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3 years ago

What would cause rain showers to continue over a particular area for several days?

Physics

2 answers:

fiasKO [112]3 years ago

8 0

<h2>Answer:<em><u> a high-pressure system</u></em></h2>

Aloiza [94]3 years ago

5 0

I think is a high-pressure system because it is only in one particular area.

You might be interested in

Light waves are electromagnetic waves that travel at 3.00 Light waves are electromagnetic waves that travel 108 m/s. The eye is

svlad2 [7]

(a) $5.45 \cdot 10^{14} Hz$

The relationship between frequency and wavelength of an electromagnetic wave is given by

$c=f \lambda$

where

$c=3.00 \cdot 10^8 m/s$ is the speed of light

$f$ is the frequency

$\lambda$ is the wavelength

In this problem, we are considering light with wavelength of

$\lambda=5.50 \cdot 10^{-7} m$

Substituting into the equation and re-arranging it, we can find the corresponding frequency:

$f=\frac{c}{\lambda}=\frac{3.00 \cdot 10^8 m/s}{5.50 \cdot 10^{-7} m}=5.45 \cdot 10^{14} Hz$

(b) $1.83\cdot 10^{-15} s$

The period of a wave is equal to the reciprocal of the frequency:

$T=\frac{1}{f}$

And using $f=5.45 \cdot 10^{14} Hz$ as we found in the previous part, we can find the period of this wave:

$T=\frac{1}{5.45 \cdot 10^{14} Hz}=1.83\cdot 10^{-15} s$

5 0

3 years ago

Two power lines run parallel for a distance of 220 m and are separated by a distance of 40.0 cm. If the current in each of the t

daser333 [38]

Answer:

The magnitude of force is 1.86 N and the direction of force is towards the other wire.

Explanation:

Given:

Current flowing through each power line, I = 130 A

Distance between the two power lines, d = 40 cm = 0.4 m

Length of power lines, L = 220 m

The force exerted by the power lines on each other is given by the relation:

$F = \frac{\mu_{0}LII }{2\pi d}$

Substitute the suitable values in the above equation.

$F = \frac{4\pi\times10^{-7}\times220\times130\times130 }{2\pi\times0.4}$

F = 1.86 N

Since the direction of current flowing through the power lines are opposite to each other, so the force is attractive in nature. Hence, the direction of force experienced by the power lines on each other is towards the each other.

5 0

3 years ago

On the way to the moon, the Apollo astronauts reach a point where the Moon’s gravitational pull is stronger than that of Earth’s

Drupady [299]

Answer:

rm = 38280860.6[m]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

$r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]$

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

$F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2} } \\$

$F_{m} =G*\frac{m_{m}*m_{a} }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]$

When we match these equations the masses cancel out as the universal gravitational constant

$G*\frac{m_{e} *m_{a} }{r_{e}^{2} } = G*\frac{m_{m} *m_{a} }{r_{m}^{2} }\\\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} }$

To solve this equation we have to replace the first equation of related with the distances.

$\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m} )^{2} } = \frac{7.36*10^{22} }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17} \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0$

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

$r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c } }{2*a}\\ where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) } }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]$