What features of the organism are preserved in Amber

Two converging lenses are placed 30 cm apart. The focal length of the lens on the right is 20 cm while the focal length of the l

Masja [62]

Answer:

a) I2 = 3 (o-10) / (o- 30) , b) h ’/h= 3 (o-10) / o (o-30)

Explanation:

The builder's equation is

1 / f = 1 / o + 1 / i

Where f is the focal length, or e i are the distance to the object and image, respectively

As the separation between the lenses is greater than the focal distances, we must work them individually and separately. Let's start with the leftmost lens with focal length f = 15 cm

Let's calculate the position of the image of this lens

1 / i1 = 1 / f - 1 / o

1 / i1 = 1/15 - 1 / o

i1 = o 15 / (o-15)

Let's calculate the distance to the image of the second lens, for this the image of the first is the distance to the object of the second

o2 = d-i1

We write the builder equation

1 / f2 = 1 / o2 + 1 / i2

1 / i2 = 1 / f2 -1 / o2

1 / i2 = 1 / f2 - 1 / (d-i1)

1 / i2 = 1/20 - 1 / (d-i1) (1)

Let's evaluate the last term

d-i1 = d - 15 o / (o-15)

d-i1 = (d (o-15) - 15 o) / (o-15)

d- i1 = (30 or -30 15 -15 o) / (o-15)

d-i1 = (15 or - 450) / (o- 15)

d-i1 = = (15 or -450) / (o-15)

replace in 1

1 / i2 = 1/20 - (or - 15) / (15 or -450)

1 / i2 = [(15 o-450) - (o-15) 20] / (15 or -150)

1 / i2 = (15 or - 450 - 20 or + 300) / (15 or - 150)

1 / i2 = (-5 or -150) / (15 or -150)

1 / i2 = (or -30) / (3 or - 30)

I2 = 3 (o-10) / (o- 30)

Part B

The height of the image, we use the magnification equation

m = h ’/ h = - i / o

h ’= - h i / o

In our case

h ’= h i2 / o

h ’= h 3 (o-10) / o (o-30)

If they give the distance to the object it is easier

5 0

3 years ago

A 1. 0 μf capacitor is being charged by a 9. 0 v battery through a 10 mω resistor.

Advocard [28]

The potential across the capacitor at t = 1.0 seconds, 5.0 seconds, 20.0 seconds respectively is mathematically given as

t=0.476v
t=1.967v
V2=4.323v

<h3>What is the potential across the capacitor?</h3>

Question Parameters:

A 1. 0 μf capacitor is being charged by a 9. 0 v battery through a 10 mω resistor.

at

t = 1.0 seconds
5.0 seconds
20.0 seconds.

Generally, the equation for the Voltage is mathematically given as

v(t)=Vmax=(i-e^{-t/t})

Therefore

For t=1

V=5(i-e^{-1/10})

t=0.476v

For t=5s

V2=5(i-e^{-5/10})

t=1.967

For t=20s

V2=5(i-e^{-20/10})

V2=4.323v

Therefore, the values of voltages at the various times are

t=0.476v
t=1.967v
V2=4.323v

Read more about Voltage

brainly.com/question/14883923

Complete Question

A 1.0 μF capacitor is being charged by a 5.0 V battery through a 10 MΩ resistor.

Determine the potential across the capacitor when t = 1.0 seconds, 5.0 seconds, 20.0 seconds.

7 0

2 years ago

A 1-kg rock is suspended from the tip of a horizontal meterstick at the 0-cm mark so that the meterstick barely balances like a

tigry1 [53]

Explanation:

Given that,

Mass if the rock, m = 1 kg

It is suspended from the tip of a horizontal meter stick at the 0-cm mark so that the meter stick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark.

We need to find the mass of the meter stick. The force acting by the stone is

F = 1 × 9.8 = 9.8 N

Let W be the weight of the meter stick. If the net torque is zero on the stick then the stick does not move and it remains in equilibrium condition. So, taking torque about the pivot.

$9.8\times 12.5=W\times (50-12.5)\\\\W=\dfrac{9.8\times 12.5}{37.5}$

W = 3.266 N

The mass of the meters stick is :

$m=\dfrac{W}{g}\\\\m=\dfrac{3.266}{9.81}\\\\m=0.333\ kg$

So, the mass of the meter stick is 0.333 kg.

5 0

3 years ago

What holds the moon in place, orbiting around the Earth

AnnZ [28]

The Earth's gravity keeps the Moon orbiting us. It keeps changing the direction of the Moon's velocity. This means gravity makes the Moon accelerate all the time, even though its speed remains constant.

4 0

2 years ago

Two charges are placed on the x axis. One of the charges (q1 = +7.7 µC) is at x1 = +3.1 cm and the other (q2 = -19 µC) is at x2

Alinara [238K]

Answer:

a) $E=50.53\times 10^{6}\ N/C$