Which of the following types of waves is not part of the electromagnetic spectrum? A) microwaves
B) gamma rays
C) ultraviolet radiation
D) radio waves
E) sound waves

Answers

Answer 1

Answer:

Answer: Sound Waves

Explanation:

Sound waves are the only waves on this list that are not part of the electromagnetic spectrum. This is because sound waves require a medium to travel (molecules to transmit the sound waves), while waves on the electromagnetic spectrum do not require a medium. They are able to travel through space for example, while sound would not be able to.

Answer 2

Answer:

Sound waves (E) are not electromagnetic at all.

Microwaves, gamma rays, ultraviolet waves, and radio waves all are.

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

Answers

Answer:

Radio waves

Explanation:

Radio wavs are electromagnetic waves.

Hope this helped!

The answer is Radio Waves because it is electromagnetic

If a person’s weight is W on the surface of the earth, calculate what it would be, in terms of W, at the surface of (a) the moon;
(b) Mars;
(c) Jupiter.

Answers

Answer:b

Explanation:

A boy throws a 15 kg ball at 4.7 m/s to a 65 kg girl who is stationary and standing on a skateboard. After catching the ball, the girl is travelling at: a) 0.88 m/s b) 0 m/s c) 1.1 m/s d) 3.2 m/s

Answers

Answer:

$a)v_(f)=0.88m/s$

Explanation:

To solve this problem we use the Momentum's conservation Law, before and after the girl catch the ball:

$\n p_(1)=p_(2)\nm_(ball)*v_(o.ball)+m_(girl)*v_(o.girl) = m_(ball)*v_(f.ball) + m_(girl)*v_(f.girl)$ (1)

At the beginning the girl is stationary:

$v_(o.girl)=0m/s$ (2)

If the girl catch the ball, both have the same speed:

$v_(f.girl)=v_(f.ball)=v_(f)$ (3)

We replace (2) and (3) in (1):

$m_(ball)*v_(o.ball) = (m_(ball)+m_(girl))*v_(f) \n$

We can now solve the equation for v_{f}:

$v_(f)=(m_(ball)*v_(o.ball))/((m_(ball)+m_(girl)))=(15*4.7)/(15+65)=0.88m/s$

A capacitor consists of two closely spaced metal conductors of large area, separated by a thin insulating foil. It has an electrical capacity of 3000.0 μF and is charged to a potential difference of 60.0 V. Calculate the amount of energy stored in the capacitor. Tries 0/20 Calculate the charge on this capacitor when the electrical energy stored in the capacitor is 6.53 J. Tries 0/20 If the two plates of the capacitor have their separation increased by a factor of 5 while the charge on the plates remains constant, by what factor is the energy stored in the capacitor increased?

Answers

Answer:

1 = 5.4 J

2 = 0.1979 C

3 = 5

Explanation:

Energy in a capacitor, E is

E = 1/2 * C * V²

E = 1/2 * 3000*10^-6 * 60²

E = 1/2 * 3000*10^-6 * 3600

E = 1/2 * 10.8

E = 5.4 J

E = Q²/2C = 6.53 J

E * 2C = Q²

Q² = 6.53 * 2 * 3000*10^-6

Q² = 13.06 * 3000*10^-6

Q² = 0.03918

Q = √0.03918

Q = 0.1979 C

The Capacitor, C is inversely proportional to the distance of separation, D. Thus, if D is increased by 5 to be 5D, then C would be C/5. And therefore, our energy stored in the capacitor is increased by a factor of 5.

20.0 moles, 1840 g, of a nonvolatile solute, C 3H 8O 3 is added to a flask with an unknown amount of water and stirred. The solution is allowed to reach 90.0°C . The vapor pressure of pure water at this temperature is 528.8 mm Hg. The vapor pressure of the solution is 423.0 mm Hg. How many kg of water was present?

Answers

Answer:

0.144 kg of water

Explanation:

From Raoult's law,

Mole fraction of solvent = vapor pressure of solution ÷ vapor pressure of solvent = 423 mmHg ÷ 528.8 mmHg = 0.8

Let the moles of solvent (water) be y

Moles of solute (C3H8O3) = 2 mole

Total moles of solution = moles of solvent + moles of solute = (y + 2) mol

Mole fraction of solvent = moles of solvent/total moles of solution

0.8 = y/(y + 2)

y = 0.8(y + 2)

y = 0.8y + 1.6

y - 0.8y = 1.6

0.2y = 1.6

y = 1.6/0.2 = 8

Moles of solvent (water) = 8 mol

Mass of water = moles of water × MW = 8 mol × 18 g/mol = 144 g = 144/1000 = 0.144 kg

A person is pushing a lawnmower of mass m D 38 kg and with h D 0:75 m, d D 0:25 m, `A D 0:28 m, and `B D 0:36 m. Assuming that the force exerted on the lawnmower by the person is completely horizontal and that the mass center of the lawnmower is at G, and neglecting the rotational inertia of the wheels, determine the minimum value of this force that causes the rear wheels (labeled A) to lift off the ground. In addition, determine the corresponding acceleration of the mower.