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3 years ago

11

The filament of an electric bulb draws a current of 0.4 ampere. Calculate the amount of charge flowing through the filament if t

he bulb glows for 2 hours. Also calculate the number of electrons passed.

1 answer:

Marina86 [1]3 years ago

3 0

Here I =0.4 amp and t=2hr or 2×60×60 sec
We known I=nQ/t where Q is charge and n is no of charge.
n=It/Q
0.4x(2x60x60)/1.6x10^-19

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A small 20-kg canoe is floating downriver at a speed of 2 m/s. What is the canoe’s kinetic energy? A. 40 J B. 80 J C. 18 J

rusak2 [61]

A small 20-kg canoe is floating downriver at a speed of 2 m/s. 40 J is the canoe’s kinetic energy.

Answer: Option A

<u>Explanation:</u>

The given canoe has the mass and is being given to move at a speed. Therefore the kinetic energy of the canoe can be calculated using the following method,

Given that mass of the canoe = 20 kg and its speed =1 m/s

As we know that the Kinetic energy has the formula,

$\text {Kinetic energy}=\frac{1}{2} \boldsymbol{m} \boldsymbol{v}^{2}$

Therefore, substituting the value into the equation, we get,

$K . E .=\frac{1}{2} \times 20 \times 2^{2}$ = 40 J

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4 years ago

Read 2 more answers

A car starts from rest and undergoes an acceleration of 4.0 m/s/s for a time of 5.0 s. What is the final velocity of the car?

Kruka [31]

Answer:

20m/s

Explanation:

acceleration=final velocity-initial velocity/time

4.0m/s²=v m/s-0m/s/5.0sec

5.0sec×4.0m/s²=v m/s-0m/s×5.0m/s/5.0m/s

20m/s=v

7 0

3 years ago

Three long parallel wires each carry 2.0-A currents in the same direction. The wires are oriented vertically, and they pass thro

blagie [28]

Answer:

$21.2\times 10^{-6}$ T

Explanation:

$i$ = magnitude of current in each wire = 2.0 A

$a$ = length of the side of the square = 4 cm = 0.04 m

$r$ = length of the diagonal of the square = $\sqrt{2}$ a = $\sqrt{2}$ (0.04) = 0.057 m

$B$ = magnitude of magnetic field by wires at A and C

$B = \left ( \frac{\mu _{o}}{4\pi } \right )\left ( \frac{2i}{a} \right )$

$B = (10^{-7}) \left ( \frac{2(2)}{0.04} \right )$

$B = 10\times 10^{-6}$ T

$B'$ = magnitude of magnetic field by wire at B

$B' = \left ( \frac{\mu _{o}}{4\pi } \right )\left ( \frac{2i}{r} \right )$

$B' = (10^{-7}) \left ( \frac{2(2)}{0.057} \right )$

$B' = 7.02\times 10^{-6}$ T

Net magnitude of the magnetic field at D is given as

$B_{net} = \sqrt{B^{2}+B^{2}} + B'$

$B_{net} = \sqrt{2} B + B'$

$B_{net} = \sqrt{2} (10\times 10^{-6}) + (7.02\times 10^{-6})$

$B_{net} = 21.2\times 10^{-6}$ T

8 0

3 years ago

How many possible combinations are there for the values of l and ml when n = 2?

Alexeev081 [22]

For n = 2, l will have two values 0,1

For l = 0, ml = 0

l = 1, ml = -1,0,1

These are the possible combinations.

The n is the principal quantum number

l is the azimuthal quantum number

ml is the magnetic quantum number.

The formula to find the azimuthal quantum number is

l = 0,1,2,3 .......n-1

If n = 2, the azimuthal quantum number,

l = 0 , 1

Then, the ml can be found using the formula,

ml = 2l +1

For l = 0 , ml = 2(0) + 1

= 1

Therefore, ml will have only one value, which is 0

For l =1 , ml = 2(1) + 1

= 3

Therefore, the ml will have three values, they are -1,0,1

Learn more about the quantum numbers in

brainly.com/question/16977590

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3 0

1 year ago

Consider the interference pattern produced by two parallel slits of width a and separation d, in which d = 3a. The slits are ill

laila [671]

Answer:

a) m =1 θ = sin⁻¹ λ / d, m = 2 θ = sin⁻¹ ( λ / 2d) , c) m = 3

Explanation:

a) In the interference phenomenon the maxima are given by the expression

d sin θ = m λ

the maximum for m = 1 is at the angle

θ = sin⁻¹ λ / d

the second maximum m = 2

θ = sin⁻¹ ( λ / 2d)

the third maximum m = 3

θ = sin⁻¹ ( λ / 3d)

the fourth maximum m = 4

θ = sin⁻¹ ( λ / 4d)

b) If we take into account the effect of diffraction, the intensity of the maximums is modulated by the envelope of the diffraction of each slit.

I = I₀ cos² (Ф) (sin x / x)²

Ф = π d sin θ /λ

x = pi a sin θ /λ

where a is the width of the slits

with the values of part a are introduced in the expression and we can calculate intensity of each maximum

c) The interference phenomenon gives us maximums of equal intensity and is modulated by the diffraction phenomenon that presents a minimum, when the interference reaches this minimum and is no longer present

maximum interference d sin θ = m λ

first diffraction minimum a sin θ = λ

we divide the two expressions

d / a = m

In our case

3a / a = m

m = 3

order three is no longer visible

7 0

3 years ago