Which law states that absolute zero cannot be reached?

What is the orbital period of a spacecraft in a low orbit near the surface of mars? The radius of mars is 3.4×106m.

valkas [14]

<h2>Answer: 56.718 min</h2>

Explanation:

According to the Third Kepler’s Law of Planetary motion<em> </em><em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. </em>

In other words, this law states a relation between the orbital period $T$ of a body (moon, planet, satellite) orbiting a greater body in space with the size $a$ of its orbit.

This Law is originally expressed as follows:

$T^{2}=\frac{4\pi^{2}}{GM}a^{3}$ (1)

Where;

$G$ is the Gravitational Constant and its value is $6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$

$M=6.39(10)^{23}kg$ is the mass of Mars

$a=3.4(10)^{6}m$ is the semimajor axis of the orbit the spacecraft describes around Mars (assuming it is a <u>circular orbit </u>and a <u>low orbit near the surface </u>as well, the semimajor axis is equal to the radius of the orbit)

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

$T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}$ (2)

$T=\sqrt{\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(6.39(10)^{23}kg)}(3.4(10)^{6}m)^{3}}$ (3)

$T=\sqrt{11581157.44 s^{2}}$ (4)

Finally:

$T=3403.1099s=56.718min$ This is the orbital period of a spacecraft in a low orbit near the surface of mars

6 0

3 years ago

How to use kinetic energy to calculate velocity

BartSMP [9]

As we know the formula of kinetic energy is

$KE = \frac{1}{2} mv^2$

here given that

KE = 150,000 J

mass = 120 kg

we can use this to find speed

$150,000 = \frac{1}{2} * 120 * v^2$

$v^2 = 2500$

$v = 50 m/s$

So speed of above object is 50 m/s

7 0

2 years ago

In an Atwood's machine, one block has a mass of 602.0 g, and the other a mass of 717.0 g. The pulley, which is mounted in horizo

Wittaler [7]

Answer:

The acceleration of the both masses is 0.0244 m/s².

Explanation:

Given that,

Mass of one block = 602.0 g

Mass of other block = 717.0 g

Radius = 1.70 cm

Height = 60.6 cm

Time = 7.00 s

Suppose we find the magnitude of the acceleration of the 602.0-g block

We need to calculate the acceleration

Using equation of motion

$s=ut+\dfrac{1}{2}at^2$

Where, s = distance

t = time

a = acceleration

Put the value into the formula

$60.0\times10^{-2}=0+\dfrac{1}{2}\times a\times(7.00)^2$

$a=\dfrac{60.0\times10^{-2}\times2}{(7.00)^2}$

$a=0.0244\ m/s^2$

Hence, The acceleration of the both masses is 0.0244 m/s².

5 0

3 years ago

Part B

Katena32 [7]

Answer:

While slavery was the major issue separating the North and South, it was not slavery itself that sparked the conflict. The South wanted to secede from the Union, and the North refused. While President Abraham Lincoln personally opposed slavery, he recognized that it was legal under the U.S. Constitution at the time. He also recognized that few in the North were ready to go to war to free the slaves. For Lincoln and the northern majority, preservation of the Union was the foremost goal.

4 0

2 years ago

HELP ME 20 POINTS!!!Assume that the average volume of an adult human body is one-tenth

Agata [3.3K]

Answer:

,Assume that the average volume of an adult human body is one-tenth

cubic meter (0.10 m) and that there are two billion (2.0 x 109)

adults in the world.

a. What would be the total volume of all the adults in the world?

b. Compute the length of one edge of a cubic container that has a

volume equal to the volume of all the adults in the world.