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padilas [110]
3 years ago
11

A converging lens has a focal length of 20 cm. An object 1 cm tall is placed 10 cm from the center of the lens. What is the heig

ht of the image? (note: solve for the image distance first)
Physics
1 answer:
SCORPION-xisa [38]3 years ago
3 0

Answer: 2 cm

Explanation:

Given , for a converging lens

Focal length : f=20\ cm

Height of object : h=1\ cm

Object distabce from lens : u=-10\ cm

Using lens formula: \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}, we get

\dfrac{1}{20}=\dfrac{1}{v}+\dfrac{1}{10}, where v = image distance from the lens.

On solving aboive equation , we get

\dfrac{1}{v}=\dfrac{1}{20}-\dfrac{1}{10}=\dfrac{1-2}{20}=\dfrac{-1}{20}\Rightarrow\ v=-20\ cm

Formula of Magnification : m=\dfrac{v}{u}=\dfrac{h'}{h} , where h' is the height of image.

Put value of u, v and h in it , we get

\dfrac{-20}{-10}=\dfrac{h'}{1}\\\\\Rightarrow\ h'=2\ cm

Hence, the height of the image is 2 cm.

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Young's experiment is performed with light of wavelength 502 nm from excited helium atoms. Fringes are measured carefully on a s
jok3333 [9.3K]

Answer:

0.88 mm

Explanation:

Given

Wavelength of the atoms, λ = 502 nm = 502*10^-9 m

Radius of the screen away from the double slit, r = 1 m

We know that Y(20) = 11.4*10^-3 m

d = (20 * R * λ) / Y(20)

d = (20 * 1 * 502*10^-9)/11.4*10^-3

d = 1*10^-5 / 11.4*10^-3

d = 0.88 mm

Therefore, the distance of separation between the two slits is 0.88 mm

7 0
4 years ago
Which of the following (with specific heat capacity provided) would show the smallest temperature change upon gaining 200.0 J of
sergij07 [2.7K]

Answer:

A) 50.0 g Al

Explanation:

We can calculate the temperature change of each substance by using the equation:

\Delta T=\frac{Q}{mC_s}

where

Q = 200.0 J is the heat provided to the substance

m is the mass of the substance

C_s is the specific heat of the substance

Let's apply the formula for each substance:

A) m = 50.0 g, Cs = 0.903 J/g°C

\Delta T=\frac{200}{(50)(0.903)}=4.4^{\circ}C

B) m = 50.0 g, Cs = 0.385 J/g°C

\Delta T=\frac{200}{(50)(0.385)}=10.4^{\circ}C

C) m = 25.0 g, Cs = 0.79 J/g°C

\Delta T=\frac{200}{(25)(0.79)}=10.1^{\circ}C

D) m = 25.0 g, Cs = 0.128 J/g°C

\Delta T=\frac{200}{(25)(0.128)}=62.5^{\circ}C

E) m = 25.0 g, Cs = 0.235 J/g°C

\Delta T=\frac{200}{(25)(0.235)}=34.0^{\circ}C

As we can see, substance A) (Aluminium) is the one that undergoes the smallest temperature change.

7 0
3 years ago
Now imagine a person dragging a 50 kg box along the ground with a rope, as
ANTONII [103]

Answer:

The coefficient of static friction between the box and floor is, μ = 0.061

Explanation:

Given data,

The mass of the box, m = 50 kg

The force exerted by the person, F = 50 N

The time period of motion, t = 10 s

The frictional force acting on the box, f = 30 N

The normal force on the box, η = mg

                                                     = 50 x 9.8

                                                     = 490 N

The coefficient of friction,

                            μ = f/ η

                               = 30 / 490

                               = 0.061

Hence, the coefficient of static friction between the box and floor is, μ = 0.061

7 0
4 years ago
Two hundred turns of (insulated) copper wire are wrapped around a wooden cylindrical core of cross-sectional area 1.20 × 10−3 m2
Novosadov [1.4K]

Answer:

The charge flows through a point in the circuit during the change is 0.044 C.

Explanation:

Given that,

Number of turns in the copper wire, N = 200

Area of cross section, A=1.2\times 10^{-3}\ m^2

Resistance of the circuit, R = 118 ohms

If an externally applied uniform longitudinal magnetic filed in the core changes from 1.65 T in one direction to 1.65 T in the opposite direction.

We need to find the charge flows through a point in the circuit during the change. Due to change in magnetic field an emf is induced in it. It is given by :

\epsilon=\dfrac{d\phi}{dt}

Using Ohm's law :

\epsilon=IR

IR=-\dfrac{d\phi}{dt}\\\\I=-\dfrac{1}{R}\dfrac{d\phi}{dt}

Electric current is equal to the rate of change of electric charge. So,

dq=\dfrac{NA(B(0)-B(t))}{R}\\\\dq=\dfrac{200\times 1.2\times 10^{-3}(1.65+1.65)}{18}\\\\dq=0.044\ C

So, the charge flows through a point in the circuit during the change is 0.044 C.

4 0
3 years ago
The purpose of teaching high levels of math is to filter out students who lack logical intelligence
Pachacha [2.7K]
Answer: False I think that is worded weird tho
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3 years ago
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