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3 years ago

The ____ is the place where position equals zero

Physics

1 answer:

Alexus [3.1K]3 years ago

7 0

The answer to your question is the Origin

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An astronaut weighs 8.00 × 102 newtons on the sur- face of Earth. What is the weight of the astronaut 6.37 × 106 meters above th

kolbaska11 [484]

Answer:

$mg=200.4 N$ .

Explanation:

This problem can be solved using Newton's law of universal gravitation: $F=G\frac{m_{1}m_{2}}{r^{2}}$ ,

where F is the gravitational force between two masses $m_{1}$ and $m_{2}$ , $r$ is the distance between the masses (their center of mass), and $G=6.674*10^{-11}(m^{3}kg^{-1}s^{-2})$ is the gravitational constant.

We know the weight of the astronout on the surface, with this we can find his mass. Letting $w_{s}$ be the weight on the surface:

$w_{s}=mg$ ,

$mg=8*10^{2}$ ,

$m=(8*10^{2})/g$ ,

since we now that $g=9.8m/s^{2}$ we get that the mass is

$m=81.6kg$ .

Now we can use Newton's law of universal gravitation

$F=G\frac{Mm}{r^{2}}$ ,

where $m$ is the mass of the astronaut and $M$ is the mass of the earth. From Newton's second law we know that

$F=ma$ ,

in this case the acceleration is the gravity so

$F=mg$ , (<u>becarefull, gravity at this point is no longer</u> $9.8m/s^{2}$ <u>because we are not in the surface anymore</u>)

and this get us to

$mg=G\frac{Mm}{r^{2}}$ , where $mg$ is his new weight.

We need to remember that the mass of the earth is $M=5.972*10^{24}kg$ and its radius is $6.37*10^{6}m$ .

The total distance between the astronaut and the earth is

$r=(6.37*10^{6}+6.37*10^{6})=2(6.37*10^{6})=12.74*10^{6}$ meters.

Now we can compute his weigh:

$mg=G\frac{Mm}{r^{2}}$ ,

$mg=(6.674*10^{-11})\frac{(5.972*10^{24})(81.6)}{(12.74*10^{6})^{2}}$ ,

$mg=200.4 N$ .

5 0

3 years ago

1. An isotope has the same ________, but a different ________ .

Vika [28.1K]

1:D(Atomic number, Mass number) 2:B(Neutrons) 3:b(18) hope it helps

3 0

3 years ago

a capacitor of unknown charged capacitance c is charged to 100v and the connected across an intially uncharged 60micro f capacit

poizon [28]

Answer:

hi thank you for your free point

6 0

3 years ago

A 10-kg mass slides down a flat hill that makes an angle of 10° with the horizontal. If friction is negligible, what is the resu

slega [8]

Answer:

Resultant force = 17.02 N

Explanation:

As we assume the coefficient of friction is negligible, the normal force won't affect the resultant force.

As a result of this, the sine of the angle times the force of gravity on the 10 kg mass is equal to the resultant force, which is, force due to gravity = m × a = 10 kg × 9.8 m/s² = 98 N

98 sin(10)=?

Sin(10)= 0.1736

98 x 0.1736= 17.02 N.

Therefore, the resultant force = 17.02 N

8 0

3 years ago

Answer all of these questions and you will get the brainlist

loris [4]

Answer:

a is potential and b is kinetic

7 0

3 years ago