Ask question

Ask question

All categories

2 years ago

14

A 3.0-cm tall statue is 48 cm in front of a convex lens having a focal length of i. (2 pts) Is the image of the statue real or v

irtual (give reason (2 pts) for your answer)? magnitude 20 cm. ii. (5 pts) How far is the image from the lens? ii. 3 pts) ls the image upright or inverted? iv. (5 pts) How tall is the image?

1 answer:

erica [24]2 years ago

6 0

Explanation:

Given that,

Height of object = 3.0 cm

Distance of object u= 48 cm

Focal length = 48 cm

We need to calculate the image distance

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{20}=\dfrac{1}{v}+\dfrac{1}{-48}$

$\dfrac{1}{v}=\dfrac{1}{20}+\dfrac{1}{48}$

$\dfrac{1}{v}=\dfrac{17}{240}$

$v=14.11\ cm$

(I). The image is real.

(II). The distance of the image from the lens is 14.11 cm.

(II). The image is inverted.

(IV). We need to calculate the height of the image

Using formula of magnification

$m = \dfrac{v}{u}=\dfrac{h'}{h}$

$\dfrac{v}{u}=\dfrac{h'}{h}$

Put the value into the formula

$\dfrac{14.11}{48}=dfrac{h'}{3.0}$

$h'=\dfrac{14.11}{48}\times3.0$

$h'=0.88\ cm$

The height of the image is 0.88 cm.

Hence, This is the required solution.

You might be interested in

in physics lab, a cube slides down a frictionless incline as shown in the figure below, and elastically strikes another cube at

Tema [17]

<span>In the physics lab, a cube slides down a frictionless incline as shown in the figure below, check the image for the complete solution:

</span>

3 0

2 years ago

A steel cylinder at sea level contains air at a high pressure. Attached to the tank are two gauges, one that reads absolute pres

marishachu [46]

Answer:

C) The pressure reading stays the same.

Explanation:

3 0

2 years ago

Read 2 more answers

Two resistors of 5.0 and 9.0 ohm are connected in parallel. A 4.0 ohm resistor is then connected in series with the parallel com

rewona [7]

Answer:

I1 = 0.772 A

Explanation:

<u>Given</u>: R1 = 5.0 ohm, R2 = 9.0 ohm, R3 = 4.0 ohm, V = 6.0 Volts

<u>To find</u>: current I = ? A

<u>Solution: </u>

Ohm's law V= I R

⇒ I = V / R

In order to find R (total) we first find R (p) fro parallel combination. so

1 / R (p) = 1 / R1 + 1/ R2 ∴(P) stand for parallel

R (p) = R1R2 / ( R1 + R2)

R (p) = (5.0 × 9.0) / (5.0 + 9.0)

R (p) = 3.214 ohm

Now R (total) = R (p) + R3 (as R3 is connected in series)

R (total) = 3.214 ohm + 4.0 Ohm

R (total) = 7.214 ohm

now I (total) = 7.214 ohm / 6.0 Volts

I (total) = 1.202 A

This the total current supplied by 6 volts battery.

as voltage drop across R (p) = V = R (p) × I (total)

V (p) = 3.214 ohm × 1.202 A = 3.864 volts

Now current through 5 ohms resister is I1 = V (P) / R1

I1 = 3.864 volts / 5 ohm

I1 = 0.772 A

3 0

2 years ago

A 250 GeV beam of protons is fired over a distance of 1 km. If the initial size of the wave packet is 1 mm, find its final size

Margarita [4]

Answer:

The final size is approximately equal to the initial size due to a very small relative increase of $1.055\times 10^{- 7}$ in its size

Solution:

As per the question:

The energy of the proton beam, E = 250 GeV = $250\times 10^{9}\times 1.6\times 10^{- 19} = 4\times 10^{- 8} J$

Distance covered by photon, d = 1 km = 1000 m

Mass of proton, $m_{p} = 1.67\times 10^{- 27} kg$

The initial size of the wave packet, $\Delta t_{o} = 1 mm = 1\times 10^{- 3} m$

Now,

This is relativistic in nature

The rest mass energy associated with the proton is given by:

$E = m_{p}c^{2}$

$E = 1.67\times 10^{- 27}\times (3\times 10^{8})^{2} = 1.503\times 10^{- 10} J$

This energy of proton is $\simeq 250 GeV$

Thus the speed of the proton, v $\simeq c$

Now, the time taken to cover 1 km = 1000 m of the distance:

T = $\frac{1000}{v}$

T = $\frac{1000}{c} = \frac{1000}{3\times 10^{8}} = 3.34\times 10^{- 6} s$

Now, in accordance to the dispersion factor;

$\frac{\delta t_{o}}{\Delta t_{o}} = \frac{ht_{o}}{2\pi m_{p}\Delta t_{o}^{2}}$

$\frac{\delta t_{o}}{\Delta t_{o}} = \frac{6.626\times 10^{- 34}\times 3.34\times 10^{- 6}}{2\pi 1.67\times 10^{- 27}\times (10^{- 3})^{2} = 1.055\times 10^{- 7}$

Thus the increase in wave packet's width is relatively quite small.

Hence, we can say that:

$\Delta t_{o} = \Delta t$

where

$\Delta t$ = final width

3 0

2 years ago

Two objects, X and Y, move toward one another and eventually collide. Object X has a mass of 2M and is moving at a speed of 2v0

Sergeu [11.5K]

Answer:

c. The force exerted by X on Y is F to the right, and the force exerted by Yon X is F to the left.

Explanation:

If we take both objects as one single system, during the collision, assuming no external forces acting, the only forces present are the one that object X exerts on object Y, and the force that object Y does on object X.

These two forces, according to Newton's 3rd Law, form an action-reaction pair, and consequently, are equal in magnitude, acting in opposite directions.

As the object X is moving to the right, the force that produces, (F), is in the same direction (on Y), while for object Y, moving to the left, the force that produces (F also in magnitude) is in the same direction (on X), so the right answer is c.

The effect of the forces is different, due to masses are different, according Newton's 2nd Law.

6 0

2 years ago