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VashaNatasha [74]

2 years ago

7

An electric heater containing two heating wires X and Y is connected to a power supply of electromotive force(emf) 9.0V and negl

igible internal resistance as shown in fig 6.1
Wire X has a resistance of 2.4 ohm and wire Y has a resistance of 1.2ohm. A voltmeter is connected in parallel with the wires. A variable resistor is used to adjust the power dissipated in wires X and Y.

The variable resistor is adjusted so that the voltmeter reads 6.0V.

Calculate the resistance of the variable resistor.

1 answer:

mina [271]2 years ago

7 0

Answer:

0.4 ohms.

Explanation:

From the circuit,

The voltage reading in the voltmeter = voltage drop across each of the parallel resistance.

1/R' = 1/R1+1/R2

R' = (R1×R2)/(R1+R2)

R' = (2.4×1.2)/(2.4+1.2)

R' = 2.88/3.6

R' = 0.8 ohms.

Hence the current flowing through the circuit is

I = V'/R'................ Equation 1

Where V' = voltmeter reading

I = 6/0.8

I = 7.5 A

This is the same current that flows through the variable resistor.

Voltage drop across the variable resistor = 9-6 = 3 V

Therefore, the resistance of the variable resistor = 3/7.5

Resistance = 0.4 ohms.

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hodyreva [135]

20) When light passes from air to glass and then to air

21) When a light ray enters a medium with higher optical density, it bends towards the normal

22) Index of refraction describes the optical density

23) Light travels faster in the material with index 1.1

24) Glass refracts light more than water

25) Index of refraction is $n=\frac{c}{v}$

26) Critical angle: $[tex]sin \theta_c = \frac{n_2}{n_1}$ [/tex]

27) Critical angle is larger for the glass-water interface

Explanation:

20)

It is possible to slow down light and then speed it up again by making light passing from a medium with low optical density (for example, air) into a medium with higher optical density (for example, glass), and then make the light passing again from glass to air.

This phenomenon is known as refraction: when a light wave crosses the interface between two different mediums, it changes speed (and also direction). The speed decreases if the light passes from a medium at lower optical density to a medium with higher optical density, and viceversa.

21)

The change in direction of light when it passes through the boundary between two mediums is given by Snell's law:

$n_1 sin \theta_1 = n_2 sin \theta_2$

with

$n_1, n_2$ are the refractive index of 1st and 2nd medium

$\theta_1, \theta_2$ are the angle of incidence and refraction (the angle between the incident ray (or refracted ray) and the normal to the boundary)

The larger the optical density of the medium, the larger the value of n, the smaller the angle: so, when a light ray enters a medium with higher optical density, it bends towards the normal.

22)

The index of refraction describes the optical density of a medium. More in detail:

A high index of refraction means that the material has a high optical density, which means that light travels more slowly into that medium
A low index of refraction means that the material has a low optical density, which means that light travels faster into that medium

Be careful that optical density is a completely different property from density.

23)

As we said in part 22), the index of refraction describes the optical density of a medium.

In this case, we have:

A material with refractive index of 1.1
A material with refractive index of 2.2

As we said previously, light travels faster in materials with a lower refractive index: therefore in this case, light travels more quickly in material 1, which has a refractive index of only 1.1, than material 2, whose index of refraction is much higher (2.2).

24)

Rewriting Snell's law,

$sin \theta_2 = \frac{n_1}{n_2}sin \theta_1$ (1)

For light moving from air to water:

$n_1 \sim 1.00$ is the index of refraction of air

$n_2 = 1.33$ is the index of refraction ofwater

In this case, $\frac{n_1}{n_2}=\frac{1.00}{1.33}=0.75$

For light moving from air to glass,

$n_2 = 1.51$ is the index of refraction of glass

And so

$\frac{n_1}{n_2}=\frac{1.00}{1.51}=0.66$

From eq.(1), we see that the angle of refraction $\theta_2$ is smaller in the 2nd case: so glass refracts light more than water, because of its higher index of refraction.

25)

The index of refraction of a material is

$n=\frac{c}{v}$

c is the speed of light in a vacuum

v is the speed of light in the material

So, the index of refraction is inversely proportional to the speed of light in the material:

The higher the index of refraction, the slower the light
The lower the index of refraction, the faster the light

26)

From Snell's law,

$sin \theta_2 = \frac{n_1}{n_2}sin \theta_1$

We notice that when light moves from a medium with higher refractive index to a medium with lower refractive index, $n_1 > n_2$ , so $\frac{n_1}{n_2}>1$ , and since $sin \theta_2$ cannot be larger than 1, there exists a maximum value of the angle of incidence $\theta_c$ (called critical angle) above which refraction no longer occurs: in this case, the incident light ray is completely reflected into the original medium 1, and this phenomenon is called total internal reflection.

