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3 years ago

What happens to some of the sun's energy that does not reach the earth's surface?

1 answer:

4 0

Hello,

Here is your answer:

The proper answer for this question is that they "get absorbed by the ozone layer".

If you need anymore help feel free to ask me!

Hope this helps!

You might be interested in

What is the energy of moving electrical charges

Simora [160]

The energy of moving electrical charges is Electrical energy

Hope its the answer you are finding and hope it helps....

3 0

3 years ago

A. Write two or three sentences to describe the conductivity of an insulator. Explain its conductivity in terms of the electrons

aliya0001 [1]

PART A)

Conductivity of insulator is very small as there is no free electrons to conduct the current trough that medium

So here number of conduction electrons are very less in insulators

PART B)

Resistance is the property of a conducting medium which will oppose the flow of current trough it

Resistance of wire directly depends on its length so resistance of long wire will be more than the resistance of short wire

Resistance inversely depends on the area so if a wire has more crossectional area then its resistance must be small

PART C)

power of light bulb is defined as rate of electrical energy

it is given by formula

P = i V

here we know that

i = 1.46 A

V = 120 volts

so power is given as

$P = 1.46 \times 120$

$P = 175.2 Watt$

3 0

3 years ago

Read 2 more answers

Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m

Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

$A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}$ .....(I)

$tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}$ ....(II)

Put the value of $\omega=0.628\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}$

$A=0.0198$

From equation (II)

$tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}$

$d=0.0023$

Put the value of $\omega=3.14\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}$

$A=0.0203$

From equation (II)

$tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}$

$d=0.0120$

Put the value of $\omega=6.28\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}$

$A=0.0209$

From equation (II)

$tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}$

$d=0.0257$

Put the value of $\omega=9.42\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}$

$A=0.0217$

From equation (II)

$tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}$

$d=0.0413$

Hence, This is the required solution.

5 0

3 years ago

A cubical block of wood, 10.0 cm on a side, floats at the interface between oil and water with its lower surface 1.50 cm below t

bezimeni [28]

Answer:

the gauge pressure at the upper face of the block is 116 Pa

Explanation:

Given the data in the question;

A cubical block of wood, 10.0 cm on a side.

height h = 1.50 cm = ( 1.50 × ( 1 / 100 ) ) m = 0.0150 m

density ρ = 790 kg/m³

Using expression for the gauged pressure;

p-p₀ = ρgh

where, p₀ is atmospheric pressure, ρ is the density of the substance, g is acceleration due to gravity and h is the depth of the fluid.

we know that, acceleration due to gravity g = 9.8 m/s²

so we substitute

p-p₀ = 790 kg/m³gh × 9.8 m/s² × 0.0150 m

= 116.13 ≈ 116 Pa

Therefore, the gauge pressure at the upper face of the block is 116 Pa

4 0

3 years ago

4. With a diameter that's 11 times larger than Earth's, _______ is the largest planet.

Komok [63]

The answer is B.Jupiter

7 0

3 years ago