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3 years ago

Wispy, feathery clouds made mostly of ice crystals that form at high altitudes, usually above 6 kilometers.

Physics

1 answer:

Oliga [24]3 years ago

4 0

Cirrus clouds are the clouds mostly made of ice becasue of the high altitudes

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One of the way atoms bond with each other would be through:

Galina-37 [17]

Answer:

ehdgywi

Explanation:

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8 0

3 years ago

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

A flywheel with radius of 0.400 mm starts from rest and accelerates with a constant angular acceleration of 0.600 rad/s2rad/s2.

charle [14.2K]

Answer: $0.00024\ m/s^2$

Explanation:

Given

Radius of flywheel is $r=0.4\ mm$

Angular acceleration $\alpha=0.6\ rad/s^2$

For no change in radius, tangential acceleration is given as

$\Rightarrow a_t=a\lpha \times r$

Insert the values

$\Rightarrow a_t=0.6\times 0.4\times 10^{-3}\ m/s^2\\\Rightarrow a_t=2.4\times 10^{-4}\ m/s^2\ \text{or}\ 0.00024\ m/s^2$

5 0

3 years ago

What is the fundamental frequency (in Hz) of a 0.632 m long tube, open at both ends, on a day when the speed of sound is 344 m/s

Greeley [361]

Answer:

$f=272.15Hz$

Explanation:

Given data

Length of tube L=0.632 m

Speed of sound v=344 m/s

To find

Fundamental frequency f

Solution

The fundamental frequency of the tube can be given as:

$f=\frac{v}{2L}\\ f=\frac{344m/s}{2(0.632m)}\\ f=272.15Hz$

4 0

4 years ago

How does an electromagnet become permanent

love history [14]

A moving electrical charge produces a magnetic field and a moving magnetic field produces an electrical field. An electromagnet works by coiling a bunch of wire and spinning a couple of magnets around that wire at high speeds. When this occurs the magnets induce an electric current in the wire and hence the electricity production. Once the magnets stop spinning, the induced electrical field dissipates and the current stops flowing through the wire.

6 0

3 years ago