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3 years ago

What factors contribute to global winds identify areas where winds are weak

2 answers:

5 0

The factors that influence wind and contribute to global winds are pressure gradient, Rossby waves and jet streams, and local weather conditions.

Gennadij [26K]3 years ago

5 0

The factors that influence wind and contribute to world winds are the pressure gradient, Rossby waves and jet streams, and native climate. additionally, the wind is controlled by the combined forces of pressure gradient, the result, and friction.
The 2 factors that cause the wind to deviate are horizontal pressure gradient that is that the propulsion behind wind, and also the result that is what shifts wind direction.
The Coriolis force caused by the Earth's rotation is what offers winds among aggressive systems their dextral circulation within the northern hemisphere because the wind moves outward and is deflected right from the middle of the air mass. The pressure begins to fall quicker whereas the wind speed at the same time will increase.

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gayaneshka [121]

Both fission and fusion are nuclear reactions that produce energy, but the applications are not the same. Fission is the splitting of a heavy, unstable nucleus into two lighter nuclei, and fusion is the process where two light nuclei combine together releasing vast amounts of energy.

6 0

3 years ago

a 1000 khz am radio station broadcasts with a power of 20 kw. how many photons does the transmitter antenna emit each second

Gemiola [76]

1000 khz am radio station broadcasts with a power of 20 kw number of photon emitted per second is 30.16 x 10^30 photon/s.

The frequency of the radio station is:
f
=
1000
k
H
z
=
1
×
10^6Hz
The transmit power is: P = 20kW = 20 X 10^3 W

The transmit power is: h = 6.63 x 10 ^-34 m^2.kg/s

The number of photon emitted per second = N = P/hf = 30.16 x 10^30 photon/s.

1000 khz am radio station broadcasts with a photon of 20 kw1000 khz am radio station broadcasts with a power of 20 kw1000 khz am radio station broadcasts with a power of 20 kw1000 khz am radio station broadcasts with a power of 20 kw1000 khz am radio station broadcasts with a power of 20 kw.1000 khz am radio station broadcasts with a power of 20 kw1000 khz am radio station broadcasts with a power of 20 kw1000 khz am radio station broadcasts with a power of 20 kw.

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6 0

1 year ago

At a playground, a child slides down a slide that makes a 42° angle with the horizontal direction. The coefficient of kinetic fr

kumpel [21]

Answer:

$a = 5.1\ m/s^2$

Explanation:

Given,

The angle of the slide= $42^\circ$

The mass of the child is= m

coefficient of friction = 0.20

when she slides down now apply Newton's law

$ma =mg \sin\theta - f$

$ma = mg\sin \theta -\mu mg \cos \theta$

therefore the acceleration

$a=g \sin\theta -\mu gcos\theta$

$a=g[\sin \theta -\mu \cos \theta]$

$a=9.8\times [\sin 42^\circ -0.2\times \cos 42^\circ]$

$a = 5.1\ m/s^2$

hence, the magnitude of acceleration during her sliding is equal to $a = 5.1\ m/s^2$

4 0

3 years ago

A reconnaissance plane flies 404 km awayfrom its base at 730 m/s, then flies back to its base at 1095 m/s.What is it’s average s

elena-14-01-66 [18.8K]

The average speed of the plane is 875.999 m/s.

Average speed can be defined as the ratio of total distanced travelled by the object to that of total time taken to cover the distance.

Mathematically, Average speed = Av = $\frac{Total Distance}{Total Time}$

According to the question,

Speed of the plane away from its base V₁ = 730 m/s

Speed of the plane when it flies back V₂ = 1095 m/s

Plane flies the distance D = 404 km

Total Distance covered by the plane S = 404 * 2 km

(because the distance travelled by the plane when going away from the base and then flying back to the base is same)

Therefore S = 808 km = 808 ˣ 10³ m

Time taken by the plane while flying away from the base T₁ = $\frac{D}{V1}$

T₁ = $\frac{404000}{730}$ = 553.425 s

Time taken by the plane while flying back to the base T₂ = $\frac{D}{V2}$

T₂ = $\frac{404000}{1095}$ = 368.949s

Total Time T = T₁ + T₂ = 922.375 s

Therefore Av = $\frac{Total Distance}{Total Time}$

= $\frac{D}{T}$ m/s

= $\frac{808000}{922.375}$ m/s

= 875.999 m/s

The average speed of the plane will be 875.999 m/s.

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7 0

1 year ago

Last night Mookie Betts hit a baseball at 32.5 m/s at a 45° angle. Betts

podryga [215]

Answer:

a) Since the height of the baseball at 99 m was 8.93 m and the fence at that distance is 3m tall, the hit was a home run.

b) The total distance traveled by the baseball was 108.7 m.

Explanation:

a) To know if the hit was a home run we need to calculate the height of the ball at 99 m:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}$

Where:

$y_{f}$ : is the final height =?

$y_{0}$ : is the initial height = 1 m

$v_{0_{y}$ : is the initial vertical velocity = v₀sin(45)

v₀: is the initial velocity = 32.5 m/s

g: is the gravity = 9.81 m/s²

t: is the time

First, we need to find the time by using the following equation:

$t = \frac{x}{v_{0_{x}}} = \frac{99 m}{32.5 m/s*cos(45)} = 4.31 s$

Now, the height is:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2} = 1m + 32.5 m/s*sin(45)*4.31 s - \frac{1}{2}9.81 m/s^{2}*(4.31 s)^{2} = 8.93 m$

Since the height of the baseball at 99 m was 8.93 m and the fence at that distance is 3m tall, the hit was a home run.

b) To find the distance traveled by the baseball first we need to find the time of flight:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}$

$0 = 1 m + 32.5m/s*sin(45)t - \frac{1}{2}9.81 m/s^{2}t^{2}$

$1 m + 32.5m/s*sin(45)t - \frac{1}{2}9.81 m/s^{2}t^{2} = 0$

By solving the above quadratic equation we have:

t = 4.73 s

Finally, with that time we can find the distance traveled by the baseball:

$x = v_{0_{x}}*t = 32.5 m/s*cos(45)*4.73 s = 108.7 m$

Hence, the total distance traveled by the baseball was 108.7 m.

I hope it helps you!

4 0

3 years ago