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3 years ago

Which electromagnetic radiation can be seen with the human eye

1 answer:

6 0

Answer:the human retina can only detect incident light that falls in waves 400 to 720 nano-meters long so we cant see microwave or ultraviolet wavelengths. this also applies to infrared lights which has wavelengths longer than visible and shorter than microwaves thus being invisible to the human eye

Explanation:

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A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 14.413

natima [27]

Answer:

The value of charge q₃ is 40.46 μC.

Explanation:

Given that.

Magnitude of net force $F=14.413\ N$

Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.

We need to calculate the distance

Using Pythagorean theorem

$r=\sqrt{x_{2}^2+y_{2}^2}$

Put the value into the formula

$r=\sqrt{(8.2\times10^{-2})^2+(3.1\times10^{-2})^2}$

$r=0.0876\ m$

We need to calculate the magnitude of the charge q₃

Using formula of net force

$F_{12}=kq_{2}(\dfrac{q_{3}}{r_{3}^2}+\dfrac{q_{1}}{r_{1}^2})$

Put the value into the formula

$14.413=9\times10^{9}\times7.7\times10^{-6}(\dfrac{q_{3}}{(0.0438)^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})$

$(\dfrac{q_{3}}{(4.38\times10^{-2})^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})=\dfrac{14.413}{9\times10^{9}\times7.7\times10^{-6}}$

$\dfrac{q_{3}}{(0.0438)^2}=207\times10^{-4}+3.909\times10^{-4}$

$q_{3}=0.0210909\times(0.0438)^2$

$q_{3}=40.46\times10^{-6}\ C$

$q_{3}=40.46\ \mu C$

Hence, The value of charge q₃ is 40.46 μC.

5 0

3 years ago

When turned on, a fan requires 5.0 seconds to get up to its final operating rotational speed of 1200 rpm. a) How large is the fi

vova2212 [387]

Answer:

125.6 rad/s

25.12 rad/s²

Explanation:

t = time required by the fan to get up to final operating speed = 5 sec

w = final operating rotational speed = 1200 rpm

we know that :

1 revolution = 2π rad

1 min = 60 sec

w = $1200\frac{rev}{min}\frac{2\pi rad}{1 rev}\frac{1 min}{60 sec}$

w = $\frac{1200\times 2\pi }{60}\frac{rad}{s}$

w = 125.6 rad/s

w₀ = initial angular speed = 0 rad/s

α = angular acceleration

using the equation

w = w₀ + α t

125.6 = 0 + α (5)

α = 25.12 rad/s²

5 0

3 years ago

Which form of the law of conservation of energy describes the motion of the block as it slides on the floor from the bottom of t

aivan3 [116]

Answer:

Explanation:

Let assume begins movement at zero point, that is, height is equal to zero. The block has an initial linear kinetic energy and no gravitational potential energy and end with no linear kinetic energy, some gravitational potential energy and work losses due to slide friction. In mathematical terms, this system can be model as follows:

$K_{1} = U_{2} + W_{loss, 1 \longrightarrow 2}$

Where $K, U, W$ are linear kinetic energy, gravitational potential energy and work, respectively.

8 0

3 years ago

..................,,,,,,,,,

Fiesta28 [93]

Answer:

.....,.,.,.,.,.,.,.,.,.,.,.,..,.,.,.,.,.,.,,.,.┌(・。・)┘♪

7 0

3 years ago

Read 2 more answers

What is meant by length of the travel path

Vanyuwa [196]

Answer:Physics. Distance, the total distance an object travels dependent on its path through space. Optical path length, the product of the distance light travels and the refractive index of the medium it travels through. Mean free path, the average distance that a particle travels before scattering.

Explanation:

6 0

3 years ago