Ask question

Ask question

All categories

3 years ago

10

12. A concave lens has a focal length of 10 cm. An object 2.5 cm high is placed 30 cm from the lens. Determine the position and

size of the image. (3)

1 answer:

Nimfa-mama [501]3 years ago

6 0

Answer:

I think 9.5

Explanation:

............

You might be interested in

What would make oppositely charged objects attract each other more?

Lady_Fox [76]

Answer:

A. increasing the positive charge of the positively charged object and increasing the negative charge of the negatively charged object

Explanation:

4 0

3 years ago

Read 2 more answers

Which example demonstrates constant speed with changing direction?

Darya [45]

Ffffffddddffffffffffc

3 0

3 years ago

2. A 1.30-m long gas column that is open at one end and closed at the other end has a fundamental resonant frequency 80.0 Hz. Wh

Elina [12.6K]

To solve this problem, it will be necessary to apply the concepts related to the fundamental resonance frequency in a closed organ pipe.

This is mathematically given as

$f_n (2n+1)(\frac{v}{4L})$

For fundamental frequency n is 0, then,

$f_0 = \frac{v}{4L}$

When,

v = Velocity of sound

L = Length,

Rearranging to find the velocity,

$v = f_0 (4L)$

$v = (80Hz)(4)(1.3m)$

$v = 416m/s$

Therefore the speed of sound in this gas is 416m/s

7 0

4 years ago

Why are renewable sources of energy better than nonrenewable sources?

Anton [14]

Because it does not produce waste, thus it doesn't harm the environment. also renewable sources are infinite.

5 0

3 years ago

Read 2 more answers

A car of mass m accelerates from speed v_1 to speed v_2 while going up a slope that makes an angle theta with the horizontal. Th

Karo-lina-s [1.5K]

Answer:

Work done by external force is given as

$Work_{external} = mgLsin\theta + \mu mgLcos(\theta) + \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$

Explanation:

As per work energy Theorem we can say that work done by all force on the car is equal to change in kinetic energy of the car

so we will have

$Work_{external} + Work_{gravity} + Work_{friction} = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$

now we have

$W_{gravity} = -mg(Lsin\theta)$

$W_{friction} = -\mu mgcos(\theta) L$

so from above equation

$Work_{external} - mgLsin\theta - \mu mgLcos(\theta) = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$

so from above equation work done by external force is given as

$Work_{external} = mgLsin\theta + \mu mgLcos(\theta) + \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$

8 0

3 years ago