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frozen [14]
3 years ago
9

50POINTS! Find the orbital speed of a satellite in a circular orbit 1700km above the surface of the Earth. M_earth 5.97e24kg, r_

avg_earth = 6.38e6m
Physics
2 answers:
crimeas [40]3 years ago
7 0

Answer:

7020.117 m/s

Explanation:

ivolga24 [154]3 years ago
6 0
<h2>Answer: 7020.117 m/s</h2>

Explanation:

The velocity of a satellite describing a circular orbit is<u> constant</u> and defined by the following expression:  

V=\sqrt{G\frac{M}{R}} (1)  

Where:  

G=6.674({10}^{-11})\frac{N{m}^{2}}{{kg}^{2}} is the gravity constant

M_{Earth}=5.97{10}^{24}kg the mass of the massive body around which the satellite is orbiting, in this case, the Earth .

R=r_{Earth}+h=8080000m the radius of the orbit (measured from the center of the planet to the satellite).  

This means the radius of the orbit is equal to <u>the sum</u> of the average radius of the Earth r_{Earth} and the altitude of the satellite above the Earth's surface h.

Note this orbital speed, as well as orbital period, does not depend on the mass of the satellite. It depends on the mass of the massive body (the Earth).

Now, rewriting equation (1) with the known values:

V=\sqrt{(6.674({10}^{-11})\frac{N{m}^{2}}{{kg}^{2}})\frac{5.97{10}^{24}kg}{8080000m}}

V=7020.117\frac{m}{s}  

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Nuetrik [128]
<h3>Answer</h3>

6.6 N pointing to the right

<h3>Explanation</h3>

Given that,

two forces acting of magnitude 3.6N

angle between them = 48°

To find,

the third force that will cause the object to be in equilibrium

<h3>1)</h3>

Find the vertical and horizontal components of the two forces

vertical force1 = sin(24)(3.6)

vertical force2= -sin(24)(3.6)

<em>(negative sign since it is acting on opposite direction)</em>

vertical force3 = sin(24)(3.6) - sin(24)(3.6)

                        = 0

<h3>2)</h3>

horizontal force1 = cos(24)(3.6)

horizontal force2= cos(24)(3.6)

horizontal force3 = cos(24)(3.6) + cos(24)(3.6)

                            = 2(cos(24)(3.6))

                            = 6.5775 N

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<em />

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4 0
3 years ago
Humans protect themselves from natural disasters by all of the following except
jeyben [28]
I'm pretty sure it's sunscreen
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3 years ago
A 5.10 kgkg watermelon is dropped from rest from the roof of a 18.5 mm-tall building and feels no appreciable air resistance.
VARVARA [1.3K]

Work done is by the change in the potential energy of the system. The work done by gravity is 924.63 J.

<h3>What is the Kinetic Energy?</h3>
  • Potential energy in physics is the energy that an item retains as a result of its position in relation to other objects, internal tensions, electric charge, or other elements.
  • The gravitational potential energy of an object, which is based on its mass and distance from another object's center of mass, the elastic potential energy of an extended spring, and the electric potential energy of an electric charge in an electric field are examples of common types of potential energy. The joule, denoted by the letter J, is the energy unit in the International System of Units (SI).

Solution:

mass = 5.10 kg

height = 18.5 mm

We know that work done by the gravity on the watermelon is the change in the potential energy of the watermelon, therefore,

Work done due to gravity = change in the potential energy of the system

W = \Delta PE

W = mg (h₀ - h₁)

W = 5.10 × 9.8 × 18.5

W = 924.63 J

know more about potential energy brainly.com/question/24284560

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7 0
1 year ago
Using 0.500 g of nichrome, you are asked to fabricate a wire with uniform cross-section. The resistance of the wire is 0.673 Ω.
mojhsa [17]

Explanation:

Given that,

Mass of Nichrome, m = 0.5 g

The resistance of the wire, R = 0.673 ohms

Resistivity of the nichrome wire, \rho=10^{-6}\ \Omega -m

Density, d=8.31\times 10^3\ kg/m^3

(A) The length of the wire is given by using the definition of resistance as :

Volume,

V=A\times l\\\\A=\dfrac{V}{l}\\\\Since, V=\dfrac{m}{d}\\\\V=\dfrac{m}{d}\\\\V=\dfrac{0.5\times 10^{-3}}{8.31\times 10^3}\\\\V=6.01\times 10^{-8}\ m^3

Area,

A=\dfrac{V}{l}\\\\A=\dfrac{6.01\times 10^{-8}}{l}....(1)

R=\rho \dfrac{l}{A}\\\\l=\dfrac{RA}{\rho}\\\\l=\dfrac{0.673\times 6.01\times 10^{-8}}{l\times 10^{-6}}\\\\l=0.201\ m

(b)  Equation (1) becomes :

A=\dfrac{6.01\times 10^{-8}}{l}\\\\A=\dfrac{6.01\times 10^{-8}}{0.201}\\\\\pi r^2=3\times 10^{-7}\\\\r=\sqrt{\dfrac{3\times 10^{-7}}{\pi}} \\\\r=3.09\times 10^{-4}\ m

Hence, this is the required solution.                                                                  

5 0
3 years ago
A horizontal aquifer is overlain by 73.5 ft of water-saturated clay with a porosity of 0.4. The solid density of the clay partic
mr_godi [17]

Answer:

Explanation:

solution is in the attachment below

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