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Sliva [168]
4 years ago
10

A shuttle on Earth has a mass of 4.5 E 5 kg. Compare its weight on Earth to its weight while in orbit at a height of 6.3 E 5 met

ers above the surface of the Earth. Approximately what percent of its weight does the space shuttle retain?
11%
44%
83%
91%
Physics
1 answer:
faltersainse [42]4 years ago
4 0

Answer:

83%

Explanation:

On the surface, the weight is:

W = GMm / R²

where G is the gravitational constant, M is the mass of the Earth, m is the mass of the shuttle, and R is the radius of the Earth.

In orbit, the weight is:

w = GMm / (R+h)²

where h is the height of the shuttle above the surface of the Earth.

The ratio is:

w/W = R² / (R+h)²

w/W = (R / (R+h))²

Given that R = 6.4×10⁶ m and h = 6.3×10⁵ m:

w/W = (6.4×10⁶ / 7.03×10⁶)²

w/W = 0.83

The shuttle in orbit retains 83% of its weight on Earth.

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Question number 11 how did we found the answer ?
BaLLatris [955]

Answer:

Option A. 57.14 Ω

Explanation:

From the question given above, the following data were obtained:

Resistor 1 (R₁) = 100 Ω

Resistor 2 (R₂) = 400 Ω

Resistor 3 (R₃) = 200 Ω

Equivalent Resistor (Rₚ) =?

The equivalent resistor in the above circuit can be obtained as follow:

1/Rₚ = 1/R₁ + 1/R₂ + 1/R₃

1/Rₚ = 1/100 + 1/400 + 1/200

Find the least common multiple (lcm) of 100, 400 and 200. The result is 400. Divide 400 by 100, 200 and 400 respectively and multiply the result with the numerator as shown

1/Rₚ = (4 + 1 + 2)/400

1/Rₚ = 7/400

Invert

Rₚ = 400/7

Rₚ = 57.14 Ω

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3 years ago
Which of the following is a source of thermal energy that is absorbed by Earth's atmosphere?
Mazyrski [523]

Answer:

100% of the energy entering earth's atmosphere comes from the sun. ~50% of the incoming energy is absorbed by the earth's surface i.e. the land and oceans.

8 0
3 years ago
Given theta = 7pi/6, find (sec theta, cos theta) ...?
uysha [10]
There is an identity of sin(2pi-x) = -sin(x) and cos(2pi-x) = cos(x). This is what we are going to use. 


<span>(7pi/6) = (2pi)-(pi/6)

 </span>
Therefore: 
<span>
1) sec(7pi/6) = 1/cos(2pi-(pi/6)) = 1/cos(pi/6) = 2sqrt(3)/3 
</span>
<span>2) cos(7pi/6) = cos(pi/6) = sqrt(3)/2


I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

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3 years ago
In each of the parts of this question, a nucleus undergoes a nuclear decay. Determine the resulting nucleus in each case.
MA_775_DIABLO [31]

A) Francium-223

In an alpha decay, a nucleus decay emitting an alpha particle, which corresponds to a nucleus of helium: so, it consists of 2 protons and 2 neutrons.

X \rightarrow X' + \alpha

This means that in the decay:

- The original nucleus loses 2 protons --> so its atomic number Z decreases by 2 units

- The original nucleus loses 2 nucleons (2 protons and 2 neutrons) --> so its mass number A decreases by 4 units

In this example, the original nucleus is Ac (Actinium), with

Z = 89

A = 227

After the decay, it must be

Z - 2 = 89 - 2 = 87

A - 4 = 227 - 4 = 223

We see from the periodict table, Z=87 corresponds to Francium (Fr), so the final nucleus will be francium-223 (the isotope of francium with 223 nucleons).

B) Polonium-211

In a beta-minus decay, a neutron in the nucleus turns into a proton, emitting a fast-moving electron (the beta particle) and an anti-neutrino.

n \rightarrow p + e^- + \bar{\nu}

Therefore, in this process:

- The original nucleus gains 1 protons, so its atomic number Z increases by 1 unit

- The original nucleus does not lose/gain nucleons, so its mass number A remains the same

In this example, the original nucleus is Bi (bismuth)-211, with

Z = 83

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So After the decay, it will be

Z + 1 = 83 + 1 = 84

A = 211

So, the nucleus will be Polonium (Z=84), isotope with 211 nucleons.

C) Neon-22

In a beta-plus decay, a proton in the nucleus turns into a neutron, emitting a fast-moving positron (the beta particle) and a neutrino.

p \rightarrow n + e^+ +\nu

Therefore, in this process:

- The original nucleus loses 1 protons, so its atomic number Z decreases by 1 unit

- The original nucleus does not lose/gain nucleons, so its mass number A remains the same

In this example, the original nucleus is Na (sodium)-22, with

Z = 11

A = 22

So After the decay, it will be

Z - 1 = 11 - 1 = 10

A = 22

So, the nucleus will be Neon (Z=10), isotope with 22 nucleons.

D) Technetium-98

In a gamma decay, an unstable nucleus emits a gamma ray:

X' \rightarrow X + \gamma

In this process, only energy is released (in the form of gamma ray), so there is no gain/loss of protons/neutrons in the process. This means that:

- The atomic number Z remains constant

- The mass number A remains constant

In this example, we have a nucleus of Tc (Technetium)-98, with

Z = 43

A = 98

These numbers will not change during the decay: this means that after the decay, we will still have a nucleus of Technetium-98.

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