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3 years ago

What is a front? for weather

Physics

2 answers:

Masteriza [31]3 years ago

3 0

Boundary separating two masses of air of different densities

Sophie [7]3 years ago

3 0

A weather front is a boundary separating two masses of air of different densities, and is the principal cause of meteorological phenomena outside the tropics. In surface weather analyses, fronts are depicted using various colored triangles and half-circles, depending on the type of front.

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Water is flowing in a pipe with a varying cross-sectional area, and at all points the water completely fills the pipe. At point

Salsk061 [2.6K]

Answer:

The fluids speed at a) $0.105\,m^{2}$ and b) $0.047\,m^{2}$ are $2.33\,\frac{m}{s^{2}}$ and $5.21\,\frac{m}{s^{2}}$ respectively

c) Th volume of water the pipe discharges is: $882\,m^{3}$

Explanation:

To solve a) and b) we should use flow continuity for ideal fluids:

$\Delta Q=0$ (1)

With Q the flux of water, but Q is $Av$ using this on (1) we have:

$A_{2}v_{2}-A_{1}v_{1}=0$ (2)

With A the cross sectional areas and v the velocities of the fluid.

a) Here, we use that point 2 has a cross-sectional area equal to $A_{2}=0.105\,m^{2}$ , so now we can solve (2) for $v_{2}$ :

$v_{2}=\frac{A_{1}v_{1}}{A_{2}}=\frac{(0.070)(3.5)}{0.105}\approx2.33\,\frac{m}{s^{2}}$

b) Here we use point 2 as $A_{2}=0.047\,m^{2}$ :

$v_{2}=\frac{A_{1}v_{1}}{A_{2}}=\frac{(0.070)(3.5)}{0.047}\approx5.21\,\frac{m}{s^{2}}$

c) Here we need to know that in this case the flow is the volume of water that passes a cross-sectional area per unit time, this is $Q=\frac{V}{t}$ , so we can write:

$A_{1}v_{1}=\frac{V}{t}$ , solving for V:

$V=A_{1}v_{1}t=(0.070m^{2})(3.5\frac{m}{s})(3600s)=882\,m^{3}$

3 0

4 years ago

A lifeguard on a beach observes that waves have a speed of 7 m/s and a distance of 4.6m between wave crests. what is the period

Pepsi [2]

Answer:

0.657 seconds

Explanation:

speed of wave= wavelength / time period

time period= wavelength / speed

= 4.6/7

=0.657 sec

4 0

3 years ago

A charge Q experiences no net force at a particular point in space. Which of the following situations described below must ALWAY

laila [671]

Answer:

-There are no other charges nearby.

Explanation:

-There are no other charges nearby.

If there is no net charge in nearby space then the force on this charge will be ZERO

-If there are other charges nearby, they must all have the same sign as Q.

There there is nearby charge of same sign then it will have repulsion force on Q

-If there are other charges nearby, they must all have the opposite sign of Q.

if there is nearby charge of opposite sign then the force must be attraction force.

-If there are other charges nearby, the total positive charge must equal the total negative charge.

If there exist two type of charges nearby then there may exist either attraction or repulsion force on it

8 0

4 years ago

When you drop a 0.37kg apple earth exerts a force on it that accelerates it at 9.8m/s^2 toward the earths surface. According to

Delvig [45]

force applied due to earth on the apple is given as

$F = ma$

given that

m = 0.37 kg

a = 9.8 m/s^2

now we have

$F = m*a$

$F = 0.37* 9.8 = 3.63 N$

now as per Newton's III law equal and opposite force is applied on earth

so now we can find acceleration of Earth using same equation as we used above

$F = ma$

here we know that

m = mass of earth

F = 3.63 N

$3.63 = 5.98 * 10^{24} * a$

$a = \frac{3.63}{5.98 * 10^{24}$

$a = 6.1 * 10^{-25} m/s^2$

so acceleration of earth is very small and given by above value

5 0

3 years ago

Why did donald hings invent the walkie talkie ?

Bumek [7]

The "walkie-talkie" is Don Hings' most well-known invention. The earliest versions of this device were designed as portable field radios for the bush pilots of Consolidated Mining and Smeltin , who had to fly their planes between remote sites in the far north of Canada.

3 0

3 years ago