All categories

3 years ago

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

1 answer:

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.

You might be interested in

a 1500 kg car accelerates uniformly from rest to 10.0 meters per secound in 3.0 secound .what is the work done on the car in thi

zubka84 [21]

Answer:

The work done on the car is, W = 75,000 J

The power delivered by the engine, P = 25,000 watts

Explanation:

Given,

The mass of the car, m = 1500 Kg

The initial velocity of the car, u = 0

The final velocity of the car, v = 10 m

The time duration of the travel, t = 3 s

Using the first equation of motion

v = u + at

a = (v - u) / t

Substituting the given values in the above equation

a = (10 - 0) / 3

= 3.33 m/s²

Using the second equations of motion

s = ut + 1/2 at²

= 0 + 0.5 x 3.33 x 3²

= 15 m

The force exerted by the car

F = m x a

= 1500 Kg x 3.33 m/s²

= 5000 N

The work done by the car,

W = F x S

= 5000 N x 15 m

= 75,000 J

Hence, the work done on the car is, W = 75,000 J

The power delivered by the engine,

P = W / t

= 75,000 J / 3 s

= 25,000 watts

The power delivered by the engine, P = 25,000 watts

5 0

4 years ago

Which of the following statements describes the law of inertia?

Drupady [299]

Answer:

B. objects remain in the same state of motion unless a force acts on them

Explanation:

Sana nakatulong

8 0

3 years ago

At a particular instant, a hot air balloon is 100 m in the air and descending at a constant speed of 2.0 m/s. at this exact inst

rewona [7]

Answer:

86.4 m horizontal from landing spot

Explanation:

Find out how long before the ball hits the ground

vertical speed of ball = -2 m/s gravity = - 9.81 m/s^2

find time to hit ground from 100 m

( height will be<u> zero</u> when it hits the ground)

<u>0 </u>= 100 - 2 t - 1/2 ( 9.81) t^2

use Quadratic Formula to find t = 4.32 seconds

horizontal speed of ball = 20 m/s

in 4.32 seconds it will travel horizontally 20 m/s * 4.32 s = 86.4 m

3 0

2 years ago

Give a calculate answer to show that the two values (English system and metric system) for the Planck Constant are equivalent.

Nataly_w [17]

Answer:

Given values of Planck Constant are equivalent in English system and metric system.

Explanation:

Value of Planck's constant is given in English system as 4.14 x 10⁻¹⁵eV s.

Converting this in to metric system .

We have 1 eV = 1.6 x 10⁻¹⁹ J

Converting

4.14 x 10⁻¹⁵eV s = 4.14 x 10⁻¹⁵x 1.6 x 10⁻¹⁹ = 6.63 x 10⁻³⁴ Joule s

So Given values of Planck Constant are equivalent in English system and metric system.

7 0

3 years ago

Two boys are sliding toward each other on a frictionless, ice-covered parking lot. Jacob, mass 45 kg, is gliding to the right at

Maurinko [17]

Answer:

$v=0.3381\ m.s^{-1}$ in the direction of motion of Jacob

Explanation:

Given:

mass of Jacob, $m_j=45\ kg$

velocity of Jacob, $v_j=8.08\ m.s^{-1}$

mass of Ethan, $m_e=31\ kg$

velocity of Ethan, $v_e=10.9\ m.s^{-1}$

Now using the conservation of linear momentum for the case:

(When the two masses in motion combine to form one after the collision then they will move together in the direction of the greater momentum.)

$(m_j+m_e).v=m_j.v_j-m_e.v_e$

$(45+31)\times v=45\times 8.08-31\times 10.9$

$v=0.3381\ m.s^{-1}$ in the direction of motion of Jacob as it was assumed to be positive.

5 0

3 years ago