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2 years ago

Narrow belts of high-speed winds that blow in the upper troposphere and lower stratosphere are known as

Physics

1 answer:

Drupady [299]2 years ago

3 0

<h2>Jet Stream</h2>

Explanation:

Jet streams are the winds that flow in the upper levels of the atmosphere.
These are relatively narrow bands of strong wind.
These high-speed winds blow from west to east in jet streams.
Sometimes, the flow of wind shifts from north to south.
Jet streams follow the boundaries between hot and cold air.

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Use the drop-down menus to complete the statement.

Nikitich [7]

Classics.

Resistance is equal to relation between voltage and current.

$R = \frac{U}{I}$

If we express current:

$I = \frac{U}{R}$

If current is in fact 0 then one of the quantities either voltage or resistance must be equal to zero. Since resistance cannot be equal 0, because that would violate mathematical law that states that division by zero is undefined the only logical conclusion is voltage.

So the answer should be C voltage and B zero.

Hope this helps!

8 0

2 years ago

Read 2 more answers

Different color ca

777dan777 [17]

Answer: The color of candy cane

Explanation:

An independent variable is the Variable that the scientist changed in order to change the dependent variable. The scientist changes the color of the candy cane to test the rate at which they dissolve.

4 0

3 years ago

You need to focus a 10 mW, 632.8 nm Gaussian laser beam that is 5.0 mm in diameter into a sample. You have access to a lens with

Anna [14]

Answer:

ee that the lens with the shortest focal length has a smaller object

Explanation:

For this exercise we use the constructor equation or Gaussian equation

$\frac{1}{f} = \frac{1}{p} + \frac{1}{q}$

where f is the focal length, p and q are the distance to the object and the image respectively.

Magnification a lens system is

m = $\frac{h'}{h}$ = - $\frac{q}{p}$

h ’= -\frac{h q}{p}

In the exercise give the value of the height of the object h = 0.50cm and the position of the object p =∞

Let's calculate the distance to the image for each lens

f = 6.0 cm

$\frac{1}{q} = \frac{1}{f } - \frac{1}{p}$

as they indicate that the light fills the entire lens, this indicates that the object is at infinity, remember that the light of the laser rays is almost parallel, therefore p = inf

q = f = 6.0 cm

for the lens of f = 12.0 cm q = 12.0 cn

to find the size of the image we use

h ’= h q / p

where p has a high value and is the same for all systems

h ’= h / p q

Thus

f = 6 cm h ’= fo 6 cm

f = 12 cm h ’= fo 12 cm

therefore we see that the lens with the shortest focal length has a smaller object

8 0

2 years ago

A single Oreo cookie provides 53 kcal of energy. An athlete does an exercise that involves repeatedly lifting (without accelerat

Sever21 [200]

Answer:

Approximately $325$ (rounded down,) assuming that $g = 9.81\; {\rm N \cdot kg^{-1}}$ .

The number of repetitions would increase if efficiency increases.

Explanation:

Ensure that all quantities involved are in standard units:

Energy from the cookie (should be in joules, ${\rm J}$ ):

$\begin{aligned} & 53\; {\rm kCal} \times \frac{1\; {\rm kJ}}{4.184\; {\rm kCal}} \times \frac{1000\; {\rm J}}{1\; {\rm kJ}} \approx 2.551 \times 10^{5}\; {\rm J} \end{aligned}$ .

Height of the weight (should be in meters, ${\rm m}$ ):

$\begin{aligned} h &= 2\; {\rm dm} \times \frac{1\; {\rm m}}{10\; {\rm dm}} = 0.2\; {\rm m}\end{aligned}$ .

Energy required to lift the weight by $\Delta h = 0.2\; {\rm m}$ without acceleration:

$\begin{aligned} W &= m\, g\, \Delta h \\ &= 100\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times 0.2\; {\rm m} \\ &= 196\; {\rm N \cdot m} \\ &= 196\; {\rm J} \end{aligned}$ .

At an efficiency of $0.25$ , the actual amount of energy required to raise this weight to that height would be:

$\begin{aligned} \text{Energy Input} &= \frac{\text{Useful Work Output}}{\text{Efficiency}} \\ &= \frac{196\; {\rm J}}{0.25} \\ &=784\; {\rm J}\end{aligned}$ .

Divide $2.551 \times 10^{5}\; {\rm J}$ by $784\; {\rm J}$ to find the number of times this weight could be lifted up within that energy budget:

$\begin{aligned} \frac{2.551 \times 10^{5}\; {\rm J}}{784\; {\rm J}} &\approx 325 \end{aligned}$ .

Increasing the efficiency (the denominator) would reduce the amount of energy input required to achieve the same amount of useful work. Thus, the same energy budget would allow this weight to be lifted up for more times.

4 0

1 year ago

The average distance of Enceladus from Saturn is 238,000 km; the average distance of Titan from Saturn is 1,222,000 km. How much

aleksklad [387]

Answer:

Titan takes 11.634 times longer to orbit Saturn as compared to Enceladus.

Explanation:

We have been given that the average distance of Enceladus from Saturn is 238,000 km; the average distance of Titan from Saturn is 1,222,000 km.

We will use Kepler's Law to solve our given problem.

$\frac{(T_1)^2}{(T_2)^2}=\frac{(r_1)^3}{(r_2)^3}$