(m/s) 20 10. 100 -10 total distance

How fast would you have to travel on the surface of earth at the equator to keep up with the sun (that is, so that the sun would

professor190 [17]

Answer: 1037 miles per hour

Explanation: In order to see the sun in the same position in the sky, you would have to travel against the speed of rotation of the earth, because this is what causes the sun to appear in a constantly changing position.

Because of this, we will have to calculate the speed of rotation of the earth. To get started, we must know the circumference of the earth. Assuming the circumference formula for a sphere,

$Circumference=2\pi R$

Where R is the radius of the earth, we find that the perimeter of the earth is approximately 24881 miles. The equation to calculate speed is given by

$v=\frac{Distance}{Time}$

Because the earth completes one rotation in 24 hours, we have to find the speed of rotation as the perimeter of the earth divided by 24 hours.

The obtained result is 1037 miles per hour.

You would have to travel at 1037 miles per hour in the direction opposite to the direction the rotation is ocurring in.

5 0

3 years ago

An elevator filled with passengers has a mass of 1603 kg. (a) The elevator accelerates upward from rest at a rate of 1.20 m/s2 f

Galina-37 [17]

Answer:

T = 17649.03 N = 17.65 KN

Explanation:

The tension in the cable must be equal to the apparent weight of the passenger. For upward acceleration:

$T = W_A = m(g+a)\\$

where,

T = Tension in cable = ?

$W_A$ = Apparent weight

m = mass = 1603 kg

g = acceleration due to gravity = 9.81 m/s²

a = acceleration of elevator = 1.2 m/s²

Therefore,

$T = (1603\ kg)(9.81\ m/s^2+1.2\ m/s^2)\\\\$

T = 17649.03 N = 17.65 KN

8 0

2 years ago

When an electron loses or gains a charge, it becomes?

suter [353]

Answer:

It becomes an Ion with either positive or negative charge

6 0

3 years ago

Read 2 more answers

calculate earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the s

IceJOKER [234]

Answer:

Hello your question is incomplete below is the complete question

Calculate Earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the sun, Take the eccentricity of Earth's orbit to be 1/60 and its Semimajor axis to be 93,000,000

answer : V = 1.624* 10^-5 m/s

Explanation:

First we have to calculate the value of a

a = 93 * 10^6 mile/m * 1609.344 m

= 149.668 * 10^8 m

next we will express the distance between the earth and the sun

$r = \frac{a(1-E^2)}{1+Ecos\beta }$ --------- (1)

a = 149.668 * 10^8

E (eccentricity ) = ( 1/60 )^2

$\beta$ = 90°

input the given values into equation 1 above

r = 149.626 * 10^9 m

next calculate the Earths velocity of approach towards the sun using this equation

$v^2 = \frac{4\pi^2 }{r_{c} }$ ------ (2)

Note :

Rc = 149.626 * 10^9 m

equation 2 becomes

( $V^2 = (\frac{4\pi^{2} }{149.626*10^9})$

therefore : V = 1.624* 10^-5 m/s

4 0

3 years ago

Hey pls help thanks a lot

galben [10]

Answer:

I am not sure about the answer as I don't have a proper calculator besides me now

Explanation:

but I used this equation:

(8.20)sin30(1-d)=10d

Idk whether it is correct or not, I'm just a student too

what is your method of doing this question?

4 0

3 years ago