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neonofarm [45]
3 years ago
9

An object is placed 9.5 cm in front of a convex spherical mirror. Its image forms 3.2 cm behind the mirror. What is the radius o

f curvature of the mirror? Show all work.
Physics
2 answers:
o-na [289]3 years ago
8 0

Using the mirror formula.

1/v + 1/u = 1/f

1/9.5 + 1/3.2 = 1/f

1/f = 3.2 + 9.5 / 9.5 * 3.2

1/f = 4.82 cm

Radius = 2f

Radius = 2 x 4.82

Radius = 9.64 cm

gayaneshka [121]3 years ago
6 0

Answer:

Radius of curvature, R =  9.64 cm              

Explanation:

It is given that,

Object distance, u = -9.5 cm

Image distance, v = 3.2 cm

We have to find the radius of curvature of the mirror. The relationship between focal length and the radius of curvature as R = 2 f where f is the focal length of the mirror.

Using Mirror's formulas as :

\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}

\dfrac{1}{3.2}+\dfrac{1}{-9.5}=\dfrac{1}{f}

f = 4.82 cm

And radius of curvature of the mirror becomes, R = 9.64 cm. Hence, this is the required solution.

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A fire engine is moving south at 35 m/s while blowing its siren at a frequency of 400 Hz.
vodomira [7]

To solve this problem we will apply the concepts related to the Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. Mathematically it can be described as

f_d= f_s \frac{(v+v_d)}{(v-v_s)}

Here,

f_d=frequency received by detector

f_s=frequency of wave emitted by source

v_d=velocity of detector

v_s=velocity of source

v=velocity of sound wave

Replacing we have that,

f_d = 400(\frac{(343+18)}{(343-35)})

f_d=422 Hz

Therefore the frequencty that will hear the passengers is 422Hz

8 0
3 years ago
Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.1 m/s. Ignore f
tatiyna

Answer:

a) h=3.16 m, b)  v_{cm }^ = 6.43 m / s

Explanation:

a) For this exercise we can use the conservation of mechanical energy

Starting point. Highest on the hill

           Em₀ = U = mg h

final point. Lowest point

           Em_{f} = K

Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere

           K = ½ m v_{cm }^{2} + ½ I_{cm} w²

angular and linear speed are related

           v = w r

           w = v / r

            K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²

            Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)

as there are no friction losses, mechanical energy is conserved

             Em₀ = Em_{f}

             mg h = ½ v_{cm }^{2} (m + I_{cm} / r²)         (1)

             h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)

for the moment of inertia of a basketball we can approximate it to a spherical shell

             I_{cm} = ⅔ m r²

we substitute

            h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)

            h = ½ v_{cm }^{2}/g    5/3

             h = 5/6 v_{cm }^{2} / g

           

let's calculate

           h = 5/6 6.1 2 / 9.8

           h = 3.16 m

b) this part of the exercise we solve the speed of equation 1

          v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)

in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia

              I_{cm} = ½ m r²

we substitute

             v_{cm } = √ [2gh / (1 + ½)]

             v_{cm } = √(4/3 gh)

let's calculate

             v_{cm } = √ (4/3 9.8 3.16)

             v_{cm }^ = 6.43 m / s

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