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4 years ago

10

Warm air aloft is normally higher pressure relative to colder air aloft, which is normally of lower pressure (at equivalent alti

tudes). (hint: think of columns of cold and warmer air standing side-by-side)a. Trueb. False

1 answer:

Allisa [31]4 years ago

6 0

Answer:

the statement given is False.

Explanation:

The pressure is defined by

P = F / A

The force of the air is its weight,

P = W_air / A

The weight of an air column on us depends on the energy of the molecules that is related to the temperature

E = 3/2 K T

When the air is warmer it has more energy so the molecules can move more distance and therefore the average density of the gas decreases, as the density decreases, the weight of the column on us also decreases, therefore the pressure low.

The correct statement is: When the air is hotter it has less pressure than when the air is cold.

It is summary the statement given is False

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Questions 14 out of 20

KiRa [710]

Answer:

B

Explanation:

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4 years ago

A particle has a charge of q = +4.9 μC and is located at the origin. As the drawing shows, an electric field of Ex = +242 N/C ex

irina1246 [14]

a)

$F_{E_x}=1.19\cdot 10^{-3}N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=0$

b)

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=3.21\cdot 10^{-3}N$ (+z axis)

c)

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=3.21\cdot 10^{-3} N$ (+y axis)

$F_{B_y}=3.21\cdot 10^{-3}N$ (-x axis)

Explanation:

a)

The electric force exerted on a charged particle is given by

$F=qE$

where

q is the charge

E is the electric field

For a positive charge, the direction of the force is the same as the electric field.

In this problem:

$q=+4.9\mu C=+4.9\cdot 10^{-6}C$ is the charge

$E_x=+242 N/C$ is the electric field, along the x-direction

So the electric force (along the x-direction) is:

$F_{E_x}=(4.9\cdot 10^{-6})(242)=1.19\cdot 10^{-3} N$

towards positive x-direction.

The magnetic force instead is given by

$F=qvB sin \theta$

where

q is the charge

v is the velocity of the charge

B is the magnetic field

$\theta$ is the angle between the directions of v and B

Here the charge is stationary: this means $v=0$ , therefore the magnetic force due to each component of the magnetic field is zero.

b)

In this case, the particle is moving along the +x axis.

The magnitude of the electric force does not depend on the speed: therefore, the electric force on the particle here is the same as in part a,

$F_{E_x}=1.19\cdot 10^{-3} N$ (towards positive x-direction)

Concerning the magnetic force, we have to analyze the two different fields:

- $B_x$ : this field is parallel to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=0^{\circ}$ , so the force due to this field is zero.

$- B_y$ : this field is perpendicular to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=90^{\circ}$ . Therefore, $\theta=90^{\circ}$ , so the force due to this field is:

$F_{B_y}=qvB_y$

where:

$q=+4.9\cdot 10^{-6}C$ is the charge

$v=345 m/s$ is the velocity

$B_y = +1.9 T$ is the magnetic field

Substituting,

$F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And the direction of this force can be found using the right-hand rule:

- Index finger: direction of the velocity (+x axis)

- Middle finger: direction of the magnetic field (+y axis)

- Thumb: direction of the force (+z axis)

c)

As in part b), the electric force has not change, since it does not depend on the veocity of the particle:

$F_{E_x}=1.19\cdot 10^{-3}N$ (+x axis)

For the field $B_x$ , the velocity (+z axis) is now perpendicular to the magnetic field (+x axis), so the force is

$F_{B_x}=qvB_x$

And by substituting,

$F_{B_x}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+x axis)

- Thumb: force (+y axis)

For the field $B_y$ , the velocity (+z axis) is also perpendicular to the magnetic field (+y axis), so the force is

$F_{B_y}=qvB_y$

And by substituting,

$F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+y axis)

- Thumb: force (-y axis)

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4 years ago

As you watch the video, notice that the size of the tidal bulges varies with the Moon's phase, which depends on its orbital posi

Naya [18.7K]

Answer:

Both

A. Low tides are lowest at both full moon and new moon.

B. High tides are highest at both full moon and new moon.

Explanation:

Tides are formed as a consequence of the differentiation of gravity due to the moon across to the Earth sphere.

Since gravity variate with the distance:

$F = G\frac{m1\cdot m2}{r^{2}}$ (1)

Where m1 and m2 are the masses of the two objects that are interacting and r is the distance Where m1 and m2 are the masses of the two objects that are interacting and r is the distance between them.

For example, see the image below, point A is closer to the moon than point b and at the same time the center of mass of the Earth will feel more attracted to the moon than point B. Therefore, that creates a tidal bulge in point A and point B.

On the other hand, a full moon it gets when Sun, the Earth and the moon are in a line and the moon is reflecting the sunlight.

When the Moon is between the Earth and the Sun it will be illuminated in its back, so it is not possible to see it from the Earth (that is called new moon).

In those two cases mentioned above, the Sun tidal force contributes to the tidal force of the moon over the earth making high tides higher and low tides lower.

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4 years ago