PHYSICS COLLEGE

Which of the following are electromagnetic waves?a. Water wavesb. Radio wavesc. Sound wavesd. Seismic waves

Answers

Answer 1

Answer:

Answer:

Radio waves

Explanation:

Radio wavs are electromagnetic waves.

Hope this helped!

Answer 2

Answer: The answer is Radio Waves because it is electromagnetic

Related Questions

Why Coulomb force is called "Mutual Force"???????????????????????????????????????
Neon signs require about 12,000 V for their operation. Consider a neon-sign transformer that operates off 120- V lines. How many more turns should be on the secondary compared with the primary?
An engineer is designing a small toy car that a spring will launch from rest along a racetrack. She wants to maximize the kinetic energy of the toy car when it launches from the end of a compressed spring onto the track, but she can make only a slight adjustment to the initial conditions of the car. The speed of the car just as it moves away from the spring onto the track is called the launch speed. Which of the following modifications to the car design would have the greatest effect on increasing the kinetic energy of the car? Explain your reasoning.Decrease the mass of the car slightly.Increase the mass of the car slightly.Decrease the launch speed of the car slightly.Increase the launch speed of the car slightly.
1. Compare and contrast the SI and the English systems of measurement.
an object down, but this is not true. If you place a box of mass 8 kg on a moving horizontal conveyor belt, the friction force of the belt acting on the bottom of the box speeds up the box. At first there is some slipping, until the speed of the box catches up to the speed of the belt, which is 5 m/s. The coefficient of kinetic friction between box and belt is 0.6. (a) How much time does it take for the box to reach this final speed

What happens to the pressure in all parts of a confined fluid if the pressure in one part is increased? The pressure in the other parts remains the same.The pressure everywhere increases by different amounts depending on the area of each part.
The pressure everywhere increases by the same amount.
The pressure everywhere decreases to conserve total pressure.

Answers

Answer:

option C

Explanation:

the correct answer is option C

When in a confined fluid the pressure is increased in one part than the pressure will equally distribute in the whole system.

According to Pascal's law when pressure is increased in the confined system then the pressure will equally transfer in the whole system.

This law's application is used in machines like hydraulic jacks.

The most soaring vocal melody is in Johann Sebastian Bach's Mass in B minor. In one section, the basses, tenors, altos, and sopranos carry the melody from a low D to a high A. In concert pitch, these notes are now assigned frequencies of 146.8 Hz and 880.0 Hz. (Use 343 m/s as the speed of sound, and 1.20 kg/m3 as the density of air.)a. Find the wavelength of the initial note.
b. Find the wavelength of the final note.
c. Assume the choir sings the melody with a uniform sound level of 70.0 dB. Find the pressure amplitude of the initial note.
d. Find the pressure amplitude of the final note.
e. Find the displacement amplitude of the initial note.
f. Find the displacement amplitude of the final note.

Answers

Answer:

Detailed step wise solution is attached below

Explanation:

(a) wavelength of the initial note 2.34 meters

(b) wavelength of the final note 0.389 meters

(d) pressure amplitude of the final note 0.09 Pa

(e) displacement amplitude of the initial note 4.78*10^(-7) meters

(f) displacement amplitude of the final note 3.95*10^(-8) meters

A 0.454-kg block is attached to a horizontal spring that is at its equilibrium length, and whose force constant is 21.0 N/m. The block rests on a frictionless surface. A 5.30×10?2-kg wad of putty is thrown horizontally at the block, hitting it with a speed of 8.97 m/s and sticking.Part AHow far does the putty-block system compress the spring?

Answers

The distance the putty-block system compress the spring is 0.15 meter.

Given the following data:

Mass = 0.454 kg

Spring constant = 21.0 N/m.

Mass of putty = $5.30* 10^(-2)\;kg$

Speed = 8.97 m/s

To determine how far (distance) the putty-block system compress the spring:

First of all, we would solver for the initialmomentum of the putty.

$P_p = mass * velocity\n\nP_p = 5.30* 10^(-2)* 8.97\n\nP_p = 47.54 * 10^(-2) \;kgm/s$

Next, we would apply the law of conservation of momentum to find the final velocity of the putty-block system:

$P_p = (M_b + M_p)V\n\n47.54* 10^(-2) = (0.454 + 5.30* 10^(-2))V\n\n47.54* 10^(-2) = 0.507V\n\nV = (0.4754)/(0.507)$

Velocity, V = 0.94 m/s

To find the compression distance, we would apply the law of conservation of energy:

$U_E = K_E\n\n(1)/(2) kx^2 = (1)/(2) mv^2\n\nkx^2 =M_(bp)v^2\n\nx^2 = (M_(bp)v^2)/(k) \n\nx^2 = ((0.454 + 5.30* 10^(-2)) * 0.94^2)/(21)\n\nx^2 = ((0.507 * 0.8836))/(21)\n\nx^2 = ((0.4480))/(21)\n\nx=√(0.0213)$

x = 0.15 meter

Answer:

Explanation:

Force constant of spring K = 21 N /m

we shall find the common velocity of putty-block system from law of conservation of momentum .

