Which are electromagnetic waves? check all that apply.earthquake waves
infrared waves
ocean waves
radio waves
untraviolet waves

Answers

Answer 1

Answer:

Since electromagnetic waves do not require a medium for their transmission, the electromagnetic waves are radio waves, ultraviolet waves and infrared waves.

What are electromagnetic waves?

Electromagnetic waves or radiations are waves which occur as a result of the interaction between the electric and magnetic fields.

Electromagnetic waves do not require a material medium for their transmission and as such can travel through a vacuum.

Some examples of electromagnetic waves are radio waves, ultraviolet waves, microwaves, infrared waves etc.

Therefore, the electromagnetic waves are radio waves, ultraviolet waves and infrared waves.

Learn more about electromagnetic waves at: brainly.com/question/25847009

Answer 2

Answer: The electromagnetic waves are:
Radio waves
Ultraviolet waves
And Infrared waves
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Diethyl ether has a H°vap of 29.1 kJ/mol and a vapor pressure of 0.703 atm at 25.0°C. What is its vapor pressure at 60.0°C?

Answers

Answer:

The vapor pressure at 60.0°C is 2.416 atm

Explanation:

To solve this problem, we use Clausius-Clapeyron equation

$ln(P_2)/(P_1) = (-\delta H)/(R)[(1)/(T_2)-(1)/(T_1)]= (\delta H)/(R)[(1)/(T_1)-(1)/(T_2)]$

where;

Initial pressure P₁ = 0.703 atm

Initial Temperature T₁ = 25+273 = 298K

Final temperature T₂ = 60+273 = 333K

Change in enthalpy of vaporization ΔH = 29.1 KJ/mol = 29100J/mol

R is Boltzman constant = 8.314 J/K.mol

$ln(P_2)/(P_1) = (29100)/(8.314)[(1)/(298)-(1)/(333)] =1.23449$

$(P_2)/(P_1) = e^(1.23449)$ ⇒ $(P_2)/(P_1) = 3.43663$

P₂ = P₁ (3.43663) = (0.703 atm)(3.43663) = 2.416 atm

P₂ = 2.416 atm

Therefore, the vapor pressure at 60.0°C is 2.416 atm.

A Michelson interferometer operating at a 400 nm wavelength has a 3.95-cm-long glass cell in one arm. To begin, the air is pumped out of the cell and mirror M2 is adjusted to produce a bright spot at the center of the interference pattern. Then a valve is opened and air is slowly admitted into the cell. The index of refraction of air at 1.00 atmatm pressure is 1.00028.Required:
How many bright-dark-bright fringe shifts are observed as the cell fills with air?

Answers

Answer:

55.3

Explanation:

The computation of the number of bright-dark-bright fringe shifts observed is shown below:

$\triangle m = (2d)/(\lambda) (n - 1)$

where

d = $3.95 * 10^(-2)m$

$\lambda = 400 * 10^(-9)m$

n = 1.00028

Now placing these values to the above formula

So, the number of bright-dark-bright fringe shifts observed is

$= (2 *3.95 * 10^(-2)m)/(400 * 10^(-9)m) (1.00028 - 1)$

= 55.3

We simply applied the above formula so that the number of bright dark bright fringe shifts could come

Express the following speeds as a function of the speed of light, c: (a) an automobile speed (93 km/h) (b) the speed of sound (329 m/s) (c) the escape velocity of a rocket from the Earth's surface (12.1 km/s) (d) the orbital speed of the Earth about the Sun (Sun-Earth distance 1.5×108 km).

Answers

Answer:

(a). An automobile speed as a function of speed of light is $8.61*10^(-8)\ m/s$

(b). The speed of sound as a function of speed of light is $10.9*10^(-8)\ m/s$

(c). The escape velocity of a rocket as a function of speed of light is $4.03*10^(-8)\ m/s$

(d). The orbital speed of the Earth about the Sun is $3*10^(8)\ m/s$

Explanation:

Given that,

Express the following speeds as a function of the speed of light.

Automobile speed = 93 km/h

We know that,

A function of speed of light is

$c=3*10^(8)\ m/s$

(a). Automobile speed = 93 km/h

Speed $v_(a)=93*(5)/(18)$

$v_(a)=25.83\ m/s$

We need to express the speed of automobile speed as a function of speed of light

Using formula of speed

$v=(v_(a))/(v_(l))$

Put the value into the formula

$v=(25.83)/(3*10^(8))$

$v=8.61*10^(-8)\ m/s$

(b). The speed of sound is 329 m/s.

