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3 years ago

What are the components of the "Earth Radiation Budget?"

Physics

1 answer:

Alex777 [14]3 years ago

4 0

Answer:

The energy entering, reflecting, absorbed, and emitted by the earth system are the components of the Earth's radiation budget.

Explanation:

I hope this helps also I hope you have a great day and a new year.

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ch4aika [34]

Answer: C.

Explanation:

Just took quiz

8 0

3 years ago

Which of the following does each different kind of atom represent?

r-ruslan [8.4K]

Answer:

Explanation:

because atoms make up an element.

5 0

3 years ago

Read 2 more answers

Christina drives his moped 7 kilometers North. She stops for lunch and then drives 5 kilometers East. What distance did she cove

Mashutka [201]

Answer:

She covers the distance is 12 km.

The magnitude of displacement is 8.6 km.

The direction of her displacement is north east.

Explanation:

Given that,

Christina drives his moped 7 kilometers North and stop for lunch and then drive 5 km east.

We need to calculate the total distance

Using formula of distance

$d=d_{1}+d_{2}$

Put the value into the formula

$d=7+5$

$d=12\ km$

We need to calculate the magnitude of displacement

Using formula of displacement

$D=xi+yj$

$D=5i+7j$

$D=\sqrt{5^2+7^2}$

$D= 8.6\ km$

The direction of her displacement is north east.

Hence, She covers the distance is 12 km.

The magnitude of displacement is 8.6 km.

The direction of her displacement is north east.

6 0

3 years ago

Leia is presenting her project about gravity to her class. She says that both mass and distance affect how strong the gravitatio

DaniilM [7]

Answer:

No, because as distance increases, gravitional force decreases.

5 0

3 years ago

Blood in a carotid artery carrying blood to the head is moving at 0.15 m/s when it reaches a section where plaque has narrowed t

sp2606 [1]

Answer:

26.9 Pa

Explanation:

We can answer this question by using the continuity equation, which states that the volume flow rate of a fluid in a pipe must be constant; mathematically:

$A_1 v_1 = A_2 v_2$ (1)

where

$A_1$ is the cross-sectional area of the 1st section of the pipe

$A_2$ is the cross-sectional area of the 2nd section of the pipe

$v_1$ is the velocity of the 1st section of the pipe

$v_2$ is the velocity of the 2nd section of the pipe

In this problem we have:

$v_1=0.15 m/s$ is the velocity of blood in the 1st section

The diameter of the 2nd section is 74% of that of the 1st section, so

$d_2=0.74d_1$

The cross-sectional area is proportional to the square of the diameter, so:

$A_2=(0.74)^2 A_1=0.548 A_1$

And solving eq.(1) for v2, we find the final velocity:

$v_2=\frac{A_1 v_1}{A_2}=\frac{A_1 (0.15)}{0.548 A_1}=0.274 m/s$

Now we can use Bernoulli's equation to find the pressure drop:

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2$

where

$\rho=1025 kg/m^3$ is the blood density

$p_1,p_2$ are the initial and final pressure

So the pressure drop is:

$p_1 - p_2 = \frac{1}{2}\rho (v_2^2-v_1^2)=\frac{1}{2}(1025)(0.274^2-0.15^2)=26.9 Pa$

8 0

3 years ago