The value of the critical angle is given by

$sin \theta_c = \frac{n_2}{n_1}$

For angles of incidence above this value, total internal reflection occurs.

27)

Using:

$sin \theta_c = \frac{n_2}{n_1}$

For the interface glass-air,

$n_1 \sim 1.51\\n_2 = 1.00$

The critical angle is

$\theta_c = sin^{-1}(\frac{n_2}{n_1})=sin^{-1}(\frac{1.00}{1.51})=41.5^{\circ}$

For the interface glass-water,

$n_1 \sim 1.51\\n_2 = 1.33$

The critical angle is

$\theta_c = sin^{-1}(\frac{n_2}{n_1})=sin^{-1}(\frac{1.33}{1.51})=61.7^{\circ}$

So, the critical angle is larger for the glass-water interface.

Learn more about refraction:

brainly.com/question/3183125

brainly.com/question/12370040

#LearnwithBrainly

7 0

2 years ago

Someone fires a 0.04 kg bullet at a block of wood that has a mass of 0.5 kg. (The block of wood is sitting on a frictionless sur

d1i1m1o1n [39]

Answer:

The speed of bullet and wooden bock coupled together, V = 22.22 m/s

Explanation:

Given that,

Mass of the bullet, m = 0.04 Kg

Mass of the wooden block, M = 0.5 Kg

The initial velocity of the bullet, u = 300 m/s

The initial velocity of the wooden block, U = 0 m/s

The final velocity of the bullet and wooden bock coupled together, V = 0 m/s

According to the conservation of linear momentum, the total momentum of the body after impact is equal to the total momentum before impact.

Therefore,

mV + MV = mu + MU

V(m+M) = mu

V = mu/(m+M)

Substituting the values in the above equation,

V = 0.04 Kg x 300 m/s / (0.04 Kg+ 0.5 Kg)

= 22.22 m/s

Hence, the speed of bullet and wooden bock coupled together, V = 22.22 m/s

8 0

2 years ago

Scotesia swims from the north end to the south end of a 50.0 m pool in 20.0 s. As she begins to make the return trip , Sean, who

slega [8]

Answer:

a) 2.5m/s

b) 0.91m/s

c) 0m/s

Explanation:

Average velocity can be said to be the ratio of the displacement with respect to time.

Average speed on the other hand is the ratio of distance in relation to time

Thus, to get the average velocity for the first half of the swim

V(average) = displacement of first trip/time taken on the trip

V(average) = 50/20

V(average) = 2.5m/s

Average velocity for the second half of the swim will be calculated in like manner, thus,

V(average) = 50/55

V(average) = 0.91m/s

Average velocity for the round trip will then be

V(average) = 0/75, [50+25]

V(average) = 0m/s

3 0

2 years ago

Because the soles of your shoes have cleats, you can exert a forward force of 100 N even on slippery ice. A 10-kg picnic cooler

Brilliant_brown [7]

Answer:

you must throw 3 snowballs

Explanation:

We can solve this exercise using the concepts of conservation of the moment, let's define the system as formed by the refrigerator and all the snowballs. Let's write the moment

Initial. Before bumping that refrigerator

p₀ = n m v₀

Where n is the snowball number

Final. When the refrigerator moves

pf = (n m + M) v

The moment is preserved because the forces during the crash are internal

n m v₀ = (n m + M) v

n m (v₀ - v) = M v

n = M/m v/(vo-v)

Let's look for the initial velocity of the balls, suppose the person throws them with the maximum force if it slides in the snow (F = 100N), let's use the second law and Newton

F = m a

a = F / m

The distance the ball travels from zero speed to maximum speed is the extension of the arm (x = 1 m), let's look kinematically for the speed of the balls when leaving the arm

v₁² = v₀² + 2 a x

v₁² = 0+ 2 (100/1) 1

v₁ = 14.14 m / s

This is the initial speed for the crash

v₀ = v = 14.14 m / s

Let's calculate

n = M/m v/ (v₀-v)

n = 10/1 3 / (14.14 -3)

n = 2.7 balls

you must throw 3 snowballs

7 0

2 years ago

What is the net force required to give an automobile with a mass of 1,600 kg an acceleration of 4.5 m/s2

Paladinen [302]

Answer:

Fnet=7200 N

Explanation:

Fnet=mass x acceleration

mass= 1600kg acceleration=4.5m/s^2

8 0

2 years ago