Ask question

Ask question

All categories

2 years ago

15

An object is 16.0cm to the left of a lens. The lens forms an image 36.0cm to the right of the lens. a) What is the focal length

of lens? (5 pts) b) Sketch the ray diagram considering an extended object? (Spts) c) Calculate the magnification? Describe the image type?

1 answer:

PSYCHO15rus [73]2 years ago

3 0

Answer:

-2.25

Real and inverted image

Explanation:

u = Object distance = 16 cm

v = Image distance = 36 cm

f = Focal length

Lens Equation

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}=\frac{1}{16}+\frac{1}{36}\\\Rightarrow \frac{1}{f}=\frac{13}{144}\\\Rightarrow f=\frac{144}{13}=11.07\ cm$

Focal length of the lens is 11.07 cm. Positive value indicates the lens is a convex lens

Magnification

$m=-\frac{v}{u}\\\Rightarrow m=-\frac{36}{16}\\\Rightarrow m=-2.25$

Since magnification is negative the image inverted

You might be interested in

When a driver presses the brake pedal, his car can stop with an acceleration of 5.4 m/s2. How far will the car travel while comi

notsponge [240]

$Given:\\a=5.4 \frac{m}{s^2} \\v_0= 25\frac{m}{s} \\\\Find:\\s=?\\\\Solution:\\\\s=v_0t- \frac{at^2}{2} \\\\a= \frac{\Delta v}{t} \\\\v_e=0\Rightarrow \Delta v=v_0\\\\t= \frac{v_0}{a} \\\\s= \frac{v_0^2}{a} - \frac{v_0^2}{2a} =\frac{v_0^2}{2a} \\\\s= \frac{ (25\frac{m}{s})2 }{2\cdot 5.4 \frac{m}{s^2} } \approx 57.9m\\\\C\rightarrow Correct\;answer$

3 0

3 years ago

To practice Problem-Solving Strategy 27.1: Magnetic Forces. A particle with mass 1.81×10−3 kgkg and a charge of 1.22×10−8 CC has

Roman55 [17]

Answer:

The magnitude of the acceleration is $a = 0.33 m/s^2$

The direction is $- \r k$ i.e the negative direction of the z-axis

Explanation:

From the question we are that

The mass of the particle $m = 1.8*10^{-3} kg$

The charge on the particle is $q = 1.22*10^{-8}C$

The velocity is $\= v = (3.0*10^4 m/s ) j$

The the magnetic field is $\= B = (1.63T)\r i + (0.980T) \r j$

The charge experienced a force which is mathematically represented as

$F = q (\= v * \= B)$

Substituting value

$F = 1.22*10^{-8} (( 3*10^4 ) \r j \ \ X \ \ ( 1.63 \r i + 0.980 \r j )T)$

$= 1.22 *10^{-8} ((3*10^4 * 1.63)(\r j \ \ X \ \ \r i) + (3*10^4 * 0.980) (\r j \ \ X \ \ \r j))$

$= (-5.966*10^4 N) \r k$

Note :

$i \ \ X \ \ j = k \\\\j \ \ X \ \ k = i\\\\k \ \ X \ \ i = j\\\\j \ \ X \ \ i = -k \\\\k \ \ X \ \ j = -i\\\\i \ \ X \ \ k = - j\\$

Now force is also mathematically represented as

$F = ma$

Making a the subject

$a = \frac{F}{m}$

Substituting values

$a =\frac{(-5.966*10^4) \r k}{1.81*10^{-3}}$

$= (-0.33m/s^2)\r k$

$= 0.33m/s^2 * (- \r k)$

6 0

3 years ago

A 3500kg suv is driving north at 36m/s what the momentum

lianna [129]

Your answer is

<span>126000</span>

4 0

2 years ago

A car travels 20 km South, then turns and travels 30 km East? What is the total displacement?

Ne4ueva [31]

Answer:

Explanation:36.05 km

Given

First car travels $r_1=20\ km$ South

then turns and travels $r_2=30\ km$ east

Suppose south as negative y axis and east as positive x axis

So, $r_1=-20\hat{j}$

$r_2=30\hat{i}$

Displacement is the shortest between initial and final point

Dispalcement $=r=r_1+r_2$

Displacement $=-20\hat{j}+30\hat{i}$

Displacement $=30\hat{i}-20\hat{j}$

Magnitude $=\sqrt{30^2+(-20)^2}$

Magnitude $=36.05\ km$

4 0

2 years ago

Consider an unknown charge that is released from rest at a particular location in an electric field so that it has some initial

sineoko [7]

Answer:

b)

Explanation:

If the charge is released at rest in an electric field, it will move along the electric field, going to regions of higher electric potential if it is a negative charge (against the field direction) and towards lower potential regions if it is positive (along the field). This means that the charge will gain kinetic energy, energy that only can come from a decrease in the electric potential energy.

For a positive charge: ΔEp = q*ΔV < 0 (as ΔV < 0)

For a negative charge: ΔEp = (-q) *ΔV < 0 (as ΔV > 0)

4 0

3 years ago