All categories

2 years ago

which type of front occurs when cold air and warm air are next to each other but are at a a stand still ?

Physics

2 answers:

vladimir1956 [14]2 years ago

7 0

Answer:

Stationary front

Explanation:

A stationary front is usually defined as a front that forms when the cold and warm air masses are next to each other but are at a standstill. This normally takes place when both the air masses push one another but neither of them is sufficiently strong to push the other. Both the air masses remain stationary because the winds blow parallel to the front instead of blowing perpendicular.

This type of front is basically responsible for the formation of cloudy weather that leads to rainfall or snowfall.

This stationary front is usually identified as the alternate red-colored semicircles and blue-colored triangles on weather maps. Both blue triangles and red semicircles are directed in the opposite direction.

algol132 years ago

3 0

A stationary front
hope this helps

You might be interested in

A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE g

7nadin3 [17]

Answer:

(a) r = 1.062·R $_E$ = $\frac{531}{500} R_E$

(b) r = $\frac{33}{25} R_E$

Explanation:

Here we have escape velocity v $_e$ given by

$v_e =\sqrt{\frac{2GM}{R_E} }$ and the maximum height given by

$\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r}$

Therefore, when the initial speed is 0.241v $_e$ we have

v = $0.241\times \sqrt{\frac{2GM}{R_E} }$ so that;

v² = $0.058081\times {\frac{2GM}{R_E} }$

v² = ${\frac{0.116162\times GM}{R_E} }$

$\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r}$ is then

$\frac{1}{2} {\frac{0.116162\times GM}{R_E} }-\frac{GM}{R_E} = -\frac{GM}{r}$

Which gives

$-\frac{0.941919}{R_E} = -\frac{1}{r}$ or

r = 1.062·R $_E$

(b) Here we have

$K_i = 0.241\times \frac{1}{2} \times m \times v_e^2 = 0.241\times \frac{1}{2} \times m \times \frac{2GM}{R_E} = \frac{0.241mGM}{R_E}$

Therefore we put $\frac{0.241GM}{R_E}$ in the maximum height equation to get

$\frac{0.241}{R_E} -\frac{1}{R_E} =-\frac{1}{r}$

From which we get

r = 1.32·R $_E$

ME = KE - PE

Where the KE = PE required to leave the earth we have

ME = KE - KE = 0

The least initial mechanical energy to leave the earth is zero.

3 0

3 years ago

Read 2 more answers

What is the net force exerted on a 28.8 kg shopping cart that accelerates at a rate of 2.88 m/s^2?

Mama L [17]

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 28.8 × 2.88 = 82.944

We have the final answer as

Hope this helps you

6 0

3 years ago

When droplets of water in the atmosphere act like prisms, the colors in sunlight undergo?

dexar [7]

White light is all the colours of light combined. When the droplets act like prisms, they split the white light into all its colours and also slightly bend the different colours. This is how a rainbow is formed.

4 0

2 years ago

In a single replacement reaction, what will a metal always replace?

Alexandra [31]

A metal have a nice day

8 0

2 years ago

Read 2 more answers

A 10 kg bear is running to the right at 10 m/s. He crashes into another 10 kg bear running to the left at -5 m/s. They collide a

faust18 [17]

Answer:

<em>Total momentum is conserved</em>

Explanation:

<u>Conservation of Momentum </u>

The momentum is a physical magnitude that measures the product of the object's velocity by its mass. The total momentum of a system is the sum of all its components' individual momentums. The two-bear system starts with a total moment of

$p=m_1v_1+m_2v_2=(10)(10)+(10)(-50)=50\ kg.m/s$

When both bears stick together, the total mass is 20 kg, and the new momentum is

$p'=(20)(2.5)=50 \ kg.m/s$

We have assumed both bears move to the right after the collision. In this situation, the total momentum is conserved

7 0

3 years ago