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olchik [2.2K]
3 years ago
12

Which elements are not likely to be formed in the sun sometime during its life cycle?

Physics
1 answer:
Sladkaya [172]3 years ago
8 0

Answer:

correct answer is iron

Explanation:

solution

as Iron is only formed in the cores of the star that is more massive than the sun because it takes very much energy  as gravitational force push inward on the star  

and the sun can form the helium and  carbon and the oxygen and also other elements  

but not the iron or any iron group element

so correct answer is iron

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What speed should a satellite with a mass of 1500 kg at 8,500 km above the center of the earth be traveling at in order to stay
hodyreva [135]

Answer:

6844.5 m/s.

Explanation:

To get the speed of the satellite, the centripetal force on it must be enough to change its direction. This therefore means that the centripetal force must be equal to the gravitational force.

Formula for centripetal force is;

F_c = mv²/r

Formula for gravitational force is:

F_g = GmM/r²

Thus;

mv²/r = GmM/r²

m is the mass of the satellite and M is mass of the earth.

Making v the subject, we have;

v = √(GM/r)

We are given;

G = 6.67 × 10^(-11) m/kg²

M = 5.97 × 10^(24) kg

r = 8500 km = 8500000

Thus;

v = √((6.67 × 10^(-11) × (5.97 × 10^(24)) /8500000) = 6844.5 m/s.

7 0
3 years ago
Read 2 more answers
What is storage of energy​
Talja [164]

Hey,

Storage of energy is known as potential energy. If you take a spring for example, and you stretch it then you hold it, it has the energy you put into it but it is potential since it's not moving. Of course once you let go of it its potential energy turns into kinetic energy.

There are other examples as well, such as a battery. I decided to use the spring example because it is simpler.

I hope it helped.

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3 0
3 years ago
8.
GuDViN [60]

This ratio (Fnet/m) is sometimes called the gravitational field strength

and is expressed as 9.8 N/kg ⇒ answer D

Explanation:

The gravitational field strength at a point is:

  • The gravitational force exerted per unit mass placed at that point.
  • This means that the gravitational field strength, g is equal to the force experienced by a mass of 1 kg in that gravitational field
  • Gravitational field strength = Weight/mass
  • Its unit is Newton per kilogram
  • Gravitational field strength ≈ 9.8 N/kg

From the notes above

The ratio \frac{F_{net}}{mass} = Gravitational field strength (g)

The answer is:

This ratio (Fnet/m) is sometimes called the gravitational field

strength and is expressed as 9.8 N/kg

Learn more:

You can learn more about gravitational field strength in brainly.com/question/6763771

LearnwithBrainly

5 0
3 years ago
The circumference of a sphere was measured to be
professor190 [17]

To solve this problem we will apply the concepts related to the calculation of the surface, volume and error through the differentiation of the formulas given for the calculation of these values in a circle. Our values given at the beginning are

\phi = 76cm

Error (dr) = 0.5cm

The radius then would be

\phi = 2\pi r \\76cm = 2\pi r\\r = \frac{38}{\pi} cm

And

\frac{d\phi}{dr} = 2\pi \\d\phi = 2\pi dr \\0.5 = 2\pi dr

PART A ) For the Surface Area we have that,

A = 4\pi r^2 \\A = 4\pi (\frac{38}{\pi})^2\\A = \frac{5776}{\pi}

Deriving we have that the change in the Area is equivalent to the maximum error, therefore

\frac{dA}{dr} = 4\pi (2r) \\dA = 4r (2\pi dr)

Maximum error:

dA = 4(\frac{38}{\pi})(0.5)

dA = \frac{76}{\pi}cm^2

The relative error is that between the value of the Area and the maximum error, therefore:

\frac{dA}{A} = \frac{\frac{76}{\pi}}{\frac{5776}{\pi}}

\frac{dA}{A} = 0.01315 = 1.31\%

PART B) For the volume we repeat the same process but now with the formula for the calculation of the volume in a sphere, so

V = \frac{4}{3} \pi r^3

V = \frac{4}{3} \pi (\frac{38}{\pi})^3

V = \frac{219488}{3\pi^2}

Therefore the Maximum Error would be,

\frac{dV}{dr} = \frac{4}{3} 3\pi r^2

dV = 2r^2 (2\pi dr)

dV = 4r^2 (\pi dr)

Replacing the value for the radius

dV = 4(\frac{38}{\pi})^2(0.5)

dV = \frac{2888}{\pi^2} cm^3

And the relative Error

\frac{dV}{V} = \frac{ \frac{2888}{\pi^2}}{ \frac{219488}{3\pi^2} }

\frac{dV}{V} = 0.03947

\frac{dV}{V} = 3.947\%

3 0
3 years ago
A 27 kg is accelerated at a rate of 1.7m/s/s . what force does the object experience
tensa zangetsu [6.8K]
The answer should be (C)46n
7 0
3 years ago
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