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Force due to gravity. While JWST is in orbit, the Earth will be at a distance of 1.494 x 109 m from the telescope, and the Sun w

Alona [7]

Answer:

the gravitational force JWST will feel from the Sun is 37.785 Newton { leftward }

Explanation:

Given that;

distance of earth from the telescope r1 = 1.494 × 10⁹ m

and the Sun will be 1.49598 × 10¹¹ m further away

so r2 = r1 + 1.49598 × 10¹¹

r2 = 1.494 × 10⁹ m + 1.49598 × 10¹¹ m = 1.51092 × 10¹¹ m

mass of sun Ms = 1.9884 × 10³⁰ kg

mass of earth Me = 5.945× 10²⁴ kg

mass of JWST Mj = 6500 kg

What is the gravitational force JWST will feel from the Sun (strength and direction)?

the gravitational force of the sun will be attractive based on Newton law of gravitational force; so

Fjs = GMjMs / r2²

constant G = 6.674 × 10⁻¹¹ Nm²/kg²

Force on the JWST by the sun will be;

Fjs = GMjMs / r2² { leftward}

we substitute

Fjs = [(6.674 × 10⁻¹¹ Nm²/kg²)(6500 kg )(1.9884 × 10³⁰ kg)] / (1.51092 × 10¹¹ m)²

= (8.62587804 × 10²³) / ( 2.28287925 × 10²² )

= 37.785 Newton { leftward }

Therefore, the gravitational force JWST will feel from the Sun is 37.785 Newton { leftward }

7 0

3 years ago

Which data set has the smallest standard deviation?

777dan777 [17]

The answer would be B.
<span>
Standard deviation basically measures how spread out the values are. Without solving, you can easily tell which one among your choices have a smaller deviation. The closer the values are to each other the smaller the standard deviation. The values of choice B are the closest together, so you can assume that they have the smallest standard deviation. </span>

8 0

4 years ago

Read 2 more answers

Three point charges, two positive and one negative, each having a magnitude of 20 C are placed at the vertices of an equilateral

Daniel [21]

The resultant force on the positive charge is mathematically given as

X=40N

<h3>What is the magnitude of the electrostatic force on the negative charge?</h3>

Question Parameters:

Three-point charges, two positive and one negative, each having a magnitude of 20

Generally, the -ve charge is mathematically given as

$Q+=\sqrt{x^2+x^2+2x.xcos120}\\\\Q+=\sqrt{2x^2+2x*(1/2)}$

Q+=X

Therefore

$x=\frac{Kq1q2}{r2}\\\\x=\frac{9*10^9*20*10^{-6}*20*10^{-6}}{(30*10^-2)^2}$

X=40N

For more information on Force

brainly.com/question/26115859

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Exercise is the answer hope i helped you

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