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julsineya [31]
2 years ago
11

You will begin with a relatively standard calculation.Consider a concave spherical mirror with a radius of curvature equal to 60

.0 centimeters. An object 6.00 centimeters tall is placed along the axis of the mirror, 45.0 centimeters from the mirror. You are to find the location and height of the image.a) What is the focal length Image for this mirror?b) Now use the spherical mirror equation to find the imagedistance s'.c) Find the magnification m,using sand s'.d) Finally, use the magnification to find the height of theimage y'.e) Solve the spherical mirror equation for s'.f)What is the magnification m? Use your answer from Part E
Physics
1 answer:
Strike441 [17]2 years ago
6 0

a) 30.0 cm

For any mirror, the radius of curvature is twice the focal length:

r = 2f

where

r is the radius of curvature

f is the focal length

For the mirror in this problem, we have

r = 60.0 cm is the radius of curvature

Therefore, solving the equation above for f, we find its focal length:

f=\frac{r}{2}=\frac{60.0 cm}{2}=30.0 cm

b) 90 cm

The mirror equation is:

\frac{1}{s'}=\frac{1}{f}-\frac{1}{s}

where

s' is the distance of the image from the mirror

f is the focal length

s is the distance of the object from the mirror

For the situation in the problem, we have

f = +30.0 cm is the focal length (positive for a concave mirror)

s = 45.0 cm is the object distance from the mirror

Solving the formula for s', we find

\frac{1}{s'}=\frac{1}{30.0 cm}-\frac{1}{45.0 cm}=0.011 cm^{-1}

s'=\frac{1}{0.011 cm^{-1}}=90 cm

c) -2

The magnification of the mirror is given by

M=-\frac{s'}{s}

where in this problem we have

s' = 90 cm is the image distance

s = 45.0 cm is the object distance

Solving the equation, we find:

M=-\frac{90 cm}{45 cm}=-2

So, the magnification is -2.

d) -12.0 cm

The magnification can also be rewritten as

M=\frac{y'}{y}

where

y' is the height of the image

y is the heigth of the object

In this problem, we know

y = 6.0 cm is the height of the object

M = -2 is the magnification

Solving the equation for y', we find

y'=My=(-2)(6.0 cm)=-12.0 cm

and the negative sign means that the image is inverted.

Part e and f are exactly identical as part b) and c).

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Nady [450]

Answer:

a)   W_total = 8240 J , b)  W₁ / W₂ = 1.1

Explanation:

In this exercise you are asked to calculate the work that is defined by

       W = F. dy

As the container is rising and the force is vertical the scalar product is reduced to the algebraic product.

       W = F dy = F Δy

let's apply this formula to our case

a) Let's use Newton's second law to calculate the force in the first y = 5 m

          F - W = m a

          W = mg

          F = m (a + g)

          F = 80 (1 + 9.8)

          F = 864 N

The work of this force we will call it W1

   We look for the force for the final 5 m, since the speed is constant the force must be equal to the weight (a = 0)

        F₂ - W = 0

        F₂ = W

        F₂ = 80 9.8

        F₂ = 784 N

The work of this fura we will call them W2

The total work is

         W_total = W₁ + W₂

         W_total = (F + F₂) y

         W_total = (864 + 784) 5

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b) To find the relationship between work with relate (W1) and work with constant speed (W2), let's use

        W₁ / W₂ = F y / F₂ y

        W₁ / W₂ = 864/784

        W₁ / W₂ = 1.1

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2 years ago
Technician A says test lights are great for quick tests on non-computerized circuits. Technician B says you can use a test light
Tpy6a [65]

Answer:

that technician A is right

Explanation:

The test lights are generally small bulbs that are turned on by the voltage and current flowing through the circuit in analog circuits, these two values ​​are high and can light the bulb. In digital circuits the current is very small in the order of milliamps, so there is not enough power to turn on these lights.

From the above it is seen that technician A is right

4 0
2 years ago
What force must be exerted on the master cylinder of a hydraulic lift to support the weight of a 2000-kg car (a large car) resti
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Archimedes principle states that

 

F1 / A1 = F2 / A2

F2 = (A2 / A1) * F1

 

Also, formula for the force is F = mg. Formula for the area of the cylinder is A = πr^2, therefore we get

 

F2 = (πr2^2 / πr1^2) * mg

 

Since the diameter of the cylinders are 2 cm and 24 cm, r1 = 12 and r2 = 1.

 

Substituting the values to the derived equation, we get

 

F2 = (π 1^2 / π 12^2) * 2400 * 9.8

F2 = 163.3333 N

 

 

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6 0
2 years ago
You are using a lightweight rope to pull a sled along level ground. The sled weighs 485 N, the coefficient of kinetic friction b
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Answer:

N=459.01N

Explanation:

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The component of the force on the y-axis can be obtained through the Pythagorean Theorem. This is because the components are the cathetus of a right triangle and its hypotenuse is the magnitude of the force:

sin12^\circ=\frac{F_y}{F}\\F_y=Fsin12^\circ

Replacing and solving for N:

N=W-Fsin12^\circ\\N=485N-(125N)sin12^\circ\\N=459.01N

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We know

\boxed{\sf Relative\:velocity(V_{AB})=V_A-V_B}

\\ \sf\longmapsto V_{AB}=2-2

\\ \sf\longmapsto V_{AB}=0m/s

6 0
2 years ago
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