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3 years ago

What causes two distinct pressure zones between the equator and the poles?

Physics

2 answers:

Brilliant_brown [7]3 years ago

7 0

<span>Ionospheric interactions with radiation </span>

Over [174]3 years ago

3 0

Answer:

Option (C)

Explanation:

The Coriolis force is defined as the force that generates due to the rotation of the earth around its imaginary axis. Due to the effect of the Coriolis force, the wind gets deviated. The wind is deviated towards the right in the northern hemisphere, whereas the wind gets deviated towards the left in the southern hemisphere.

Due to this changing wind direction, there occurs the low pressure and the high-pressure zone between the polar and the equatorial zone. The low pressure zones are formed at the equator and in temperate regions. Whereas, the high-pressure zones are formed at the tropics and the polar region.

Hence, the correct answer is option (C).

You might be interested in

What is the frequency of microwaves of wavelength 3 cm?

tiny-mole [99]

Answer:

10 GHz

Explanation:

Applying,

v = λf.................... Equation 1

Where v = speed of microwave, λ = wavelength, f = frequency.

make f the subject of the equation

f = v/λ................ Equation 2

Note: Microwave is an electromagnetic wave, and all electromagnetic wave have the same speed, which is 3×10⁸ m/s

From the question,

Given: λ = 3 cm = 0.03 m

Constant; v = 3×10⁸ m/s

Substitute these values into equation 2

f = 3×10⁸/(0.03)

f = 10¹⁰ Hz

f = (10¹⁰/10⁹) GHz

f = 10 GHz

4 0

2 years ago

An X-Ray machine delivers a radiation dose of 5mRem/hr. at 3ft from the machine. How far will the X-Ray technician have to move

kupik [55]

Answer:

4.7ft

Explanation:

Pls see attached file

8 0

3 years ago

Two charges, each of 2.9 microC are placed at two corners of a square 50cm on a side, If the charges are on one side of the squa

anyanavicka [17]

Answer:

The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

Explanation:

Given that,

First charge $q_{1}= 2.9\mu C$

Second charge $q_{2}= 2.9\mu C$

Distance between two corners r= 50 cm

We need to calculate the electric field due to other charges at one corner

For E₁

Using formula of electric field

$E_{1}=\dfrac{kq}{r'^2}$

Put the value into the formula

$E_{1}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\sqrt{2}\times10^{-2})^2}$

$E_{1}=52200=52.2\times10^{3}\ N/C$

For E₂,

Using formula of electric field

$E_{1}=\dfrac{kq}{r^2}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\times10^{-2})^2}$

$E_{2}=104400=104.4\times10^{3}\ N/C$

We need to calculate the horizontal electric field

$E_{x}=E_{1}\cos\theta$

$E_{x}=52.2\times10^{3}\times\cos45$

$E_{x}=36910.97=36.9\times10^{3}\ N/C$

We need to calculate the vertical electric field

$E_{y}=E_{2}+E_{1}\sin\theta$

$E_{y}=104.4\times10^{3}+52.2\times10^{3}\sin45$

$E_{y}=141310.97=141.3\times10^{3}\ N/C$

We need to calculate the net electric field

$E_{net}=\sqrt{E_{x}^2+E_{y}^2}$

Put the value into the formula

$E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}$

$E_{net}=146038.69\ N/C$

$E_{net}=146.03\times10^{3}\ N/C$

We need to calculate the direction of electric field

Using formula of direction

$\tan\theta=\dfrac{141.3\times10^{3}}{36.9\times10^{3}}$

$\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})$

$\theta=75.36^{\circ}$

Hence, The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

4 0

3 years ago

A 3.00 kg mass is traveling at an initial speed of 25.0 m/s. What is the

Nitella [24]

Answer:

The magnitude of the force required to bring the mass to rest is 15 N.

Explanation:

Given;

mass, m = 3 .00 kg

initial speed of the mass, u = 25 m/s

distance traveled by the mass, d = 62.5 m

The acceleration of the mass is given as;

v² = u² + 2ad

at the maximum distance of 62.5 m, the final velocity of the mass = 0

0 = u² + 2ad

-2ad = u²

-a = u²/2d

-a = (25)² / (2 x 62.5)

-a = 5

a = -5 m/s²

the magnitude of the acceleration = 5 m/s²

Apply Newton's second law of motion;

F = ma

F = 3 x 5

F = 15 N

Therefore, the magnitude of the force required to bring the mass to rest is 15 N.

4 0

3 years ago

the angular speed of an automobile engine is increased at a constant rate from 1300rev/min to 2000rev/min in 3s (a) what is its

GalinKa [24]

Answer:

please find attached pdf

Explanation:

Download pdf

8 0

2 years ago