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Wittaler [7]
3 years ago
8

Light rays from an object after passing through a convex lens form an image at the focal point behind the lens (opposite side of

the object). What is the actual position of the object? at the focal point in front of the lens at twice the focal length between the focal point and twice the focal length at a great distance, effectively infinite between the focal point and the lens NextReset
Physics
2 answers:
Colt1911 [192]3 years ago
8 0
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.


The actual position of the object is between the focal point and twice the focal length and <span>at the focal point in front of the lens. I think it would be A and C. </span>
BaLLatris [955]3 years ago
3 0

Answer:

effectively infinite between the focal point and the lens

Explanation:

Light rays from an object after passing through a convex lens form an image at the focal point behind the lens (opposite side of the object).

Now from lens formula we know that

\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}

now we know that image forms at focal point

so we will have

d_i = f

now from the equation above

\frac{1}{f} + \frac{1}{d_o} = \frac{1}{f}

so we will have

\frac{1}{d_o} = 0

so position of object is at infinite distance from lens

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3 years ago
If I exert a 203 N force on a 58 kg box, what will the acceleration of the box be?
Anit [1.1K]

Answer:

<h2>3.5 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

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a =  \frac{203}{58} =  \frac{7}{2}   \\

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<h3>3.5 m/s²</h3>

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7 0
3 years ago
A 2.0 kg, 20-cm-diameter turntable rotates at 100 rpm ons tionless bearings. Two 500 g blocks fall from above, hit the tum ble s
Verdich [7]

There are mistakes in the question.The correct question is here

A 2.0 kg, 20-cm-diameter turntable rotates at 100 rpm on frictionless bearings. Two 500 g blocks fall from above, hit the turntable simultaneously at opposite ends of a diameter, and stick. What is the turntable’s angular velocity, in rpm, just after this event?

Answer:

w=50 rpm

Explanation:

Given data

The mass turntable M=2kg

Diameter of the turntable d=20 cm=0.2 m

Angular velocity ω=100 rpm= 100×(2π/60) =10.47 rad/s

Two blocks Mass m=500 g=0.5 kg

To find

Turntable angular velocity

Solution

We can find the angular velocity of the turntable as follow

Lets consider turntable to be disk shape and the blocks to be small as compared to turntable

I_{turntable}w=I_{block1}w^{i}+I_{turntable}w^{i}+I_{block2}w^{i}

where I is moment of inertia

w^{i}=\frac{I_{turntable}w}{I_{block1}w^{i}+I_{block2}w^{i}+I_{turntable}w^{i}}\\   So\\I_{turntable}=M\frac{r^{2} }{2}\\I_{turntable}=2*(\frac{(0.2/2)}{2} )\\ I_{turntable}=0.01 \\And\\I_{block1}=I_{block2}=mr^{2}\\I_{block1}=I_{block2}=(0.5)*(0.2/2)^{2} \\ I_{block1}=I_{block2}==0.005\\so\\w^{i}=\frac{I_{turntable}w}{I_{block1}w^{i}+I_{block2}w^{i}+I_{turntable}w^{i}}\\w^{i}=\frac{0.01*(10.47)}{0.005+0.005+0.01} \\w^{i}=5.235 rad/s\\w^{i}=5.235*(60/2\pi )\\w^{i}=50 rpm

7 0
3 years ago
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