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kipiarov [429]
2 years ago
12

What is the magnitude of the speed of a person at the equator due to rotational speed of the Earth?

Physics
1 answer:
notka56 [123]2 years ago
4 0
The circumference of the Earth at the equator is listed as 24,901 miles.
So his speed is

                     24,901 miles per day.

Convert it to units that we have a better feel for:

                   (24,901 mi/da) x (1 da / 24 hrs)

             =    (24,901 / 24)  (miles/hour)

             =  about  1,038 miles per hour.

You'll find a huge number of people on the internet these days,
telling you that you could not be moving at that speed and not
feel it, so therefore the Earth is not spinning, and it's not a globe.

I have a lot of feelings and comments about those people, their
lines of reasoning, and their levels of education and intelligence,
so don't get me started.

I just want to guarantee you that everything you're learning about
the Earth and the solar system in school is well founded, and it's
all based on the life's work of some of the smartest people of the
past 300 years of human history.  Everything you're taught about
the Earth has good reasons behind it, whereas those other people
have nothing.

A person on Earth's equator is moving from west to east at roughly
1,038 miles per hour, relative to any point on the Earth's rotation axis.
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SP 15The magnetic field in a region of space is measured to be:This field is known to be caused by a cluster of long-straight wi
tino4ka555 [31]

Answer:

 i = 0.477 10⁴ B

the current flows in the  counterclockwise

Explanation:

For this exercise let's use the Ampere law

                    ∫ B . ds = μ₀ I

Where the path is closed

Let's start by locating the current vines that are parallel to the z-axis, so it must be exterminated along the x-axis and as the specific direction is not indicated, suppose it extends along the y-axis.

From BiotSavart's law, the field must be perpendicular to the direction of the current, so the magnetic field must go in the x direction.

We apply the law of Ampere the segment parallel to the x-axis is the one that contributes to the integral, since the other two have an angle of 90º with the magnetic field

Segment on the y axis

        L₀ = (y2-y1)

        L₀ = 3-0 = 3 cm

Segment on the point x = 2 cm

        L₁ = 3-0

        L₁ = 3cm

       B L = μ₀ I

       B 2L = μ₀ I

        i = 2 L B /μ₀

        i= 2 0.03 / 4π 10⁻⁷   B

        i = 4.77 10⁴  B

The current is perpendicular to the magnetic field whereby the current flows in the  counterclockwise

8 0
3 years ago
Is this charging by induction or conduction?
sineoko [7]
Conduction i believe
3 0
2 years ago
A short-wave radio antenna is supported by two guy wires, 155 ft and 175 ft long. Each wire is attached to the top of the antenn
Vladimir79 [104]

Answer:

163.8 ft

Explanation:

In triangle ABD

AB = 155 ft

Cos63 = \frac{BD}{AB} = \frac{BD}{155}\\BD = 155 Cos63 \\BD = 70.4 ft

Sin63 = \frac{AD}{AB} = \frac{AD}{155} \\AD = 166 Sin63\\AD = 148 ft

Using Pythagorean theorem in triangle ADC

AC^{2} = AD^{2} + DC^{2} \\175^{2} = 148^{2} + DC^{2} \\DC = 93.4 ft

d = distance between the anchor points

distance between the anchor points is given as

d = BD + CD = 70.4 + 93.4\\d = 163.8 ft

5 0
3 years ago
Read 2 more answers
When ultra violets lights shine on glass what does it do to electrons in the glass structure?
andreev551 [17]

Answer:

No

Explanation:

7 0
2 years ago
Problem 4: a long wire carries current towards east. a positive charge moves westward and just north from the wire. what is the
Alex787 [66]
The direction of the force experienced by the positive charge is upward.

We can use the right-hand rule to understand the direction of the Lorentz force acting on the charge: let's put the thumb in the same direction of the current in the wire (eastward), while the other fingers "wrap themselves" around the wire. These other fingers give the direction of the Lorentz force in every point of the space around the wire. Since the charge is located north of the wire, in that point the fingers are directed upward, so the positive charge experiences a force directed upward.
(if it was a negative charge, we should have taken the opposite direction)
4 0
3 years ago
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