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USPshnik [31]
3 years ago
5

An object is placed at 4cm distance in front of a concave mirror of radius of curvature 24cm.find the position of image.is the i

mage magnified? i want it with the graph digram
Physics
1 answer:
n200080 [17]3 years ago
6 0
Radius of curvature, r = 2f

24 cm = 2f

2f = 24 

f = 24/2 = 12

Hence the focal length is = 12 cm.

Object distance, u = 4 cm.

Since the object is placed between the focus and the pole of the mirror, at this position for a concave mirror, the resulting image would be virtual and would be behind the mirror.

1/u - 1/v = 1/f

1/4 - 1/v = 1/12

1/4 - 1/12 = 1/v

1/v = 1/4 - 1/12 =   (3 - 1)/12 = 2/12 = 1/6

1/v = 1/6

v = 6

Therefore the image is 6cm behind the mirror. 

Magnification = v/u = 6/4 = 3/2 = 1.5

Yes, the image is magnified by 1.5 times.

Simply copy the diagram for an object placed between the focus and the pole of a mirror from your textbook.   
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Mary is an avid game show fan and one of the contestants on a popular game show. She spins the wheel, and after 5.5 revolutions,
spin [16.1K]

Answer:

The angle is 23.2 radians, equivalent to 3.69 revolutions.

Explanation:

First, we need to find the angular acceleration of the wheel. This can be done using one of the kinematic formulas:

\omega^{2}=\omega_0^{2}+2\alpha\theta\\\\\implies \alpha=\frac{\omega^{2}-\omega_0^{2}}{2\theta}

Since the final angular velocity is zero after 5.5 revolutions (equivalent to 11π radians) we have that:

\alpha=\frac{-(3.15rad/s)^{2}}{2(11\pi rad)}\\\\\alpha=-0.144rad/s^{2}

Now, using the same equation, we can solve for the requested angle:

\theta=\frac{\omega^{2}-\omega_0^{2}}{2\alpha}\\\\\theta=\frac{(1.80rad/s)^{2}-(3.15rad/s)^{2}}{2(-0.144rad/s^{2})}\\\\\theta=23.2rad

Finally, it means that the angle through which the wheel has turned when the angular speed reaches 1.80 rad/s is 23.2 radians, equivalent to 3.69 revolutions.

8 0
3 years ago
you place a 0.13kg can of soup and a 0.34kg jar of pickles on the kitchen counter separated by a distance of 0.42m. what is the
Nonamiya [84]

Answer:

1.67×10^-11 N

Explanation:

m1 = 0.13kg

m2 = 0.34kg

d = 0.42m

G = 6.674 × 10 ^−11

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F= (6.674 × 10^ −11×0.13×0.34)/(0.42)²

F = 1.67×10^-11 N

4 0
4 years ago
A rock is thrown horizontally from a high building at 33.8 m/s. What is the magnitude of its velocity 4.25 s later?
Alex17521 [72]
<h2>Answer:53.63ms^{-2}</h2>

Explanation:

The equations of motion used in this question is v=u+at

When a object is projected horizontally from a sufficiently height,the x-component of acceleration remains zero because there is no force that drags the object in x direction.

But,due to gravity,the object accelerates downward at a rate of 9.8ms^{-2}.

In X-Direction,

Given that initial velocity=u_{x}=33.8ms^{-1}

Using v=u+at,

v_{x}=33.8+(0)4.25=33.8ms^{-1}

In Y-Direction,

Given that initial velocity=u_{x}=0ms^{-1}

Using v=u+at,

v_{y}=0+(9.8)4.25=41.65ms^{-1}

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Answer:

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