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fredd [130]
3 years ago
5

On the Em spectrum visible light between

Physics
1 answer:
Nesterboy [21]3 years ago
7 0
Hello, Thank you for posting your question here on brainly! feel free to ask as many as needed!

<span> the portion </span>of the electromagnetic spectrum<span> that is </span>visible <span>to the human eye. </span>Electromagnetic radiation<span> in this range of wavelengths is called </span>visible light<span> or simply </span>light<span>. A typical human eye will respond to wavelengths </span>from <span>about 390 to 700 nm. 
</span>I hope i helped you.
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the weight of an object on earth is 350 newtons. on mars the same object would weigh 134 newtons. what is the acceleration due t
Eddi Din [679]
Mass will remain constant on both planet, let mass of the object be "m".
let x be the <span>acceleration due to gravity on the surface of mars.
Weight of object on earth = m *g , where g is </span><span>acceleration due to gravity on the surface of earth
</span>⇒350 =  m *  9.8
⇒m = 350  /  9.8                        .............................equation(1)
Weight of object on mars = m * x , where x is acceleration due to gravity on the surface of mars
134  =   m   *  x                           .............................equation(2)
putting the value of m from equation (1) in equation(2) , we get,
x     =   (134  *  9.8) / 350
⇒  x   =   3.572 m/s²
7 0
4 years ago
What factors the weight of an object?
MAVERICK [17]

Answer:

Gravity

Explanation:

5 0
4 years ago
What are the characteristics of S waves?
Reptile [31]
I think the answer is D
6 0
4 years ago
The polar coordinates of the collar A are given as functions of time in seconds by r = 2+ 0.7 t2 ft and ????= 3.5t rad. What are
r-ruslan [8.4K]

Answer with explanation:

Part a)

v_{radial}=\frac{dr}{dt}=\frac{d(2+0.7t^{2})}{dt}\\\\v_{radial}=1.4t\\\\\therefore v_{radial}|_{t=4}=1.4\times 4=5.6ft/s\\\\v_{angular}=r|_{t=4}\times \frac{d\theta }{dt}=13.2\frac{3.5t}{dt}=46.2fts^{-1}\\\\\therefore v=\sqrt{v_{radial}^{2}+v_{angular}^{2}}\\\\v=46.53ft/s

Part b)

a_{radial}=\frac{d^{2}r}{dt^{2}}=\frac{d^{2}(2+0.7t^{2})}{dt^{2}}\\\\a_{radial}=1.4ft/s^{2}\\\\a_{angular}=r\times \frac{d^{2}\theta }{dt^{2}}\\\\a_{angular}=r\times \frac{d^{2}(3.5t) }{dt^{2}}\\\\\therefore a_{angular}=0\\\\\therefore Accleration=1.4ft/s^{2}

8 0
3 years ago
Receiver maxima problem. When the receiver moves through one cycle, how many maxima of the standing wave pattern does the receiv
shusha [124]

Answer:

Two.

Explanation:

  • A standing wave or stationary wave are the type of wave in which it oscillates with time but the amplitude does not move.
  • One complete wave consists of two loops.
  • It means that when the receiver moves through one cycle, two maxima of standing wave pattern the receiver pass through.
8 0
3 years ago
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