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Vadim26 [7]
3 years ago
5

How is precipitation related to high- and low-pressure air

Physics
1 answer:
pentagon [3]3 years ago
7 0

Answer:

The high-pressure region refers to the condition where the cold and dense air sinks and makes the atmosphere stable, forming clear sky or presence of low clouds in the sky.

On the other hand, the low-pressure region refers to the condition where the hot and less dense air rises upward, and with the presence of a sufficient amount of water content in the air, it gives rise t the formation of the clouds and eventually leads to the occurrence of rainfall.

This is how the process of precipitation is related to the high and the low-pressure air, where the low-pressure zone results in precipitation and high-pressure zone results in a clear sky.

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A boat traveling across a river has a resultant velocity of 10 km/h and travels 34 degrees with respect to the shore. A) What is
vodka [1.7K]

Answer:

a) 1.55 m/s

b) 2.3 m/s

Explanation:

We know that the boat travels across the river, if we define the river as the x-axis, then the velocity of the boat is only on the y-axis.

Then we can write the velocity of the boat in still water as:

S = (0, B)

Now, when the boat is on the river, the velocity of the boat will be equal to the velocity of the boat in still water plus the velocity of the river.

The velocity of the river is:

v = (R, 0).

Then the velocity of the boat in that river is:

V' = (0, B) + (R, 0) = (R, B)

Now, we know that the velocity of the boat is 10km/h, and it travels at an angle of 34° with respect to the shore.

We can use the Pythagoreans theorem to write the components of this velocity as:

x-axis component = 10km/h*cos(34°) = 8.29 km/h

y-axis component = 10km/h*sin(34°) =  5.59 km/h

Then the velocity of the boat can be written in components as:

velocity = ( 8.29 km/h,  5.59 km/h)

And we knew that the velocity of the boat was written as  (R, B)

Then we must have:

R = 8.29 km/h

B = 5.59 km/h

a) The speed of the boat in m/s:

We know that the speed of the boat is 5.59 km/h.

First, we know that:

1km = 1000m, then:

5.59 km/h = 5.59*(1000m)h = 5,590 m/h

And we know that:

1h = 3600s

Then we can write:

5,590 m/h = 5,590 m/(3600s) = 1.55 m/s

b) The speed of the river in m/s:

We know that the speed of the river is 8.29 km/h

Using the same reasoning as above, we can do the change of units as follows:

8.29 km/h = 8.29 (1000m)/(3600s) = 2.3 m/s

6 0
3 years ago
Two long wires hang vertically. Wire 1 carries an upward current of 1.50 A . Wire 2,20.0cm to the right of wire 1, carries a dow
Inessa [10]

The magnitude of the current in wire 3  is 2.4 A and in a direction pointing in the downward direction.

  • The force per unit length between two parallel thin current-carrying I_1 and I_2  wires at distance ' r ' is given by  f=\frac{u_0I_1I_2}{2\pi r}   ....(1) .
  • If the current is flowing in both wires in the same direction, and  the force between them will be the attractive force and if the current is flowing in opposite direction in wires then the force between them will be the repulsive force.

A schematic of the information provided in the question can be seen in the image attached below.

From the image, force on wire 2 due to wire 1 = force on wire 2 due to wire 3

F_2_1=F_2_3

Using equation (1) , we get

\frac{u_0I_2I_1}{0.2} =\frac{u_0I_2I_3}{0.32} \\\\\frac{I_1}{0.2} =\frac{I_3}{0.32} \\\\\frac{1.50}{0.2} =\frac{I_3}{0.32} \\\\0.48=0.2I_3\\\\I_3=2.4A

I₃ = 2.4 A and the current is pointing in the downward direction

Learn more about the magnitude and direction of forces here:

brainly.com/question/14879801?referrer=searchResults

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5 0
1 year ago
G to ml,density=3.291 g/ml
olya-2409 [2.1K]
The density value 3.291 g/ml (gram / milliliter) in words is "three point two nine one g/ml (gram / milliliter)".
=)
7 0
3 years ago
A projectile is fired at v 0 = 381.0 m/s at an angle of θ = 73.5 ∘ , with respect to the horizontal. Assume that air friction wi
Ipatiy [6.2K]

Answer:

Explanation:

velocity of projection, vo = 381 m/s

angle of projection, θ = 73.5°

The formula for the range is

R=\frac{u^{2}Sin2\theta }{g}

R=\frac{381^{2}Sin147 }{9.8}

R = 8067.4 m

Range in shorten by 34.1 %

So, the new range is

R' = 8067.4 - 34.1 x 8067.4/100

R' = 5316.4 m

5 0
3 years ago
The LR5 is the specialist submarine for underwater rescue. The average density of seawater is 1028 kg/ m3.
sladkih [1.3K]

Answer:

P = 7196 [kPa]

Explanation:

We can solve this problem using the expression that defines the pressure depending on the height of water column.

P = dens*g*h

where:

dens = 1028 [kg/m^3]

g = 10 [m/s^2]

h = 700 [m]

Therefore:

P = 1028*10*700

P = 7196000 [Pa]

P = 7196 [kPa]

5 0
3 years ago
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