Ask question

Ask question

All categories

andreyandreev [35.5K]

3 years ago

13

An object of height 2.7 cm is placed 29 cm in front of a diverging lens of focal length 16 cm. Behind the diverging lens, and 12

cm from it, there is a converging lens of the same focal length.
A. Find the location of the final image, in centimeters beyond the converging lens.
B. What is the magnification of the final image?

1 answer:

kati45 [8]3 years ago

4 0

Answer:

A) q = -8.488 cm , B) m = 0.29

Explanation:

A) For this exercise in geometric optics, we will use the equation of the constructor

$\frac{1}{f} = \frac{1}{p} + \frac{1}{q}$

where p and q are the distance to the object and image, respectively and f is the focal length

in our case the distance the object is p = 29 cm the focal length of a diverging lens is negative and indicates that it is f = - 12 cm

$\frac{1}{q} = \frac{1}{f} - \frac{1}{p}$

we calculate

$\frac{1}{q} = - \frac{1}{12} - \frac{1}{29}$

$\frac{1}{q}$ = - 0.1178

q = -8.488 cm

the negative sign indicates that the image is virtual

B) the magnification is given

$m = \frac{h'}{h} = - \frac{q}{p}$

we substitute

m = $- \frac{-8.488}{29}$

m = 0.29

the positive sign indicates that the image is right

You might be interested in

A car parked on level pavement exerts a force of 10,000 newtons on the ground. What force does the pavement exert back on the ca

natka813 [3]

Answer:

Normal force of 10,000N

Explanation:

From the question, the weight the car exerts on the pavement is 10,000N.

The pavement exerts upward and perpendicular contact force called normal force on the car to support its weight. Also, the normal force is equal and opposite to the weigh of the car.

Hence the pavement exerts normal force of 10,000N back on the car to prevent it from passing through it.

7 0

3 years ago

How many changes of state are shown?<br> 0<br> 1<br> 2<br> 3

ivann1987 [24]

Answer: 2

I just took the test.

You’re welcome

7 0

2 years ago

Read 2 more answers

You are a member of a geological team in Central Africa. Your team comes upon a wide river that is flowing east. You must determ

Dovator [93]

Answer:

(a). The width of the river is 90.5 m.

The current speed of the river is 3.96 m.

(b). The shortest time is 15.0 sec and we would end 59.4 m east of our starting point.

Explanation:

Given that,

Constant speed = 6.00 m/s

Time = 20.1 sec

Speed = 9.00 m/s

Time = 11.2 sec

We need to write a equation for to travel due north across the river,

Using equation for north

$v^2-c^2=\dfrac{w^2}{t^2}$

Put the value in the equation

$6.00^2-c^2=\dfrac{w^2}{(20.1)^2}$

$36-c^2=\dfrac{w^2}{404.01}$ ....(I)

We need to write a equation for to travel due south across the river,

Using equation for south

$v^2-c^2=\dfrac{w^2}{t^2}$

Put the value in the equation

$9.00^2-c^2=\dfrac{w^2}{(11.2)^2}$

$81-c^2=\dfrac{w^2}{125.44}$ ....(II)

(a). We need to calculate the wide of the river

Using equation (I) and (II)

$45=\dfrac{w^2}{125.44}-\dfrac{w^2}{404.01}$

$45=w^2(0.00549)$

$w^2=\dfrac{45}{0.00549}$

$w=\sqrt{\dfrac{45}{0.00549}}$

$w=90.5$

We need to calculate the current speed

Using equation (I)

$36-c^2=\dfrac{(90.5)^2}{(20.1)^2}$

$36-c^2=20.27$

$c^2=20.27-36$

$c=\sqrt{15.73}$

$c=3.96\ m/s$

(b). We need to calculate the shortest time

Using formula of time

$t=\dfrac{d}{v}$

$t=\dfrac{90.5}{6}$

$t=15.0\ sec$

We need to calculate the distance

Using formula of distance

$d=vt$

$d=3.96\times15.0$

$d=59.4\ m$

Hence, (a). The width of the river is 90.5 m.

The current speed of the river is 3.96 m.

(b). The shortest time is 15.0 sec and we would end 59.4 m east of our starting point.

3 0

3 years ago

Object A has mass mA = 9 kg and initial momentum vector pA,i = < 20, -6, 0 > kg · m/s, just before it strikes object B, wh

love history [14]

Answer:

$p= kg m/s$

Explanation:

Momentum is a vector quantity that represents the "amount of motion" of an object.

Mathematically, the momentum of an object is given by

$p=mv$

where

m is the mass of the object

v is the velocity

Since momentum is a vector, it also has a direction, which is the same as the velocity.

Therefore, if we have two objects, the total momentum of the two objects will be obtained from the vector sum of the individual momenta of the two objects.

In this problem we have:

$p_A= kg m/s$ is the momentum of object A

$p_B = kg m/s$ is the momentum of object B

Therefore, the total momentum of objects A and B can be obtained by adding each components of A to the corresponding component of B, so:

$p_x = 20 +6 = 26 kg m/s\\p_y = -6 +6 = 0 kg m/s\\p_z = 0 + 0 = 0 kg m/s$

So the total initial momentum is

$p= kg m/s$

6 0

3 years ago

Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1050 kg and was approaching at

Soloha48 [4]

Answer:

v = 11.0 m/s at 198.6° (18.6° south of west)

ΔKE = -145 kJ

Explanation:

I assume you want to find the final velocity and the change in kinetic energy.

Take east to be +x and north to be +y.

Momentum is conserved in the x direction:

(1050 kg) (0 m/s) + (750 kg) (-25.0 m/s) = (1050 kg + 750 kg) vₓ

vₓ = -10.4 m/s

Momentum is conserved in the y direction:

(1050 kg) (-6.00 m/s) + (750 kg) (0 m/s) = (1050 kg + 750 kg) vᵧ

vᵧ = -3.50 m/s

The magnitude of the final velocity is:

v² = (-10.4 m/s)² + (-3.50 m/s)²

v = 11.0 m/s

The direction of the final velocity is:

θ = atan(-3.50 m/s / -10.4 m/s)

θ = 198.6°

The initial kinetic energy is:

KE₀ = ½ (1050 kg) (6.00 m/s)² + ½ (750 kg) (25.0 m/s)²

KE₀ = 253,275 J

The final kinetic energy is:

KE = ½ (1800 kg) (11.0 m/s)²

KE = 108,682 J

The change in kinetic energy is:

ΔKE = 108,682 J − 253,275 J

ΔKE ≈ -145,000 J

7 0

3 years ago