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melamori03 [73]
3 years ago
6

Which part of the electromagnetic spectrum is nearest to radio waves?

Physics
2 answers:
Sav [38]3 years ago
8 0

Answer:

i believe it is the microwaves that is closest to the radio waves.

Explanation:

AlladinOne [14]3 years ago
5 0

Answer:

b

Explanation:

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Heres a random question just for fun and out. of boredom ok so who started the nasa program? and why? what made him want to do t
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Eisenhower started the NASA project to develop technology for military application.
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3 years ago
A test tube of length L and cross-sectional area A is submerged in water with the open end down so that the edge of the tube is
Margaret [11]

Answer:

if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.

Explanation:

The air in the tube can be considered an ideal gas,

           P V = nR T

In that case we have the tube in the air where the pressure is P1 = P_atm, then we introduce the tube to the water to a depth H

For pressure the open end of the tube is

         P₂ = P_atm + ρ g H

Let's write the gas equation for the colon

            P₁ V₁ = P₂ V₂

            P_atm V₁ = (P_atm + ρ g H) V₂

             V₂ = V₁    P_atm / (P_atm + ρ g h)

If the air obeys Boyle's law e; volume within the had must decrease due to the increase in pressure, if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.

The main assumption is that the temperature during the experiment does not change

6 0
3 years ago
Two loudspeakers are 1.60 m apart. A person stands 3.00 m from one speaker and 3.50 m from the other. (a) What is the lowest fre
VMariaS [17]

Answer:

Explanation:

Given

Distance between two loud speakers d=1.6\ m

Distance of person from one speaker x_1=3\ m

Distance of person from second speaker x_2=3.5\ m

Path difference between the waves is given by

x_2-x_1=(2m+1)\cdot \frac{\lambda }{2}

for destructive interference m=0 I.e.

x_2-x_1=\frac{\lambda }{2}

3.5-3=\frac{\lambda }{2}

\lambda =0.5\times 2

\lambda =1\ m

frequency is given by

f=\frac{v}{\lambda }

where v=velocity\ of\ sound\ (v=343\ m/s)

f=\frac{343}{1}=343\ Hz

For next frequency which will cause destructive interference is

i.e. m=1 and m=2

3.5-3=\frac{2\cdot 1+1}{2}\cdot \lambda

\lambda =\frac{1}{3}\ m

frequency corresponding to this is

f_2=\frac{343}{\frac{1}{3}}=1029\ Hz

for m=2

3.5-3=\frac{5}{2}\cdot \lambda

\lambda =\frac{1}{5}\ m

Frequency corresponding to this wavelength

f_3=\frac{343}{\frac{1}{5}}

f_3=1715\ Hz                        

8 0
3 years ago
Un alambre de teléfono de 120m de largo y de 2.2mm de diámetro se estira debido a una fuerza de 380 N cual es el esfuerzo longit
sergij07 [2.7K]

Respuesta: verifique amablemente la explicación

Explicación:

Dado lo siguiente:

Longitud (L) del cable = 120 m

Diámetro (d) = 2,2 mm (2,2 / 1000) = 2,2 * 10 ^ -3 m

Fuerza (F) = 380 N

Esfuerzo longitudinal = Fuerza / Área

Área = πd² / 4 = (π * (2.2 * 10 ^ -3) ^ 2) / 4

Área = (3.142 * 4.84 * 10 ^ -6)

Área = 0.00000380132 m²

Estrés = Fuerza / Área

Estrés = 380 / 0.00000380132

Esfuerzo longitudinal = 99952128.12 = 9.9952128 * 10^7 Nm^-2

Deformación longitudinal: extensión / longitud

Extensión = 0.10 m

Longitud = 120 m

Deformación longitudinal = 0,1 m / 120 m

Deformación longitudinal = 0.0008333 = 8.33 × 10 ^ -4

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