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A 5.75 mm high firefly sits on the axis of, and 11.3 cm in front of, the thin lens A, whose focal length is 5.77 cm . Behind len

weeeeeb [17]

Answer

given,

focal length of lens A = 5.77 cm

focal length of lens B= 27.9 cm

flies distance from mirror = 11.3 m

now,

Using lens formula

$\dfrac{1}{f} = \dfrac{1}{p} + \dfrac{1}{q}$

$\dfrac{1}{5.77} = \dfrac{1}{11.3} + \dfrac{1}{q}$

q =11.79 cm

image of lens A is object of lens B

distance of lens = 59.9 - 11.79 = 48.11

now, Again applying lens formula

$\dfrac{1}{f} = \dfrac{1}{p} + \dfrac{1}{q'}$

$\dfrac{1}{27.9} = \dfrac{1}{48.11} + \dfrac{1}{q'}$

q' =66.41 cm

hence, the image distance from the second lens is equal to q' =66.41 cm

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3 years ago

n object moves with a constant speed of 30 m/s on a circular track of radius 150 m. What is the acceleration of the object

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Acceleration is zero because it is travelling at a constant speed.

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Se suelta un cuerpo desde la azotea de un edificio de 180m de alto. Luego de 3s se lanza verticalmente hacia abajo otro cuerpo.

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Answer:CcCcccccc

Explanation:

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if the strength of the magnetic feild at a is 200 units and the strength of the magnetic field at b is 50 units what is the dist

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It would be 100 units.

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3 years ago

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The edge of a flying disc with a radius of 0.13 m spins with a tangential speed of 3.3 m/s.

Aleks [24]

By definition, centripetal acceleration is given by:

$a = \frac{v ^ 2}{r}$

Where,

v: tangential disk speed

r: disk radius

Substituting values in the given equation we have:

$a =\frac{3.3^2}{0.13}\\a = 83.76923077$

Rounding the result we have:

$a = 83.8 \frac{m}{s^2}$

Answer:

The centripetal acceleration of the disc edge in m/s^2 is:

$a = 83.8 \frac{m}{s^2}$

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4 years ago

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