calculate the pressure exerted on the floor when a elephant who weighs 6000N stands on 1 foot which has a area of 20m

A satellite, moving in an elliptical orbit, is 368 km above Earth's surface at its farthest point and 164 km above at its closes

Marat540 [252]

Answer:

a) 6636 km

b) 0.0154

Explanation:

The height above the earth at its furthest point is 368 km

The height above the earth at its closest point is 164 km

Radius of the Earth is 6370 km

The distance of the satellite from the center of the earth to the furthest point is 6370 + 368 km = 6738 km

The distance of the satellite from the center of the earth to the closest point is 6370 + 164 = 6534 km

If we add together the sum of the distance of the satellite from the furthest and its closest distance, it is equal to the 2 major semi axis.

Basically,

2a = R + r

a = (R + r) / 2

a = (6738 + 6534) / 2

a = 13272 / 2

a = 6636 km

Eccentricity, e = (a - r) / a

Eccentricity, e = (6636 - 6534) / 6636

Eccentricity, e = 102 / 6636

Eccentricity, e = 0.0154

3 0

2 years ago

Which describes one event that causes an eclipse?

seraphim [82]

Answer:

earth's shadow covering the moon,thats lunar eclipse

5 0

3 years ago

A closed, uninsulated system fitted with movable piston, so no matter is exchanged with the surroundings, was assembled. Introdu

xeze [42]

Answer: You do not specify what is being asked for. ∆E? ∆H?

∆E = (430 - 238) J = 192 J

∆H = 430 J

Explanation:

If asked for the value of ∆H the answer is simply the change in heat, and in the question, it states introduction of 430 J of heat is causing the system to expand.

Therefore ∆H = 430 J

If asked for ∆E, we know that ∆E = ±q (heat) + work (-P∆V) = ±q + w

The question states that 238 J of work are done AND the system expanded

(work is negative because expansion means work is done BY the system, releasing energy/heat... Conversely, if the system were compressed, work is done ON the system, absorbing heat/energy)

Therefore, ∆E = (430 - 238) J = 192 J

8 0

2 years ago

A physics student throws a softball straight up into the air. The ball was in the air for a total of 3.56 s before it was caught

meriva

Answer:

The initial velocity of the softball is 14.711 meters per second.

Explanation:

This is a case of an object which experiments a free fall, that is, an uniform accelerated motion due to gravity and in which effects from air friction and Earth's rotation can be neglected.

From statement we must understand that the student threw the softball upwards and it is caught at original position 3.56 seconds later. Initial and final heights, time and gravitational acceleration are known and initial speed is unknown. The following equation of motion is used:

$y = y_{o} + v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2}$ (Eq. 1)