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Karolina [17]
3 years ago
9

What is the range of wavelengths that our eyes can see on the spectrum(this is called visible light spectrum)

Physics
1 answer:
Nookie1986 [14]3 years ago
6 0

Answer:

380 to 700 nanometers

Explanation:

The visible light spectrum is the segment of the electromagnetic spectrum that the human eye can view. More simply, this range of wavelengths is called visible light. Typically, the human eye can detect wavelengths from 380 to 700 nanometers.

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A string of mass 60.0 g and length 2.0 m is fixed at both ends and with 500 N in tension. a. If a wave is sent along this string
Darya [45]

Answer:

a

The  speed of  wave is   v_1  = 129.1 \ m/s

b

The new speed of the two waves is v =  129.1 \ m/s

Explanation:

From the question we are told that

    The mass of the string is  m  =  60 \ g  =  60 *10^{-3} \ kg

    The length is  l  =  2.0 \ m

    The tension is  T  = 500 \ N

Now the velocity of the first wave is mathematically represented as

     v_1  = \sqrt{ \frac{T}{\mu} }

Where  \mu is the linear density which is mathematically represented as

      \mu  =  \frac{m}{l}

substituting values    

     \mu  =  \frac{ 60 *10^{-3}}{2.0 }

     \mu  =  0.03\ kg/m

So

   v_1  = \sqrt{ \frac{500}{0.03} }

   v_1  = 129.1 \ m/s

Now given that the Tension, mass and length are constant the velocity of the second wave will same as that of first wave (reference PHYS 1100 )

     

8 0
4 years ago
The instantaneous emf resulting from magnetic induction equals the rate of change of flux is
katen-ka-za [31]

Lenz's Law: The polarity of the induced emf is such that it produces a current whose magnetic field opposes the change in magnetic flux through the loop.

7 0
3 years ago
Hey guys! Am I right? Thanks!
worty [1.4K]

Answer:

the correct answer is reduce friction

5 0
3 years ago
A certain freely falling object, released from rest, requires 1.95 s to travel the last 23.5 m before it hits the ground. (a) Fi
Ratling [72]

Answer:

(a). The velocity of the object is -2.496 m/s.

(b).  The total distance of the object travels during the fall is 23.80 m.

Explanation:

Given that,

Time = 1.95 s

Distance = 23.5 m

(a). We need to calculate the velocity

Using equation of motion

s = ut+\dfrac{1}{2}gt^2

Put  the value into the formula

-23.5=u\times1.95+\dfrac{1}{2}\times(-9.8)\times(1.95)^2

u=\dfrac{-23.5+4.9\times(1.95)^2}{1.95}

u=-2.496\ m/s

(b). We need to calculate the total distance the object travels during the fall

Using equation of motion

v = u+gt

Put the value in the equation

-2.496=0-9.8\times t

t =\dfrac{2.496}{9.8}

t=0.254\ sec

The total time is

t'=t+1.95

t'=0.254+1.95

t'=2.204\ sec

We need to calculate the distance

Using equation of motion

s = ut+\dfrac{1}{2}gt^2

Put the value into the formula

s=0+\dfrac{1}{2}\times9.8\times(2.204)^2

s=23.80\ m

Hence, (a). The velocity of the object is -2.496 m/s.

(b).  The total distance of the object travels during the fall is 23.80 m.

4 0
3 years ago
What is the rocks potential energy when it is 2000 meters above the ground?
zhuklara [117]

Answer:  

PE = 2032 J

m = 20 kg

g = 2 in/s2

h = 2000 m

8 0
3 years ago
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