Ask question

Ask question

All categories

3 years ago

7

A concave spherical mirror has a radius of curvature of magnitude 20.0 cm. (a) Find the location of the image for object distanc

es of (i) 40.0 cm, (ii) 20.0 cm, and (iii) 10.0 cm. For each case, state whether the image is (b) real or virtual and (c) upright or inverted. (d) Find the magnification in each case.

2 answers:

sp2606 [1]3 years ago

8 0

Explanation:

It is given that,

Radius of curvature of the mirror, R = 20 cm

So, focal length of the mirror, f = -10 cm

(i) Object distance, u = -40 cm

Using the mirror's formula as :

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

v is the image distance

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

$\dfrac{1}{v}=\dfrac{1}{-10}-\dfrac{1}{-40}$

v = -13.33 cm

Magnification of mirror is calculated as :

$m=\dfrac{-v}{u}$

$m=\dfrac{-(-13.33)}{(-40)}$

m = -0.33

Since, the magnification is negative, image is real and inverted.

(ii) Object distance, u = -20 cm

Using the mirror's formula as :

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

v is the image distance

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

$\dfrac{1}{v}=\dfrac{1}{-10}-\dfrac{1}{-20}$

v = -20 cm

Magnification of mirror is calculated as :

$m=\dfrac{-v}{u}$

$m=\dfrac{-(-20)}{(-40)}$

m = -0.5

Since, the magnification is negative, image is real and inverted.

(iii) Object distance, u = -10 cm

Using the mirror's formula as :

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

v is the image distance

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

$\dfrac{1}{v}=\dfrac{1}{-10}-\dfrac{1}{-10}$

v = infinity

Magnification of mirror is calculated as :

magnification = infinity

Hence, this is the required solution.

Tems11 [23]3 years ago

7 0

Answer:

Answered

Explanation:

The radius of curvature of the mirror R = 20 cm

then the focal length f = R/2 = 10 cm

(a) From mirror formula

1/f = 1/di + /1do

then the image distance

di = fd_o / d_o - f

= (10)(40) / 40-10

= 30.76 cm

since the image distance is positive so the image is real

ii) when the object distance d_0=20 cm

di = 10×20/ 20-10

= 20

Hence, the image must be real

iii)when the object distance d_0 = 10

di = 10×10 / 10-10 = ∞ (infinite)

the image will be formed at ∞

here also image will be real but diminished.

You might be interested in

An empty rubber balloon has a mass of 12.5 g. The balloon is filled with helium at a density of 0.181 kg/m3. At this density the

Eduardwww [97]

Answer: 1.14 N

Explanation :

As any body submerged in a fluid, it receives an upward force equal to the weight of the fluid removed by the body, which can be expressed as follows:

Fb = δair . Vb . g = 1.29 kg/m3 . 4/3 π (0.294)3 m3. 9.8 m/s2

Fb = 1.34 N

In the downward direction, we have 2 external forces acting upon the balloon: gravity and the tension in the line, which sum must be equal to the buoyant force, as the balloon is at rest.

We can get the gravity force as follows:

Fg = (mb +mhe) g

The mass of helium can be calculated as the product of the density of the helium times the volume of the balloon (assumed to be a perfect sphere), as follows:

MHe = δHe . 4/3 π (0.294)3 m3 = 0.019 kg

Fg = (0.012 kg + 0.019 kg) . 9.8 m/s2 = 0.2 N

Equating both sides of Newton´s 2nd Law in the vertical direction:

T + Fg = Fb

T = Fb – Fg = 1.34 N – 0.2 N = 1.14 N

6 0

2 years ago

In ionic bonds what happens to electrons?

OverLord2011 [107]

I think the correct answer from the choices would be that metals donate electrons to nonmetals. Ionic bonding involves transfer of valence electrons. The metal looses its valence electrons which makes it a cation while the nonmetal accepts these electrons.

6 0

3 years ago

Read 2 more answers

Using Gauss's law, calculate the electric field at a point distance s from a long wire bearing uniform charge density. i need he

11111nata11111 [884]

Answer:

E = 2k $\frac{\lambda}{ r}$

Explanation:

Gauss's law states that the electric flux equals the wax charge between the dielectric permeability.

We must define a Gaussian surface that takes advantage of the symmetry of the problem, let's use a cylinder with the faces perpendicular to the line of charge. Therefore the angle between the cylinder side area has the same direction of the electric field which is radial.

Ф = ∫ E . dA = E ∫ dA = q_{int} /ε₀

tells us that the linear charge density is

λ = q_ {int} /l

q_ {int} = l λ

we substitute

E A = l λ /ε₀

is area of cylinder is

A = 2π r l

we substitute

E = $\frac{ l \ \lambda}{ \epsilon_o \ 2\pi \ r \ l }$

E = $\frac{\lambda}{ 2\pi \epsilon_o \ r}$

the amount

k = 1 / 4πε₀

E = 2k $\frac{\lambda}{ r}$

5 0

3 years ago

Which vector correctly indicates the direction of centripetal acceleration experienced by the car?

gogolik [260]

The answer is C.

I hope this helped!

5 0

3 years ago

Read 2 more answers

A dog, with a mass of 10.0 kg, is standing on a flatboat so that he is 22.5 m from the shore. He walks 7.8 m on the boat toward

marissa [1.9K]

Answer:16.096

Explanation:

Given

mass of dog $\left ( m_d\right )=10kg$

mass of boat $\left ( m_b\right )=46kg$

distance moved by dog relative to ground= $x_d$

distance moved by boat relative to ground= $x_b$

Distance moved by dog relative to boat=7.8m

There no net force on the system therefore centre of mass of system remains at its position

0= $m_d\times x_d+m_b\dot x_b$

0= $10\times x_d+46\dot x_b$

$x_d=-4.6x_b$

i.e. boat will move opposite to the direction of dog

Now

$|x_d|+|x_b|=7.8$

substituting $x_d$ value

$5.6|x_b|=7.8$

$|x_b|=1.392m$

$|x_d|=6.4032m$

now the dog is 22.5-6.403=16.096m from shore

4 0

3 years ago