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dedylja [7]
2 years ago
11

An excited hydrogen atom releases an electromagnetic wave to return to its normal state. You use your futuristic dual electric/m

agnetic field tester on the electromagnetic wave to find the directions of the electric field and magnetic field. Your device tells you that the electric field is pointing in the positive y direction and the magnetic field is pointing in the positive x direction. In which direction does the released electromagnetic wave travel?
Physics
1 answer:
aniked [119]2 years ago
7 0

Answer: The released electromagnetic wave will travel in +y direction

Explanation:

It should be noted that, in a situation, whereby an excited hydrogen atom releases an electromagnetic wave to return to its normal state. And it's also evident that the futuristic dual electric/magnetic field tester on the electromagnetic wave to find the directions of the electric field and magnetic field is used. Eventually, your device tells you that the electric field is pointing in the positive y direction and the magnetic field is pointing in the positive x direction. Therefore, the released electromagnetic wave will travel in +y direction.

You might be interested in
Which optical device can focus light to a point through reflection?
nekit [7.7K]

Answer:

It is A

Explanation:

6 0
2 years ago
Compute the mean and maximum velocities for a liquid with a flow rate of 20 L/min in a 1.5-in nominal diameter sanitary pipeline
vlada-n [284]

Answer:

Mean velocity = 0.292 m/s

Maximum velocity = 0.584 m/s

The flow is laminar as Re = 229.2

Explanation:

D = 1.5 inches = 0.0381 m

Q = volumetric flow rate = 20 L/min = 0.000333 m³/s

Q = A × v

A = Cross sectional Area = πD²/4 = π(0.0381)²/4 = 0.00114 m²

v = average velocity

v = Q/A = 0.000333/0.00114 = 0.292 m/s

For flow in circular pipes, maximum velocity = 2 × average velocity = 2 × 0.292 = 0.584 m/s

To check if the flow is laminar or turbulent, we need its Reynolds number

Re = (ρvD)/μ

ρ = 1030 kg/m

v = 0.292 m/s

D = 1.5 inches = 0.0381 m

μ = 50 cP = 0.5 poise = 0.05 Pa.s

Re = (1030 × 0.292 × 0.0381)/0.05 = 229.2

For laminar flow, Re < 2100

For turbulent flow, Re > 4000

And 229.2 < 2100, hence, this flow is laminar.

7 0
2 years ago
A uniform 1.4-kg rod that is 0.75 m long is suspended at rest from the ceiling by two springs, one at each end of the rod. Both
svetlana [45]

Answer:

7 deg

Explanation:

m = mass of the rod = 1.4 kg

W = weight of the rod = mg = (1.4) (9.8) = 13.72 N

k_{L} = spring constant for left spring = 59 Nm^{-1}

k_{R} = spring constant for right spring = 33 Nm^{-1}

x_{L} = stretch in the left spring

x_{R} = stretch in the right spring

L = length of the rod = 0.75 m

\theta = Angle the rod makes with the horizontal

Using equilibrium of force in vertical direction for left spring

k_{L} x_{L} = (0.5) W\\(59) x_{L} = (0.5) (13.72)\\x_{L} = 0.116 m

Using equilibrium of force in vertical direction for right spring

k_{R} x_{R} = (0.5) W\\(33) x_{R} = (0.5) (13.72)\\x_{R} = 0.208 m

Angle made with the horizontal is given as

\theta = tan^{-1}(\frac{(x_{R} - x_{L})}{L} )\\\theta = tan^{-1}(\frac{(0.208 - 0.116)}{0.75} )\\\theta = 7 deg

3 0
2 years ago
A monkey weighs 6.00 x 102 N and swings from vine to vine. As the monkey grabs a new vine, both vines make an angle of 35.0° wit
zmey [24]

Answer:

T=366.23\ N

Explanation:

Given:

  • mass of monkey, w=600\ N
  • angle of vine from the vertical, \theta=35^{\circ}

Now follow the schematic to understand the symmetry and solution via Lami's theorem.

<u>The weight of the monkey will be balanced equally by the tension in both the vines:</u>

Using Lami's Theorem:

\frac{w}{sin\ 70^{\circ}} =\frac{T}{sin\ 145^{\circ}}

\frac{600}{sin\ 70^{\circ}} =\frac{T}{sin\ 145^{\circ} }

T=366.23\ N

4 0
2 years ago
Calculate the current through a 10.0-m long 22-gauge nichrome wire with a radius of 0.321 mm if it is connected across a 12.0-V
Kipish [7]

Answer:

Therefore,

Current through Nichrome wire is 0.3879 Ampere.

Explanation:

Given:

Length = l = 10 meter

Radius = r = 0.321\ mm =0.321\times 10^{-3}\ meter

Resistivity=\rho=1.00\times 10^{-6}\ ohm\ meter

V = 12 Volt

To Find:

Current, I =?

Solution:

Resistance for 0.0-m long 22-gauge nichrome wire with a radius of 0.321 mm if it is connected across a 12.0-V battery given as

R=\dfrac{\rho\times l}{A}

Where,

R = Resistance

l = length

A = Area of cross section = πr²

\rho=Resistivity=1.00\times 10^{-6}\ ohm\ meter

Substituting the values we get

R=\dfrac{1\times 10^{-6}\times 10}{3.14\times (0.321\times 10^{-3})^{2}}

R=\dfrac{1\times 10^{-5}}{3.23\times 10^{-7}}

R=\dfrac{1\times 10^{2}}{3.23}

R=30.95\ ohm

Now by Ohm's Law,

V= I\times R

Substituting the values we get

I=\dfrac{V}{R}=\dfrac{12}{30.95}=0.3876\ Ampere

Therefore,

Current through Nichrome wire is 0.3879 Ampere.

4 0
2 years ago
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