What is the specific heat capacity of water?

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 “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

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 circular orbit and a low orbit near the surface 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

2 Ten identical lengths of wire are laid closely side-by-

Mice21 [21]

Answer:

(a) 0.71 mm

(b) 0.158 cubic cm

Explanation:

The width of one wire is the diameter of the wire.

(a) Let the diameter of each wire is d.

So, 10 d = 14.2 mm

d = 1.42 mm

radius of each wire, r = d/2 = 1.42/2 = 0.71 mm

(b) Length, L = 10 cm

The volume of the single wire is given by

$V =\pi\times r^2\times h \\\\V =3.14\times 0.071^2\times 10\\\\V =0.158 cm^3$

7 0

2 years ago

Write any two importance of gravitational force

saul85 [17]

Answer:

plz mark me as brainliest plz

Explanation:

The gravitational force of the earth keeps us bound to the earth. Gravitational force between earth and sun makes the earth move around the sun. Gravitational force between moon and earth makes the moon go around the earth.

5 0

2 years ago

When a mass of 0.350 kg is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The mass is now displ

Ray Of Light [21]

Answer:

period of oscillations is 0.695 second

Explanation:

given data

mass m = 0.350 kg

spring stretches x = 12 cm = 0.12 m

to find out

period of oscillations

solution

we know here that force

force = k × x .........1

so force = mg = 0.35 (9.8) = 3.43 N

3.43 = k × 0.12

k = 28.58 N/m

so period of oscillations is

period of oscillations = 2π × $\sqrt{\frac{m}{k} }$ ................2

put here value

period of oscillations = 2π × $\sqrt{\frac{0.35}{28.58} }$

period of oscillations = 0.6953

so period of oscillations is 0.695 second

4 0

3 years ago

In which labeled portion of the curve would you use the heat of vaporization to calculate the heat absorbed? (image attached ins

sergij07 [2.7K]

In the labeled portion of the curve ,you use the heat of vaporization to calculate the heat absorbed in the 4th portion. It is indicated in the picture that it is the region where vaporization occurs, that is why you need to consider this portion to calculate.

3 0

2 years ago