Ask question

Ask question

All categories

2 years ago

13

if an object 18mm high is placed 12mm from a diverging lens and the image is formed 4mm in front of the lens what is the height

of the image

1 answer:

brilliants [131]2 years ago

5 0

To solve this problem we use an amplification formula for divergent lenses

$M = \frac{i}{o} = \frac{h'}{h}$

Where:

i: distance of the image to the lens

o: Distance from the object to the lens

h = height of the object

h '= height of the image

$M = \frac{4mm}{12mm} = \frac{h'}{18mm}$

$h'= 18mm\frac{4mm}{12mm}$

h '= 6 mm

The height is 6 mm

You might be interested in

Charge g is distributed in a spherically symmetric ball of radius a. (a) Evaluate the average volume charge density p. (b) Now a

nasty-shy [4]

Answer:

Explanation:

The volume of a sphere is:

V = 4/3 * π * a^3

The volume charge density would then be:

p = Q/V

p = 3*Q/(4 * π * a^3)

If the charge density depends on the radius:

p = f(r) = k * r

I integrate the charge density in spherical coordinates. The charge density integrated in the whole volume is equal to total charge.

$Q = \int\limits^{2*\pi}_0\int\limits^\pi_0 \int\limits^r_0 {k * r} \, dr * r*d\theta* r*d\phi$

$Q = k *\int\limits^{2*\pi}_0\int\limits^\pi_0 \int\limits^r_0 {r^3} \, dr * d\theta* d\phi$

$Q = k *\int\limits^{2*\pi}_0\int\limits^\pi_0 {\frac{r^4}{4}} \, d\theta* d\phi$

$Q = k *\int\limits^{2*\pi}_0 {\frac{\pi r^4}{4}} \, d\phi$

$Q = \frac{\pi^2 r^4}{2}}$

Since p = k*r

Q = p*π^2*r^3 / 2

Then:

p(r) = 2*Q / (π^2*r^3)

3 0

2 years ago

Sheila weighs 60 kg and is riding a bike. Her momentum on the bike is 340 kg • m/s. The bike hits a rock, which stops it complet

Vikki [24]

Answer:

v₂ = 5.7 m/s

Explanation:

We will apply the law of conservation of momentum here:

$Total\ Initial\ Momentum = m_{1}v_{1} + m_{2}v_{2}\\$

where,

Total Initial Momentum = 340 kg.m/s

m₁ = mass of bike

v₁ = final speed of bike = 0 m/s

m₂ = mass of Sheila = 60 kg

v₂ = final speed of Sheila = ?

Therefore,

$340\ kg.m/s = m_{1}(0\ m/s) + (60\ kg)v_{2}\\v_{2} = \frac{340\ kg.m/s}{60\ kg}\\\\$

<u>v₂ = 5.7 m/s </u>

6 0

2 years ago

A 150-N box is being pulled horizontally in a wagon accelerating uniformly at 3.00 m/s2. The box does not move relative to the w

Zepler [3.9K]

Answer:

Frictional force, F = 45.9 N

Explanation:

It is given that,

Weight of the box, W = 150 N

Acceleration, $a=3\ m/s^2$

The coefficient of static friction between the box and the wagon's surface is 0.6 and the coefficient of kinetic friction is 0.4.

It is mentioned that the box does not move relative to the wagon. The force of friction is equal to the applied force. Let a is the acceleration. So,

$m=\dfrac{W}{g}$

$m=\dfrac{150}{9.8}$

$m=15.3\ kg$

Frictional force is given by :

$F=ma$

$F=15.3\times 3$

F = 45.9 N

So, the friction force on this box is closest to 45.9 N. Hence, this is the required solution.

8 0

3 years ago

Calculate the temperature change when 1000J of heat is supplied to 100g of water.<br> Please explain

jasenka [17]

Answer:

It is sensible heat- the amount of heat absorbed by 1 kg of water when heated at a constant pressure from freezing point 0 degree Celsius to the temperature of formation of steam i.e. saturation temperature

So it is given as - mass× specific heat × rise in temperature

i.e. 4.2 × T

4.2 × (100–0)

So it is 420kj

If you ask how much quantity of heat is required to convert 1 kg of ice into vapour then you have to add latent heat of fusion that is 336 kj/kg and latent heat of vaporization 2257 kj/kg (these two process occur at constant temperature so need to add rise in tempeature)

So it will be

Q= 1×336 + 1× 4.18 ×100 + 1× 2257

Q = 3011 kj

Or 3.1 Mj

Hope you got this!!!!!!

5 0

2 years ago

Read 2 more answers

Meteor Infrasound A meteor that explodes in the atmosphere creates infrasound waves that can travel multiple times around the gl

ladessa [460]

Answer:

The frequency of these waves is $4.27\times10^{-2}\ Hz$

Explanation:

Given that,

Wavelength = 6.6 km

Distance = 8810 km

Time t = 8.67 hr

We need to calculate the velocity of sound

Using formula of velocity

$v = \dfrac{D}{T}$

Where, D = distance

T = time

Put the value into the formula

$v =\dfrac{8810}{8.67}$

$v=1016\ km/hr$

We need to calculate the frequency

Using formula of frequency

$v=n\lambda$

$n=\dfrac{v}{\lambda}$

Put the value into the formula

$n=\dfrac{1016}{6.6}$

$n=153.93\ hr$

$n=\dfrac{153.93}{60\times60}$

$n=0.0427\ Hz$

$n=4.27\times10^{-2}\ Hz$

Hence, The frequency of these waves is $4.27\times10^{-2}\ Hz$

8 0

3 years ago