Average wavelength of radio waves

Answers

Answer 1

Answer:

The average wavelength of radio waves ranges from roughly two millimeters to more than 150 kilometers. The wavelengths of radio waves are the longest in the electromagnetic spectrum

What is Wavelength?

It can be understood in terms of the distance between any two similar successive points across any wave for example wavelength can be calculated by measuring the distance between any two successive crests.

It is the total length of the wave for which it completes one cycle.

The wavelength is inversely proportional to the frequency of the wave as from the following relation.

C = νλ

They also have the lowest frequencies, ranging from around 4,000 cycles per second, or 3 kilohertz, to roughly 280 billion hertz, or 280 gigahertz.

The wavelengths of radio waves are the longest in the electromagnetic spectrum, ranging from roughly two millimeters to more than 150 kilometers.

To learn more about wavelength from here, refer to the link given below;

brainly.com/question/7143261

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Answer 2

Answer:

Answer:

Radio waves have frequencies as high as 300 gigahertz(GHz)to as low as 30 hertz(Hz).At 300 GHz the corresponding wavelength is 1mm and at 30Hz is 10,000 km

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The knot at the junction is in equilibrium under the influence of four forces acting on it. The F force acts from above on the left at an angle of α with the horizontal. The 5.7 N force acts from above on the right at an angle of 50◦ with the horizontal. The 6.2 N force acts from below on the right at an angle of 44◦ with the horizontal. The 6.7 N force acts from below on the left at an angle of 43◦ with the horizontal.1. What is the magnitude of the force F?
2. What is the angle a of the force F in the figure above?

Answers

(a) The magnitude of the force F acting on the knot is 5.54 N.

(b) The angle α of the force F is 54.4⁰.

The given parameters:

F force at α
5.7 N force at 50⁰
6.2 N force at 44⁰
6.7 N force at 43⁰

The net vertical force on the knot is calculated as follows;

$F_y = Fsin(\alpha) + 5.7 sin(50) - 6.2 sin(44) - 6.7 sin(43)\n\nF_y = F sin(\alpha) -4.51\n\nFsin(\alpha) = 4.51$

The net horizontal force on the knot is calculated as follows;

$F_x = -F cos(\alpha) + 5.7 cos(50) + 6.2cos(44) - 6.7cos(43)\n\nF_x = -Fcos(\alpha) + 3.22\n\nFcos(\alpha) = 3.22$

From the trig identity;

$sin^2 \theta + cos^ 2 \theta = 1\n\n$

$(Fsin(\alpha))^2 + (Fcos(\alpha))^2 = (4.51)^2 + (3.22)^2\n\nF^2(sin^ 2\alpha + cos^2 \alpha) = 30.71\n\nF^2(1) = 30.71\n\nF = √(30.71) \n\nF = 5.54 \ N$

The angle α of the force F is calculated as follows;

$Fsin(\alpha) = 4.51\n\nsin(\alpha) = (4.51)/(F) \n\nsin(\alpha ) = (4.51)/(5.54) \n\nsin(\alpha ) = 0.814\n\n\alpha = sin^(-1)(0.814)\n\n\alpha = 54.5 \ ^0$

Find the image uploaded for the complete question.

Learn more about net force here:brainly.com/question/12582625

The knot is in equilbrium, so there is no net force acting on it. Starting with the unknown force and going clockwise, denote each force by F₁, F₂, F₃, and F₄, respectively. We have

F₁ + F₂ + F₃ + F₄ = 0

Decomposing each force into horizontal and vertical components, we have

F cos(180º - α) + (5.7 N) cos(50º) + (6.2 N) cos(-44º) + (6.7 N) cos(-137º) = 0

F sin(180º - α) + (5.7 N) sin(50º) + (6.2 N) sin(-44º) + (6.7 N) sin(-137º) = 0

Recall that cos(180º - x) = - cos(x) and sin(180º - x) = sin(x), so these equations reduce to

F cos(α) ≈ - 3.22 N

F sin(α) ≈ 4.51 N

(1) Recall that for all x, sin²(x) + cos²(x) = 1. Use this identity to solve for F :

(F cos(α))² + (F sin(α))² = F ² ≈ 30.73 N² → F ≈ 5.5 N

(2) Use the definition of tangent to solve for α :

tan(α) = sin(α) / cos(α) ≈ 1.399 → α ≈ 126º

or about 54º from the horizontal from above on the left of the knot.

The cycle is a process that returns to its beginning, but it does not repeatitself.
True
False

Answers

False. It does repeat itself

What is the weight on Earth of an object with mass 45 kg. Hint gravity = 10 N/kg *1 point
45 N
450 N
450 kg
10N

Answers

Answer:

450N

Explanation:

weight= m*g

weight=45*10

weight=450N

An emf is induced by rotating a 1207 turn, 20.0 cm diameter coil in the Earth's 4.13 10-5 T magnetic field. What average emf is induced, given the plane of the coil is originally perpendicular to the Earth's field and is rotated to be parallel to the field in 10.0 ms

Answers

Answer:

0.157 V

Explanation:

Parameters given:

Number of turns, N = 1207

Diameter of coil = 20 cm = 0.2 m

Radius of coil, r = 0.2/2 = 0.1 m

Magnetic field strength, B = $4.13 * 10^(-5) T$

Time interval, t = 10 ms = $10 * 10^(-3) = 0.01 s$

The average EMF induced in a coil due to a magnetic field is given as:

EMF = $(- N * A * B)/(t)$

where A = Area of coil

A = π $r^(2)$

Therefore, EMF will be:

$EMF = (- 1207 * 3.142 * 0.1^2 * 4.13 * 10^(-5))/(0.01) \n\n\nEMF = -0.157 V\n$

A uniform, solid sphere of radius 3.75 cm and mass 4.00 kg starts with a purely translational speed of 1.75 m/s at the top of an inclined plane. The surface of the incline is 3.00 m long, and is tilted at an angle of 26.0∘ with respect to the horizontal. Assuming the sphere rolls without slipping down the incline, calculate the sphere's final translational speed ????2 at the bottom of the ramp.

