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3 years ago

The black board appears black because:

Physics

1 answer:

marishachu [46]3 years ago

8 0

Answer:

I think its C All visible light wavelengths (ROYGBV) are absorbed by the board

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Trucks can be run on energy stored in a rotating flywheel, with an electric motor getting the flywheel up to its top speed of 20

Genrish500 [490]

Answer:

The time it can operate between chargins in minutes is

$t=102.8 minutes$

Explanation:

Given: $m=500kg$ , $r=1.0m$ , $w=200\pi rad/s$

a). The rotational kinetic energy

$K_R=\frac{1}{2}*I*w^2$

$I=\frac{1}{2}*m*r^2$

$I=\frac{1}{2}*500kg*(1.0m)^2$

$I=250 kg*m^2$

$K_R=\frac{1}{2}*250kg*m^2*(200\pi rad/s)^2$

$K_R=49.348x10^6J$

b). The power average 0.8kW un range time can be find

$P=\frac{K_R}{t'}$

Solve to t'

$t=\frac{K_R}{P}$

$t=\frac{49.348x10^6}{0.8x10^3w}=6168.5s$

$t=6168.5s\frac{1minute}{60s}=102.8 minutes$

3 0

3 years ago

Why were the rings of Uranus not observed directly from telescopes on the ground on Earth? How were they discovered?

leonid [27]

Answer:Explained below.

Explanation:

Uranus rings is made up of jet black, coal-like particles in small bands, making them difficult to perceive from Earth.This indicates that they are probably composed of a mixture of the ice and a dark material. The nature of material is dismal, but it might be some organic compounds greatly darkened by the charged particle irradiation from the Uranian magnetosphere. Rings were discovered by using a infrared telescope throughout the occultation of a star as Uranus passed in front of it. The light from the star dimmed many times before it was obstructed by the disk of Uranus and subsequently, showing the presence of various distinct rings.

6 0

3 years ago

During your summer internship for an aerospace company, you are asked to design a small research rocket. The rocket is to be lau

Viktor [21]

Answer:

6.75 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration = 16 m/s²

g = Acceleration due to gravity = 9.81 m/s²

Let y be the distance the rocket is accelerating

960-y is the distance traveled in free fall

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 16\times y+0^2}\\\Rightarrow v^2=32y\ m/s$

In free fall

$v^2-u^2=2g(960-y)\\\Rightarrow 0-32y=2g(960-y)\\\Rightarrow -32y=2\times -9.81(960-y)\\\Rightarrow 960-y=\dfrac{-32}{2\times -9.81}y\\\Rightarrow 960-y=1.63098878695y\\\Rightarrow 960=2.63098878695y\\\Rightarrow y=\dfrac{960}{2.63098878695}\\\Rightarrow y=364.881828749\ m$

The distance the rocket will keep accelerating is 364.881828749 m

After which it will travel 960-364.881828749 = 595.118171251 m in free fall

$s=ut+\frac{1}{2}at^2\\\Rightarrow 364.881828749=0t+\frac{1}{2}\times 16\times t^2\\\Rightarrow t=\sqrt{\frac{364.881828749\times 2}{16}}\\\Rightarrow t=6.75353452598\ s$