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3 years ago

12

On the average, about what percentage of the solar energy that strikes the outer atmosphere eventually reaches the earth's surfa

ce

1 answer:

almond37 [142]3 years ago

5 0

I think the answer to your question is twenty-four percent.

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Under what condition will kinetic friction slow a sliding object?

matrenka [14]

Explanation: Slowing Down (or Stopping) occurs when the force of kinetic friction is greater than that of the external force. This also follows Newton's first law of motion as there exists a net force on the object.

6 0

3 years ago

A buoy is a floating device that can have many purposes, but often as a locator for ships. Collin constructs a hollow metal buoy

Yuri [45]

Answer:

A) since the density of the buoy is half the density of sea water when the buoy is at rest on an ocean half of the buoy will be submerged in water

B )angular frequency ( w ) = $\sqrt{\frac{3g}{h} }$

c ) h = 1.34 m ( 4 $\pi ^{2}$ / 3g )

Explanation:

A) since the density of the buoy is half the density of sea water when the buoy is at rest on an ocean half of the buoy will be submerged in water this is because the substances with lesser density floats when placed in a substance with a higher density

B ) when the Buoy is at rest ( t = 0 ) and is then pushed down calculate angular frequency ( w ) of small oscillations in terms of the given variables

volume of cone = hA /3.

h = height of cone, A = Base Area.

therefore the total volume of the Buoy above water level ( at rest )

= ( $\pi r^{2} * \frac{h}{3}$ ) * 2 = $\frac{\pi r^{2} h }{6}$

The part of the buoy immersed in water = x

then the net upward force will be

fb ( force of Buoy ) = fg ( force of gravity )

note force = Mass * Acceleration

force of buoy = $\frac{\alpha }{2} *$ $\frac{\pi r^{2} }{h}$ * a = ( force of gravity ) $\alpha * T\frac{r^{2} }{4} * x * g$

therefore a = $\frac{3g}{h} * x$

angular frequency ( w ) = $\sqrt{\frac{3g}{h} }$

C ) height of the each cone

$\frac{2\pi }{w} = 1$ therefore w = 2 $\pi$

back to the angular frequency : $\frac{3g}{h} = 4\pi ^{2}$

therefore h = 1.34 m ( 4 $\pi ^{2}$ / 3g )

4 0

3 years ago

If the mass of a planet is 0.231 mE and its radius is 0.528 rE, estimate the gravitational field g at the surface of the planet.

crimeas [40]

Answer:

$8.1 m/s^2$

Explanation:

The strength of the gravitational field at the surface of a planet is given by

$g=\frac{GM}{R^2}$ (1)

where

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

For the Earth:

$g_E = \frac{GM_E}{R_E^2}=9.8 m/s^2$

For the unknown planet,

$M_X = 0.231 M_E\\R_X = 0.528 R_E$

Substituting into the eq.(1), we find the gravitational acceleration of planet X relative to that of the Earth:

$g_X = \frac{GM_X}{R_X^2}=\frac{G(0.231M_E)}{(0.528R_E)^2}=\frac{0.231}{0.528^2}(\frac{GM_E}{R_E^2})=0.829 g_E$

And substituting g = 9.8 m/s^2,

$g_X = 0.829(9.8)=8.1 m/s^2$

3 0

3 years ago

What is the drawback to using superconductors?

Fittoniya [83]

Answer:

Option A

The cost of keeping the semiconductor below the critical temperature is unreasonable

Explanation:

First of all, we need to understand what superconductors are. Superconductors are special materials that conduct electrical current with almost zero resistance. This means that there is little or no need for a voltage source to be connected to them. As a matter of fact, once a superconductor is connected to a power supply, one can remove the power supply and the current will still flow.

However, most superconducts can only conduct at very low temperatures up to -200 degrees Celcius. This is because, at that temperature, their atoms and molecules are relatively settled, hence they pose little or no resistance to the flow of current.

This as you can guess is extremely difficult to do, as you will need a lot of effort to cool it to that temperature and maintain it.

This makes option a the answer:

The cost of keeping the semiconductor below the critical temperature is unreasonable.

7 0

3 years ago

john gottman, the eminent marriage counselor put the number of conflicts which cannot be resolved and needed to be worked on con

USPshnik [31]

He called the counterproductive.

6 0

3 years ago