Initial momentum of putty

= 5.3 x 10⁻² x 8.97

= 47.54 x 10⁻² kg m/s

If common velocity after collision be V

47.54 x 10⁻² = ( 5.3x 10⁻² + .454) x V

V = .937 m/s

If x be compression on hitting the putty

1/2 k x² = 1/2 m V²

21 x² = ( 5.3x 10⁻² + .454) x .937²

x² = .0212

x = .1456 m

14.56 cm

On a 30degrees day, there is an explosion. The sound is heard 3.4s after seeing the flash. How far away was the explosion

Answers

Answer:

Distance = 1.2 km (Approx)

Explanation:

Given:

Temperature (T) = 30°C

Total Time taken (t) = 3.4 Sec

We know that sound increases in the air sound increase nearly 0.60 m/s for every sound with temperature.

Speed of sound = 331 m/s

So,

V = (331 + 0.60T) m/s

V = 331+(0.60×300C) m/s

V = 349 m/s

Distance = speed × time

Distance = 349 × 3.4

Distance = 1,186.6 m

Distance = 1.2 km (Approx)

A bicycle racer is going downhill at 11.0 m/s when, to his horror, one of his 2.25 kg wheels comes off when he is 75.0 m above the foot of the hill. We can model the wheel as a thin-walled cylinder 85.0 cm in diameter and neglect the small mass of the spokes. (a) How fast is the wheel moving when it reaches the bottom of hill if it rolled without slipping all the way down? (b) How much total kinetic energy does it have when it reaches bottom of hill?

Answers

Answer:

a.) Speed V = 29.3 m/s

b.) K.E = 1931.6 J

Explanation: Please find the attached files for the solution

Final answer:

The wheel's speed at the bottom of the hill can be found through the conservation of energy equation considering both translational and rotational kinetic energy, while the total kinetic energy at the bottom of the hill is a sum of translational and rotational kinetic energy.

Explanation:

These two questions address the physics concepts of conservation of energy, kinetic energy, and rotational motion. To answer the first question, (a) How fast is the wheel moving when it reaches the bottom of the hill if it rolled without slipping all the way down?, we need to consider the potential energy the wheel has at the top of the hill is completely converted into kinetic energy at the bottom. This includes both translational and rotational kinetic energy. Solving for the final velocity, vf, which would be the speed of the wheel, we get vf = sqrt((2*g*h)/(1+I/(m*r^2))), where g is the acceleration due to gravity, h is the height of the hill, I is the moment of inertia of the wheel, m is the mass of the wheel, and r is the radius of the wheel.

For the second question, (b) How much total kinetic energy does it have when it reaches bottom of the hill?, we use the formula for total kinetic energy at the bottom of the hill, K= 0.5*m*v^2+0.5*I*(v/r)^2. Substituting the value of v found in the first part we find the kinetic energy which we can use the formula provided in the reference information.

Learn more about Conservation of Energy and Rotational Motion here:

brainly.com/question/13897993

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A long copper cylindrical shell of inner radius 5 cm and outer radius 8 cm surrounds concentrically a charged long aluminum rod of radius 1 cm with a charge density of 7 pC/m. All charges on the aluminum rod reside at its surface. The inner surface of the copper shell has exactly opposite charge to that of the aluminum rod while the outer surface of the copper shell has the same charge as the aluminum rod. Find the magnitude and direction of the electric field at points that are at the following distances from the center of the aluminum rod: (a) 0.5 cm,
(b) 1.5 cm,
(c) 2.5 cm,
(d) 3.5 cm,
(e) 7 cm.

Answers

Answer:

a. 0

b. 8.4N/C

c. 5.04N/C

d. 3.6 N/C

e. 1.8N/C

Explanation:

The following data are given

inner cylindrical radius,r=5cm

outer cylindrical radius R=8cm

Charge density,p=7pc/m

radius of rod= 1cm

a. at distance 0.5cm from the center of the rod, this point falls on the rod itself and since the charge spread out on the surface of the rod, there wont be any electric field inside the rod itself

Hence E=0 at 0.5cm

b. at 1.5cm i.e 0.015m

the electric field is expressed as

$E=(2*charge density)/(4\pi E_(0)r)\nE=(2*7*10^(-12) *9*10^(9))/(0.015)\nE=8.4N/C$

The direction of the field depends on the charge on the rod

c. at 2.5cm i.e 0.025m

the electric field is expressed as

$E=(2*charge density)/(4\pi E_(0)r)\nE=(2*7*10^(-12) *9*10^(9))/(0.025)\nE=5.04N/C$

The direction of the field depends on the charge on the rod

d. at 3.5cm i.e 0.035m this point is still within the rod and the inner cylinder

the electric field is expressed as

$E=(2*charge density)/(4\pi E_(0)r)\nE=(2*7*10^(-12) *9*10^(9))/(0.035)\nE=3.6N/C$

The direction of the field depends on the charge on the rod

e. at 7cm which is a point outside the rod and the cylinder, the electric field is

$E=(2*charge density)/(4\pi E_(0)r)\nE=(2*7*10^(-12) *9*10^(9))/(0.07)\nE=1.8N/C$

The direction of the field depends on the charge on the rod

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