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3 years ago

Electromagnetic radiation is:

Physics

1 answer:

ruslelena [56]3 years ago

5 0

Hello there.

<span>Electromagnetic radiation is:

</span><span>A- energy that is electric and magnetic
</span>

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What’s the power if a student does 2240j of work for 2.8 seconds

Alex787 [66]

Answer:

800 Watts

Explanation:

Power = Work/time

Working in SI units, Power = Watts, Work = Joules, Time = seconds.

Power = 2240J/2.8s = 800 Watts.

7 0

3 years ago

the electric current in a wire is 1.5 A. How many electrons flow past a given point in a time of 2 s?

34kurt

Answer:

The quantity of electrons that flows past a given point is 3.0 C.

Explanation:

An electric current (I) is the ratio of the quantity of charges (Q) that flows through a point to the time taken (t).

i.e I = $\frac{Q}{t}$

It is measured in Ampere's by the use of an ammeter in the laboratory. The quantity of charge that flow through a given point is measured in Coulombs, while time is measured in seconds.

Given that; I = 1.5A and t = 2s, find Q.

Q = It

= 1.5 × 2

= 3.0 C

The quantity of electrons that flows past a given point is 3.0 C.

6 0

3 years ago

Red light of wavelength 633 nmnm from a helium-neon laser passes through a slit 0.360 mmmm wide. The diffraction pattern is obse

777dan777 [17]

Answer:

Δy= 5,075 10⁻⁶ m

Explanation:

The expression that describes the interference phenomenon is

d sin θ = (m + ½) λ

As the observation is on a distant screen

tan θ = y / x

tan θ= sin θ/cos θ

As in ethanes I will experience the separation of the vines is small and the distance to the big screen

tan θ = sin θ

Let's replace

d y / x = (m + ½) λ

The width of a bright stripe at the difference in distance

y₁ = (m + ½) λ x / d

m = 1

y₁ = 3/2 λ x / d

Let's use m = 1, we look for the following interference,

m = 2

y₂ = (2+ ½) λ x / d

The distance to the screen is constant x₁ = x₂ = x₀

The width of the bright stripe is

Δy = λ x / d (5/2 -3/2)

Δy = 630 10⁻⁹ 2.90 /0.360 10⁻³ (1)

Δy= 5,075 10⁻⁶ m

8 0

3 years ago

The electron gun in a television tube is used to accelerate electrons with mass 9.109 × 10−31 kg from rest to 3 × 107 m/s within

zaharov [31]

Answer:

Electric field, E = 40608.75 N/C

Explanation:

It is given that,

Mass of electrons, $m=9.1\times 10^{-31}\ kg$

Initial speed of electron, u = 0

Final speed of electrons, $v=3\times 10^7\ m/s$

Distance traveled, s = 6.3 cm = 0.063 m

Firstly, we will find the acceleration of the electron using third equation of motion as :

$a=\dfrac{v^2-u^2}{2s}$

$a=\dfrac{(3\times 10^7)^2}{2\times 0.063}$

$a=7.14\times 10^{15}\ m/s^2$

Now we will find the electric field required in the tube as :

$ma=qE$

$E=\dfrac{ma}{q}$

$E=\dfrac{9.1\times 10^{-31}\times 7.14\times 10^{15}}{1.6\times 10^{-19}}$

E = 40608.75 N/C

So, the electric field required in the tube is 40608.75 N/C. Hence, this is the required solution.

3 0

3 years ago

A copper wire has a radius of 3.5 mm. When forces of a certain equal magnitude but opposite directions are applied to the ends o

denpristay [2]

Answer:

The tensile stress on the wire is 550 MPa.

Explanation:

Given;

Radius of copper wire, R = 3.5 mm

extension of the copper wire, e = 5.0×10⁻³ L

L is the original length of the copper wire,

Young's modulus for copper, Y = 11×10¹⁰Pa.

Young's modulus, Y is given as the ratio of tensile stress to tensile strain, measured in the same unit as Young's modulus.

$Y =\frac{Tensile \ stress}{Tensile \ strain} \\\\Tensile \ stress = Y*Tensile \ strain\\\\But, Tensile \ strain = \frac{extension}{original \ Length} = \frac{5.0*10^{-3} L}{L} = 5.0*10^{-3}\\\\Tensile \ stress = 11*10^{10} *5.0*10^{-3} \ = 550*10^6 \ Pa$

Therefore, the tensile stress on the wire is 550 MPa.

8 0

3 years ago