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3 years ago

a passing warm front changes air pressure. does a passing warm front increase or decrease air pressure

Physics

2 answers:

meriva3 years ago

7 0

Warmer air has molecules more spread out, so I'd say a decrease in air pressure.

Nadusha1986 [10]3 years ago

4 0

Warm air molecules are less dense and spread out more than cool air molecuels so they decreace

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Air passing over an airplane's wing travels ____________________, and therefore exerts ____________________ pressure than air tr

Aleksandr [31]

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3 years ago

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During convection, heat warms up a fluid, making the fluid less dense so it begins to RISE. As the warm fluid moves, it begins t

Fiesta28 [93]

Answer:

True

Explanation:

Convection is a form of heat transfer that is predominantly common in fluids especially liquid and gas.

It occurs by the movement of a part of substance from one place to another based on density and temperature differences.

A typical convection cell is made up of a liquid that is heated. The liquid part close to the heat source becomes warmer and rises due to its low density. The part away from the heat source is more dense and begins to sink.

This analogy is commonly demonstrated in a boiling pot of water.

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3 years ago

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A particle with a charge of +4.20nC is in a uniform electric field E⃗ directed to the left. It is released from rest and moves t

Morgarella [4.7K]

Answer:

(A). The work done is $1.50\times10^{-6}\ J$ .

(B). The potential of the starting point with respect to the endpoint is 357.14 V.

(C). The magnitude of E is 5952.38 N/C.

Explanation:

Given that,

Charge = 4.20 nC

Distance = 6.00 cm

Kinetic energy $K.E=1.50\times10^{-6}\ J$

The particle start from rest.

So, the initial kinetic energy i zero.

(A). We need to calculate the work by the electric force

Using formula of work done

$W = \Delta K.E$

$W=K.E_{f}-K.E_{i}$

Put the value into the formula

$W= 1.50\times10^{-6}-0$

$W=1.50\times10^{-6}\ J$

The work done is $1.50\times10^{-6}\ J$ .

(B). We need to calculate the potential of the starting point with respect to the endpoint

We know that.

Change in potential energy = change in kinetic energy

$\Delta P.E=\Delta K.E$

So, $U = 1.50\times10^{-6}$

Using formula of potential

$V=\dfrac{U}{q}$

Put the value into the formula

$V=\dfrac{1.50\times10^{-6}}{4.20\times10^{-9}}$

$V=357.14\ V$

The potential of the starting point with respect to the endpoint is 357.14 V.

(C). We need to calculate the magnitude of E

Using formula of work done

$W=F\times r$ ....(I)

Using formula of force

$F=qE$

Put the value in the equation (I)

$W=qE\times r$

$E=\dfrac{W}{q\times r}$

Put the value into the formula

$E=\dfrac{1.50\times10^{-6}}{4.20\times10^{-9}\times6.00\times10^{-2}}$

$E=5952.38\ N/C$

The magnitude of E is 5952.38 N/C.

Hence, This is the required solution.

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3 years ago

A ball player catches a ball 3.2 s after throwing it vertically upward. what height did it reach?

cluponka [151]

At the top of the height, the velocity is zero and acceleration is negative of acceleration due to gravity ( i.e $-9.8 m/s^2$ ).

The time of the ball in air is 3.2 s, so ascending time is $3.2/2=1.6 s$ .

Therefore from kinematic equation,

$v = u + gt$

Substituting the values we get,

$0= u - 9.8 (1.6)\\\\u=15.7 m/s$ , Here v = 0 at top.

Now from equation,

$h=ut+\frac{1}{2} gt^2$ , here h is the height .

So,

$h=(15.7 m/s) (1.6s)-\frac{1}{2} 9.8m/s^2(1.6)^2\\\\ h=25.12m-12.54m\\\\h=12.48 m$ .

Thus, the ball reached at its maximum height of 12.48 m.

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