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s344n2d4d5 [400]
3 years ago
12

What kind of wave is light?

Physics
2 answers:
Aloiza [94]3 years ago
6 0
Electromagnetic waves<span> are made of oscillating magnetic and electric fields, and like all </span>waves<span>, they carry energy</span>
Setler [38]3 years ago
6 0
Light is a type of electromagnetic radiation, it is a transverse wave
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Nervous tissue makes up most of the
LiRa [457]

Answer:

, Nervous tissue is composed of two types of cells, neurons and glial cells. Neurons are the primary type of cell that most anyone associates with the nervous system. They are responsible for the computation and communication that the nervous system provides.

8 0
3 years ago
What is the speed of a light ray (f 5.09 x10 14 Hz) in corn oil?
oee [108]

Answer:

it a

Explanation:

5 0
3 years ago
Read 2 more answers
A water-skier is moving at a speed of 14.3 m/s. When she skis in the same direction as a traveling wave, she springs upward ever
kiruha [24]

Answer:

a) 1.95 m/s

b) 5.56 m

Explanation:

Given that:

Velocity of the skier (V_s) = 14.3 m/s

For the skier moving in the direction of the wave, we have:

Period (T) = 0.450 s

Relative velocity (V) of the skier in regard with the wave =  (V_s - V_w)

where:

V_s = velocity of the skier

V_w = velocity of the wave

The wavelength (\lambda) can be written as:

\lambda = (V_s-V_w)T

\lambda = (V_s-V_w) 0.450m ---------------> Equation (1)

For the skier moving opposite in the direction of the wave, we have:

Period (T) = 0.342 s

Relative velocity (V) of the skier in regard with the wave = (V_s + V_w)

The wavelength (\lambda) can be written as:

\lambda = (V_s+V_w)T

\lambda = (V_s+V_w) 0.342m   ------------------> Equation 2

Equating equation (1) and equation (2) and substituting  V_s  = 14.3 m/s ; we have:

(V_s-V_w) 0.450m  =  (V_s-V_w) 0.342m

0.450m(V_s)-0.450m(V_w)   =  0.342m(V_s)+0.342m(V_w)

Collecting the like terms; we have:

0.450m(V_s) - 0.342m(V_s) =  0.342m(V_w)+0.450m(V_w)

(V_s)(0.450m - 0.342m) =  (V_w)0.342m+0.450m

14.3m/s(0.450m - 0.342m) =  (V_w)0.342m+0.450m

14.3m/s(0.108m =  (V_w)0.792m

1.5444m^2/s =  (V_w)0.792m

(V_w) = \frac{1.5444m^2/s}{ 0.792m}

(V_w) = 1.95 m/s

b)

The Wavelength of the wave can be calculated using :  ( \lambda }) = (V_s-V_w) 0.450m

({\lambda}) = (14.3 m/s -1.95 m/s)(0.450)

(\lambda) = (12.35)0.450m

(\lambda)= 5.5575 m

λ ≅ 5.56 m

5 0
3 years ago
styrofoam has a density of 150 kg/m3 . what is the maximum mass that can hang without sinking from a 50-cm-diameter styrofoam sp
Bad White [126]

The correct answer is 9.45 kg.

The formula for calculating the density of an object is expressed as:

Density = mass / volume

Given :

Density = 150kg/m³

Volume of the sphere = negligible

r is the radius = 50/2 = 25cm = 0.25m

We can find the volume of the Styrofoam sphere by

=(4/3)πr³

Volume (V) = (4/3)*π*(0.25)³

Volume = 0.063 cm³

Now let us find the required mass:

Mass = density * volume                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                            

Mass = 150 * 0.063

Mass = 9.45 kg

Therefore, the maximum mass that can hang without sinking from a 50.0 cm -diameter Styrofoam sphere in water is 9.45 kg.

To learn more about mass, refer: brainly.com/question/18386490

#SPJ4

4 0
1 year ago
Two point charges of equal magnitude (and opposite sign) are 7.5 cm apart. At the midpoint of the line connecting them, their co
Shkiper50 [21]

Comment

The only reason you can do this is that the charges are the same. If they were not, the problem would not be possible.

Equation

The field equation is, in its simplest form,

E = kq/r^2

So each of the charges are pulling / pushing in the same direction. The equation becomes.

kq/r^2 - (-kq/r^2) = Field magnitude in N/C

Givens

  • K = 9 * 10^9 N m^2 / c^2
  • E = 45 N/C
  • r = 7.5/2 = 3.75 cm * ( 1 m / 100 cm) = 0.0375 m
  • Find Q

Solution

k*q/0.0375 ^2 - (-kq/0.0375^2) = 45 N/C           Combine

2*k*q / 0.0375^2 = 45 N/C                                  Divide by 2

kq /(0.0375^2) = 22.5 N/C                                   Multiply by 0.0375^2

kq = 22.5 * 0.0375 ^2                                           Find d^2

kq = 22.5 * 0.001406                                            Combine

kq = 0.03164 N/C * m^2                                        Divide by k

q = 0.03164 N * m^2 /C / 9*10^9 N m^2 / c^2

q = 2.84760 * 10 ^8 C

I've left the cancellation of the units for you. Notice that only 1 C is left and it is in the numerator as it should be.


7 0
4 years ago
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