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3 years ago

What is the magnification of a virtual image if the image is 10.0 cm from a lens and the object is 30 cm from the lens

Physics

2 answers:

7nadin3 [17]3 years ago

5 0

Answer:

The magnification of a virtual image is -0.33

Explanation:

We can calculate the magnification as the ratio of height of the image to the height of the object. It can be also stated as the ratio of size of the image obtained to the original size of the object. Also this ratio can be equated to the ratio of image distance to object distance. So magnification can be termed as ratio of image distance to object distance. As in the present case, the image distance or the distance of the formation of image from the lens is 10 cm and the distance of object from the lens or the object distance is 30 cm. Then, $Magnification =-\frac{Image distance}{Object distance}=-\frac{10}{30}=-0.33$ .

So, the image will be diminished to 0.33 times the original size as the magnification of the virtual image is obtained as -0.33.

babymother [125]3 years ago

4 0

Answer:

The answer is 0.33

Explanation:

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1) The hoop and a solid disc rolling without slipping down an incline plane.

Their final velocities are proportional to their moment of inertia.

The condition for moment of inertia: v = ωR

We will use conservation of energy.

For the hoop:

$K_1 + U_1 = K_2 + U_2\\0 + m_hgh = \frac{1}{2}m_hv_h^2 + \frac{1}{2}I\omega_h^2 + 0$

They are released from rest, so their initial kinetic energy is zero. And when they reach the bottom, their final potential energy is also zero.

The moment of inertia of a hoop is

$I_h = m_hR^2$

Let's continue with the energy equations:

$m_h gh = \frac{1}{2}m_hv_h^2 + \frac{1}{2}(m_hR^2)(\frac{v_h^2}{R^2})\\m_hgh = \frac{1}{2}m_hv_h^2 + \frac{1}{2}m_hv_h^2\\m_hgh = m_hv_h^2\\v_h = \sqrt{gh}$

Similarly for the solid disk with a moment of inertia of (1/2)mR^2:

$K_1 + U_1 = K_2 + U_2\\m_dgh = \frac{1}{2}m_dv_d^2 + \frac{1}{2}I_d\omega_d^2\\m_dgh = \frac{1}{2}m_dv_d^2 + \frac{1}{2}(\frac{1}{2}m_dR^2)(\frac{v_d^2}{R^2})\\m_dgh = \frac{1}{2}m_dv_d^2 + \frac{1}{4}m_dv_d^2\\m_dgh = \frac{3}{4}m_dv_d^2\\v_d = \sqrt{\frac{4gh}{3}}$

Comparing the final velocities, we can conclude that the solid disk reaches the bottom first.

2) The angular acceleration of the pebble is equal to the angular acceleration of the tire, since they stuck together. We can deduce the angular acceleration of the tire from the linear acceleration of the bicycle.

The kinematics equations states that

$v = v_0 + at\\4.47 = 0 + 2a\\a = 2.235 ~m/s^2$

where a is the linear acceleration.

The relation with the angular and linear acceleration is

$a = \alpha R$

where R is the radius of the tire. Since it is not given in the question, we will leave it as R.

The angular acceleration of the small pebble is

$\alpha = 2.235/R ~m/s^2$

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