Answers

Answer:

v_2=4.53m/s $v_2=4.53m/s$

Explanation:

In order to solve the exercise it is necessary to apply the energy conservation equation,

The equation says the following,

$mgdsin(\theta)+(1)/(2)mv^2_1=(1)/(2)mv^2_2+(1)/(2)Iw^2$

Replacing the formula for I of a sphere, we have

$mgdsin(\theta)+(1)/(2)mv^2_1=(1)/(2)mv^2_2+(1)/(2)(2)/(5)mr^2((v_2)/(r))^2$

$mgdsin(\theta)+(1)/(2)mv^2_1=(1)/(2)mv^2_2+(1)/(5)mv^2_2=(7)/(10)mv^2_2$

$(10)/(7)gdsin(\theta)+(5)/(7)v^2_1=v^2_2$

In this way we get the expression

$v_2=\sqrt{(10)/(7)gdsin(\theta)+(5)/(7)v^2_1}$

We proceed to replace with the given values and obtain that

$v_2=\sqrt{(10)/(7)*9.8*3sin(26))+(5)/(7)*1.75^2}$

$v_2=4.53m/s$

v_2=4.53m/sv_2=4.53m/s

The equation says the following,

mgdsin(0) + 1/2mv2/1 = 1/2mv2/2 + 1/2Iw^2

Replacing the formula for I of a sphere,

mgdsin(0) + 1/2mv2/1 = 1/2mv2/2 + 1/2 2/5mr^2 (v2/r)^2

mgdsin(0) + 1/2mv2/1 = 1/2mv2/2 + 1/5mv2/2 = 7/10mv2/2

10/7gdsin(0) + 5/7v2/1 = v2/2

In this way, we get the expression

v2 = sqrt(10/7gdsin(0) + 5/7v2/1)

v2 = sqrt(10/7 * 9.8 * 3sin(26)) + 5/7 * 1.75^2

v2 = 4.53m/s

Further Explanation

The ball that rolls on the plane will experience two movements at once, namely the rotation of the axis of the ball and the translational field being traversed. Therefore, objects that do rolling motion have a rotational equation and a translational equation. The amount of kinetic energy possessed by the rolling body is the amount of rotational kinetic energy and translational kinetic energy. You will here learn about the ball rolling on a plane and incline.

An object can experience translational motion or rotational motion. Translational motion is the motion of objects whose direction is straight or curved. In translational motion using the concept of Newton II's law. While the rotational motion is the motion that has a rotation of a particular shaft. Rotational motion is caused by the torque, which is the tendency of a force to rotate a rigid body against a particular pivot point.

Learn More

Object Experience brainly.com/question/13696852

The ball that rolls brainly.com/question/13707126

Details

Grade: College

Subject: Physics

Keyword: object, ball, roll

An ac source of period T and maximum voltage V is connected to a single unknown ideal element that is either a resistor, and inductor, or a capacitor. At time t = 0 the voltage is zero. At time t = T/4 the current in the unknown element is equal to zero, and at time t = T/2 the current is I = -Imax, where Imax is the current amplitude. What is the unknown element?

Answers

Answer:

This satisfy the above given condition so we can say that this capacitor.

Explanation:

Let's take one by one option and check whether is wrong or right

For inductor:

$I=I_osin(wt-(\pi )/(2))$

Given that at t=T/4 ,I=0 and we know that

$w=(2\pi )/(T)$

So at T/4

$I=I_osin((2\pi )/(T)* (T)/(4)-(\pi )/(2))$

I=0 A

At t=T/2

$I=I_osin((2\pi )/(T)* (T)/(2)-(\pi )/(2))$

$I=I_o$

It means that this not a indutor.

For capacitor:

$I=I_osin(wt+(\pi )/(2))$

At T/4, I=0

At t=T/2

$I=I_osin((2\pi )/(T)* (T)/(2)+(\pi )/(2))$

$I= -I_o$

This satisfy the above given condition so we can say that this capacitor.

Final answer:

The nature of the unknown ideal element in the given AC circuit can be determined based on the phase difference between the current and voltage. In this case, since the current is zero at T/4 and a maximum at T/2, it suggests the current is lagging the voltage, indicating that the element in the circuit is a capacitor.

Explanation:

The question relates to an alternating current (AC) circuit connected to an unknown ideal element, and from the given conditions, it appears this element is a capacitor. Let us understand why.

Firstly, in an AC circuit, we can determine the nature of the circuit elements—resistor, inductor, or capacitor—based on the phase difference between the current and voltage. In a resistor, the current and voltage are in phase. In an inductor, the current lags behind the voltage by 90 degrees (or π/2 radians), whereas in a capacitor, the current leads the voltage by 90 degrees (or π/2 radians).

Based on the given problem, at time t = T/4, the current is zero. Considering that in one period of the AC voltage, it starts from zero, reaches a maximum, comes back to zero (at T/2), goes to a minimum (at 3T/4), and returns back to zero (at T), the current would reach its peak either at T/4 (if it's leading, a capacitor) or at 3T/4 (if it's lagging, an inductor). Here, since the current is zero at T/4 and it is a maximum at T/2 (albeit negative), it suggests the current is lagging the voltage, and hence, it suggests the element in the circuit is a capacitor.