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3 years ago

How do air mass conditions ahead of the squall line support the development of new cell?

1 answer:

7 0

<span>Storm cells in a squall line typically move from the southwest to the northeast, and as the mature cells in the northeast begin to die off, new ones are formed at the opposite end to advance the line. The air in the southwest corner has strong vertical updrafts that allow new cells to grow and develop into thunderstorms.</span>

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Suppose an object’s initial velocity is 10 m/s and its final velocity is 4 m/s. Mass is constant.

galina1969 [7]

The negative work done states that the work is done by the object and not on the object.

<u>Explanation: </u>

According to work energy theorem, the work done is equal to change in kinetic energy exhibited by the body. As the mass of the object is constant, and the velocity is decreased from 10 m/s to 4 m/s, the work done will be

$W=\frac{1}{2} \times\left(u^{2}-v^{2}\right)$

Here W is the work done, m is the mass and v is the final and u is the initial velocity of the object. As the initial velocity is greater than the final velocity. So

$W=\frac{1}{2} \times\left(4^{2}-10^{2}\right)=\frac{1}{2} \times(16-100)=\frac{1}{2} \times(-84)= - 42 J$

So the work done is negative for the given situation. The negative work done states that the work is done by the object and not on the object.

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2 years ago

Which tools would be use to find an irregularly shaped object’s mass and volume?

madreJ [45]

Scales for weight
Any beaker to measure the volume of liquid displaced

3 0

2 years ago

La corriente que fluye por una resistencia es

kirill [66]

Si I think the answer is lster

4 0

3 years ago

Oil flows upward in the wick of a lantern because of the liquid property called

Firdavs [7]

The answer is A because the lighter or heavier the liquid is (in this case it depends) it flows up.

5 0

3 years ago

Read 2 more answers

A long coaxial cable consists of an inner cylindrical conductor with radius a and an outer coaxial cylinder with inner radius b

Natasha_Volkova [10]

Answer:

Part a)

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part b)

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part d)

As we know that due to induction of charge there will be same charge appear on the inner and outer surface of the cylinder but the sign of the charge must be different

On the inner side of the cylinder there will be negative charge induce on the inner surface and on the outer surface of the cylinder there will be same magnitude charge with positive sign.

Explanation:

Part a)

By Guass law we know that

$\int E. dA = \frac{q}{\epsilon_0}$

$E. 2\pi rL = \frac{\lambda L}{\epsilon_0}$

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part b)

Outside the outer cylinder we will again use Guass law

$\int E. dA = \frac{q}{\epsilon_0}$

$E. 2\pi rL = \frac{\lambda L}{\epsilon_0}$

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part d)

As we know that due to induction of charge there will be same charge appear on the inner and outer surface of the cylinder but the sign of the charge must be different

On the inner side of the cylinder there will be negative charge induce on the inner surface and on the outer surface of the cylinder there will be same magnitude charge with positive sign.

4 0

3 years ago