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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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Ana walks from 4 m to 200 cm. Which of the following statements is true about
Gemiola [76]
Distance=2m

because 200cm = 2m
so 4m-2m=2m
3 0
3 years ago
Joe runs 10 m north, 20 m south, 9m south, and then 15 m north. What is Joe's<br> displacement?
tresset_1 [31]

Answer:

Joes displacement is 54m

4 0
1 year ago
When a metal bonds with a nonmetal, they form a(n)
MArishka [77]
I believe it would be B.
5 0
3 years ago
A 69-kg student sitting on a hardwood floor does not slide until pulled by a 260-N horizontal force. find coefficient of frictio
salantis [7]

Answer:

μ = 0.385

Explanation:

Given that,

The mass of the student, m = 69 Kg

The horizontal force applied, F = 260 N

The normal force acting on the body, weight = mg = 69g  N

                                                                                    = 676.2 N

The coefficient of friction acting on a body is equal to the force acting on the body to the normal force acting on the body due to gravitation.

The formula for coefficient of friction,

                                    μ =  F / N

Substituting the values in the above equation,

                                     μ = 260 N / 676.2 N

                                        = 0.385

Hence, the  coefficient of friction, μ = 0.385

6 0
3 years ago
If the force of gravity between a book of mass 0.50 kg and a calculator of 0.100 kg is 1.5 × 10-10 N, how far apart are they?  (
valkas [14]
The gravitational force between two masses m₁ and m₂ is
F=G \frac{m_{1} m_{2}}{d^{2}}
where
G = 6.67408 x 10⁻¹¹ m³/(kg-s²), the gravitational constant
d =  distance between the masses.

Given:
F = 1.5 x 10⁻¹⁰ N
m₁ = 0.50 kg
m₂ = 0.1 kg

Therefore
1.5 x 10⁻¹⁰ N = (6.67408 x 10⁻¹¹ m³/(kg-s²))*[(0.5*0.1)/(d m)²]
d² = [(6.67408x10⁻¹¹)*(0.5*0.1)]/1.5x10⁻¹⁰
     = 0.0222
d = 0.1492 m = 149.2 mm

Answer: 149.2 mm
8 0
3 years ago
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