the vibrations.
The cochlea changes vibrations into electric signals.
7 0
3 years ago
Equation for inclined spring
zalisa [80]

Answer:

dbdbdheh eewr h eahehwGFTT5Q3JFX

Explanation:

FJGXJDGTFJSRRXAGFEWFWDdQDE

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What's the distance over which a wave's shape repeats
gulaghasi [49]
That's called the wave's "wavelength" .
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