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2 years ago

Sugar is made of carbon, hydrogen, and oxygen atoms. Sugar is

Physics

2 answers:

Aneli [31]2 years ago

7 0

D. A solution because it dissolves when mixed with water

ZanzabumX [31]2 years ago

6 0

D because it dissolves

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A beaker of vegetable oil contains a beam of light that is aimed at a surface at an angle of 34 degrees as shown. If the index o

Over [174]

Answer:

Angle of reflection of light is 34 degree

Explanation:

As per law of reflection of light we know that

angle of incidence of light = angle of reflection of light

So here we know that

angle of incidence on the surface of oil is given as

$\theta_i = 34 degree$

so we know that

$\theta_i = \theta_r$

so here we can say that reflection angle of light will be same as angle of incidence

$\theta_r = 34 degree$

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2 years ago

What kind of reaction typically has a carbon compound and oxygen as

Anna11 [10]

Answer:

Answer would be C because combustion does the process of heat. Oxygen is a reactant in a combination reaction, which always produces energy in the form of heat and light. Carbon burns in the presence of oxygen to give carbon dioxide.

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2 years ago

Preston goes on a camel safari. There, he travels 5 km north, then 3 km east.

sammy [17]

7777Answer:

Explanation:

6 0

2 years ago

Read 2 more answers

A Ferris wheel with a radius of 5 m is rotating at a rate of one revolution every 2 minutes. How fast is a rider rising when the

Olenka [21]

Explanation:

Below is an attachment containing the solution.

5 0

2 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<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

2 years ago