A convex lens has a positive focal length and the object placed in front of it produce both virtual and real image <em>(image distance can be negative or positive depending on the nature of the image</em>).

According to the lens equation

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$ where;

f is the focal length of the lens

u is the object distance

v is the image distance

If the magnification is - 0.6

mag = v/u = -0.5

v = -0.5u

since v = 10cm

10 = -0.5u

u = -10/0.5

u =-20 cm

Substitute u = -20cm ( due to negative magnification)and v = 10cm into the lens formula to get the focal length f

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\\\frac{1}{f} = \frac{1}{-20} + \frac{1}{10}\\\frac{1}{f} = \frac{1}{-20} + \frac{1}{10}\\\frac{1}{f} = \frac{-1+2}{20} \\\frac{1}{f} = \frac{1}{20} \\cross \ multiply\\f = 20\\f = 20 cm$

Hence the focal length of the convex lens is 20cm

7 0

3 years ago

A majorette in the Rose Bowl Parade tosses a baton into the air with an initial angular velocity of 2.5 rev/s. If the baton unde

oksian1 [2.3K]

Answer:

14 rev

Explanation:

$w_{o}$ = initial angular velocity = 2.5 revs⁻¹

$w$ = final angular velocity = 0.8 revs⁻¹

$\alpha$ = Angular acceleration = - 0.2 revs⁻²

$\theta$ = Angular displacement

Using the equation

$w^{2} = w_{o}^{2} + 2 \alpha \theta\\0.8^{2} = 2.5^{2} + 2 (- 0.2) \theta\\ \theta = 14 rev$

So the number of revolutions are 14

4 0

3 years ago

The diagram below shows a closed system of two tanks that each contain water.

lidiya [134]

Lmalemwlsnlenekenfndelenekf

4 0

2 years ago

When a cup is placed on a table, which force prevents the cup from falling to the ground?

zhuklara [117]

Answer:

B. normal force

Explanation:

Because there is no frictional or resistance force. However gravitational force is applied downroad from the center of the cup thus the contact force that is perpendicular to the surface that an object contacts which is the normal force exerted upward from the table that prevents an object from falling.

6 0

3 years ago