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2 years ago

7

Near the poles, more energy is reflected back into space than is absorbed. near the poles, more energy is reflected back into sp

ace than is absorbed.
a. True
b. False

1 answer:

VladimirAG [237]2 years ago

7 0

Correct answer: a) True.

In fact, the soil at the poles consists mainly of snow. Snow has an albedo of 90%: albedo is the fraction of light (and so, of energy) that is reflected back to the space. This means that at the poles, about 90% of the light is reflected back to the space, while only 10% is absorbed.

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A tank has the shape of an inverted circular cone with height 16m and base radius 3m. The tank is filled with water to a height

rewona [7]

Answer:

$W=17085KJ$

Explanation:

From the question we are told that:

Height $H=16m$

Radius $R=3$

Height of water $H_w=9m$

Gravity $g=9.8m/s$

Density of water $\rho=1000kg/m^3$

Generally the equation for Volume of water is mathematically given by

$dv=\pi*r^2dy$

$dv=\frac{\piR^2}{H^2}(H-y)^2dy$

Where

y is a random height taken to define dv

Generally the equation for Work done to pump water is mathematically given by

$dw=(pdv)g (H-y)$

Substituting dv

$dw=(p(=\frac{\piR^2}{H^2}(H-y)^2dy))g (H-y)$

$dw=\frac{\rho*g*R^2}{H^2}(H-y)^3dy$

Therefore

$W=\int dw$

$W=\int(\frac{\rho*g*R^2}{H^2}(H-y)^3)dy$

$W=\rho*g*R^2}{H^2}\int((H-y)^3)dy)$

$W=\frac{1000*9.8*3.142*3^2}{9^2}[((9-y)^3)}^9_0$

$W=3420.84*0.25[2401-65536]$

$W=17084965.5J$

$W=17085KJ$

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4 0

2 years ago

Sound with a frequency of 1250 Hz leaves a room through a doorway with a width of 1.05 m.At what minimum angle relative to the c

kiruha [24]

Answer:

15.19°, 31.61°, 51.84°

Explanation:

We need to fin the angle for m=1,2,3

We know that the expression for wavelenght is,

$\lambda = \frac{c}{f}$

Substituting,

$\lambda = \frac{344}{1250}$

$\lambda = 0.2752m$

Once we have the wavelenght we can find the angle by the equation of the single slit difraction,

$sin\theta = \frac{m \lambda}{W}$

Where,

W is the width

m is the integer

$\lambda$ the wavelenght

Re-arrange the expression,

$\theta = sin^{-1} \frac{m\lambda}{W}$

For m=1,

$\theta = sin^{-1} \frac{1 (0.2752)}{1.05}= 15.19\°$

For m=2,

$\theta = sin^{-1} \frac{2 (0.2752)}{1.05}= 31.61\°$

For m=3,

$\theta = sin^{-1} \frac{3 (0.2752)}{1.05}= 51.84\°$

<em>The angle of diffraction is directly proportional to the size of the wavelength.</em>

5 0

2 years ago

What is the action force of a flying bird?

kirill [66]

It is called the reaction force of a bird flying.

7 0

3 years ago

How would you compare and contrast the 3 types of friction?

vlabodo [156]

Rolling friction .<span> the force that slows down the movement of a rolling object</span>
sliding friction.
Sliding friction : The opposing force that comes into play when one body is actually sliding over the surface of the other body is called sliding friction. e.g. A flat block is moving over a horizontal table.

Kinetic or dynamic friction: If the applied force is increased further and sets the body in motion, the friction opposing the motion is called kinetic friction

7 0

3 years ago

Read 2 more answers

Which statement is true for a car that first goes around a curve of radius r at a constant speed v, and then goes around the sam

sasho [114]

Centripetal force is equal to (mv^2)/r
The way I use to answer these question is to set every variable to 1
m=1
v=1
r=1
so centripetal force =1
then change the variable we're looking at
and since we're find when it's half we could either change it to 1/2 or 2, but 2 is easier to use
m=1
v=2
r=1
((1)×(2)^2)/1=4
So the velocity in the 1st part is half the velocity in the 2nd part and the centripetal force is 4× less
The answer is the centripetal force is 1/4 as big the second time around

3 0

3 years ago