All categories

2 years ago

What is the magnification when an object is placed at 2f from the pole of the convex mirror?

Physics

1 answer:

Gekata [30.6K]2 years ago

5 0

Answer:

Linear magnification = 1/3

Explanation:

Given:

Convex mirror

Object's distance from pole = 2f

Find:

Linear magnification

Computation:

Object distance, u = −2f

So,

1/v + 1/u = 1/f

1/v + 1/(-2f) = 1/f

1/v = 1/f + 1/2f

BY taking LCM

1/v = 3 / 2f

v = 2f / 3

Magnification, M = -v / u

So,

Magnification, M = (2f / 3) / 2f

Magnification, M = 2f / 6f

Magnification, M = 2 / 6

Linear magnification = 1/3

You might be interested in

Can someone help me and put them in order, I numbered them down so it can be easier to say.

Alexxx [7]

Answer:

the answer to this question is 2,4,3,1

5 0

3 years ago

Como estan formados los musculos?

VikaD [51]

Answer:

Explanation: El músculo esquelético está formado por fibras musculares, rodeadas de una capa de tejido conjuntivo, denominada endomisio. Las fibras se reúnen en fascículos primarios, que también están rodeados por otra capa de tejido conjuntivo, esta vez, más grueso, denominada perimisio.

3 0

3 years ago

Read 2 more answers

The Steamboat Geyser in Yellowstone National Park shoots water into the air at 48.0 m/s. How

qwelly [4]

Answer:

The maximum height reached by the water is 117.55 m.

Explanation:

Given;

initial velocity of the water, u = 48 m/s

at maximum height the final velocity will be zero, v = 0

the water is going upwards, i.e in the negative direction of gravity, g = -9.8 m/s².

The maximum height reached by the water is calculated as follows;

v² = u² + 2gh

where;

h is the maximum height reached by the water

0 = u² + 2gh

0 = (48)² + ( 2 x -9.8 x h)

0 = 2304 - 19.6h

19.6h = 2304

h = 2304 / 19.6

h = 117.55 m

Therefore, the maximum height reached by the water is 117.55 m.

7 0

2 years ago

Read 2 more answers

Un alumno menciona que al abrir la ventana de su casa sintió cómo el frío ingresaba a su cuerpo. Menciona cuál es la verdadera r

stepan [7]

Answer:

My believe the answer is

A.) or B.)

Explanation:

Here is why I think A is the answer.

If we use the process of elimination, it would look like this,

a) Porque el aire tiene una temperatura menor que la de su cuerpo; por eso se propaga más rápido.

<em>This makes sense because we all know in winter the weather is very cold and freezing.</em>

b) Porque la temperatura de su cuerpo, siente el aire frio que entra por la ventana.

<em>I feel like this answer is the question, but it could also be an answer, sorry, I'm a little uncertain.</em>

c) Porque el calor de su cuerpo se propaga al medio ambiente, al ser la temperatura del niño mayor que la del aire exterior.

<em>This answer has nothing to do with the question, plus it is very false, our body heat is not enough to overcome the very cold temperature from outside.</em>

d) Porque la temperatura del aire es igual a la temperatura del cuerpo.

<em>This is false because again our body heat is not even compared to the freezing cold temperatures from the winter.</em>

<em />

<h2>Well, have a nice rest of the day!</h2><h3>ba baiii!</h3>

3 0

3 years ago

Please help me with this physics prooblem

zaharov [31]

Take the missile's starting position to be the origin. Assuming the angles given are taken to be counterclockwise from the positive horizontal axis, the missile has position vector with components

$x=v_0\cos20.0^\circ t+\dfrac12a_xt^2$

$y=v_0\sin20.0^\circ t+\dfrac12a_yt^2$

The missile's final position after 9.20 s has to be a vector whose distance from the origin is 19,500 m and situated 32.0 deg relative the positive horizontal axis. This means the final position should have components

$x_{9.20\,\mathrm s}=(19,500\,\mathrm m)\cos32.0^\circ$

$y_{9.20\,\mathrm s}=(19,500\,\mathrm m)\sin32.0^\circ$

So we have enough information to solve for the components of the acceleration vector, $a_x$ and $a_y$ :

$x_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\cos20.0^\circ(9.20\,\mathrm s)+\dfrac12a_x(9.20\,\mathrm s)^2\implies a_x=21.0\,\dfrac{\mathrm m}{\mathrm s^2}$

$y_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\sin20.0^\circ(9.20\,\mathrm s)+\dfrac12a_y(9.20\,\mathrm s)^2\implies a_y=110\,\dfrac{\mathrm m}{\mathrm s^2}$

The acceleration vector then has direction $\theta$ where

$\tan\theta=\dfrac{a_y}{a_x}\implies\theta=79.2^\circ$

5 0

3 years ago