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2 years ago

The visible light spectrum ranges between

Physics

2 answers:

natulia [17]2 years ago

8 0

Answer: 380 nm (violet-blue) and 780 nm (red)

Explanation:

The portion visible by the human eye of the electromagnetic spectrum is between 380 nm (violet-blue) and 780 nm (red) approximately.

This part of the spectrum is located between ultraviolet light and infrared light.

It should be noted that the fact only part of the whole electromagnetic spectrum is visible to humans is because the receptors in our eyes are only sensitive to these wavelengths.

aleksandr82 [10.1K]2 years ago

4 0

Red goes in the first blank and violet goes in the second blank if you're using plato

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A child sits 1.6 m from the center of a merry go round that makes one complete revolution in 4.7 s. What is his angular accelera

olga2289 [7]

Answer: angular acceleration = $2.86\ rad/sec^{2}$

Given:

Distance from center of axis = 1.6 m

Time taken to complete one revolution = 4.7 sec

Therefore, we can evaluate the angular acceleration using the following formula:

$\alpha = r\times \omega^{2}$

$\alpha = r\times (\frac{2\pi}{T})^{2}$

$\alpha = 1.6\times (\frac{2\times3.14}{4.7})^{2}$

$\alpha$ = $2.86\ rad/sec^{2}$

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3 years ago

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Answer:

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Explanation:

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2 years ago

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Difference between dry cell and simple cell

lakkis [162]

Simple cells have liquid chemicals, making it harder for it to carry. While as dry cells have no liquid chemicals, making it easier to carry.

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2 years ago

A crowbar 2727 in. long is pivoted 66 in. from the end. What force must be applied at the long end in order to lift a 600600 lb

NikAS [45]

To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.

$\sum \tau = F*d$

If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,

$F*(27-6)= 6*600$

$F = \frac{6*600}{21}$

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So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb

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2 years ago

You can use any coordinate system you like in order to solve a projectile motion problem. To demonstrate the truth of this state

posledela

Answer:

a) y₂ = 49.1 m , t = 1.02 s , b) y = 49.1 m , t= 1.02 s

Explanation:

a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero

$v_{y}$ ² = $v_{oy}$ ² - 2 g (y –yo)

The origin of the coordinate system is on the floor and the ball is thrown from a height

y-yo = $v_{oy}² /2 g
 y- 0 = 10.0²/2 9.8
 y - 0 = 5.10 m
 
The height from the ground is the height that rises from the reference system plus the depth of the ground from the reference system
 y₂ = 5.1 + 44
 y₂ = 49.1 m
Let's use the other equation to find the time
 [tex]v_{y}$ = $v_{oy}$ - g t

t = $v_{oy}$ / g

t = 10 / 9.8

t = 1.02 s

b) the maximum height

y- 44.0 = $v_{y}$ ² / 2 g

y - 44.0 = 5.1

y = 5.1 +44.0

y = 49.1 m

The time is the same because it does not depend on the initial height

t = 1.02 s

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3 years ago