We need to express the speed of sound as a function of speed of light

Using formula of speed

$v=(v_(s))/(v_(l))$

Put the value into the formula

$v=(329)/(3*10^(8))$

$v=10.9*10^(-8)\ m/s$

(c). The escape velocity of a rocket from the Earth's surface is 12.1 m/s

We need to express the speed of sound as a function of speed of light

Using formula of speed

$v=(v_(e))/(v_(l))$

Put the value into the formula

$v=(12.1)/(3*10^(8))$

$v=4.03*10^(-8)\ m/s$

(d). The orbital speed of the Earth about the Sun

Distance = 1.5\times10^{8}[/tex]

We know that,

The sun rays reached on the earth in 8 min 20 sec.

We need to express the speed of sound as a function of speed of light

Using formula of speed

$v=(d)/(t)$

Put the value into the formula

$v=(1.5*10^(8)*1000)/(500)$

$v=3*10^(8)\ m/s$

Hence, (a). An automobile speed as a function of speed of light is $8.61*10^(-8)\ m/s$

(b). The speed of sound as a function of speed of light is $10.9*10^(-8)\ m/s$

(c). The escape velocity of a rocket as a function of speed of light is $4.03*10^(-8)\ m/s$

(d). The orbital speed of the Earth about the Sun is $3*10^(8)\ m/s$

What is displacement?

Answers

Displacement means when you move something from its original position. Let's say you want to sit on a chair. You move the chair from where it was originally placed. That's displacement.

While the block hovers in place, is the density of the block (top left) or the density of the liquid (bottom center) greater?

Answers

Answer:

for the body to float, the density of the body must be less than or equal to the density of the liquid.

Explanation:

For a block to float in a liquid, the thrust of the liquid must be greater than or equal to the weight of the block.

Weight is

W = mg

let's use the concept of density

ρ_body = m / V

m = ρ_body V

W = ρ_body V g

The thrust of the body is given by Archimedes' law

B = ρ_liquid g V_liquid

as the body floats the submerged volume of the liquid is less than or equal to the volume of the block

ρ_body V g = ρ_liquid g V_liquid

ρ_body = ρ liquid Vliquido / V_body

As we can see, for the body to float, the density of the body must be less than or equal to the density of the liquid.

A uniform 190 g rod with length 43 cm rotates in a horizontal plane about a fixed, vertical, frictionless pin through its center. Two small 38 g beads are mounted on the rod such that they are able to slide without friction along its length. Initially the beads are held by catches at positions 10 cm on each sides of the center, at which time the system rotates at an angular speed of 12 rad/s. Suddenly, the catches are released and the small beads slide outward along the rod. Find the angular speed of the system at the instant the beads reach the ends of the rod. Answer in units of rad/s.

Answers

Answer:

The angular speed of the system at the instant the beads reach the ends of the rod is 14.87 rad/s

Explanation:

Moment of inertia is given as;

I = ¹/₁₂×ML² + 2mr²

where;

I is the moment of inertia

M is the mass of the rod = 0.19 kg

L is the length of the rod = 0.43 m

m is the mass of the bead = 0.038 kg

r is the distance of one bead

Initial moment of inertial is given as;

$I_i = (1)/(12)ML^2 +2mr_1^2$

Final moment of inertia is also given as

$I_f= (1)/(12)ML^2 +2mr_2^2$

Angular momentum is the product of angular speed and moment of inertia;

= Iω

From the principle of conservation of angular momentum;

$I_i \omega_i = I__f } \omega_f$

$((1)/(12)ML^2 +2mr_1^2) \omega_i = ((1)/(12)ML^2 +2mr_2^2) \omega_f$

Given;

ωi = 12 rad/s

r₁ = 10.0 cm = 0.1 m

r₂ = 10.0cm/4 = 2.5 cm = 0.025 m

Substitute these values in the above equation, we will have;

$((1)/(12)*0.19*(0.43)^2 +2*0.038(0.1)^2) 12 = ((1)/(12)*0.19*(0.43)^2 +2*0.038*(0.025)^2) \omega_f\n\n0.04425 =0.002975\ \omega_f\n\n\omega_f = (0.04425)/(0.002975) = 14.87\ rad/s$