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MrMuchimi
3 years ago
8

Show that the units N/kg can be written using only units of meters (m) and second (s). Is this a unit of mass ,acceleration or f

orce.
Physics
1 answer:
nikdorinn [45]3 years ago
3 0
<span>When the difference between two results is larger than the estimates error, the result is</span>
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What two elements are used with other metals to lower their melting points?
vampirchik [111]

Answer:

C) antimony and bismuth

Explanation:

6 0
3 years ago
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What is the net force necessary for a 1.6 x 103 kg car to accelerate forward at 2.0<br> m/s?
erastova [34]

Answer:31245.132

Explanation:

yeah

8 0
4 years ago
The ________________ of a machine is equal to the ratio of the work output to its input.
erastova [34]
Answer: mechanical efficieny.

Efficieny is also expressend as percent. The formula for mechanical efficiency as percent is the ratio work output to wor input times 100.

The ideal mechanical efficiency for a machine would be 1 or 100% which means that all the input work is converted into output work. But this is just an idealization as the friction and other losses of energy make it imposible to reach 100% efficiency in reality, so the mechanical efficiency of real machinces is less than 100% or 1.
5 0
3 years ago
Why does the investigator in an expirement change only one variable
jenyasd209 [6]
To see what causes a change in experiment. If they kept everything the same what would be the point of experiment? One thing has to be different to test and get a result ! They have to see how that one variable changes the experiment. If they change so many things then the experiment will be messed up. They have to focus on that one variable during experiment to get the proper result. Hope this helps you. Think about it if you want to know how antibiotic effects bodies. You choose two people and you give antibiotic to one. And you don't give any other drug to the people you are testing. Because you truly want to know the effect of antibiotic alone.
8 0
4 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
3 years ago
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