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Dovator [93]
2 years ago
15

I decided to take a trip to the International Space Station (ISS), which is currently orbiting the Earth at an altitude (distanc

e from surface to ISS) of 408 km. When I get to the ISS, I decide to go for a spacewalk outside (disconnected from ISS). Let us say that I have a mass of 60 kg, and the mass of the Earth is 5.97 x 1024 kg. Newton’s gravitational constant G = 6.67 ∗ . The radius (distance from center to surface) of the ∗2 Earth = 6731 km.
Physics
1 answer:
Mrac [35]2 years ago
5 0

The gravitational force I’d experience while I am floating outside the ISS will be 4.47 × 10⁻¹⁵ N.

<h3>What is gravitational force?</h3><h3 />

Newton's law of gravity states that each particle having mass in the universe attracts each other particle with a force known as the gravitational force.

Gravitational force is proportional to the product of the masses of the two bodies and inversely proportional to the square of their distance.

\rm F=G\frac{mM_E}{R^2} \\\\ \rm F=6.67 \times 10^{-34}\times \frac{60 \times 5.97 \times 10^{24}}{(6731)^2} \\\\ F=4.47 \times 10^{-15} \ N

Hence, the gravitational force I’d experience while I am floating outside the ISS will be 4.47 × 10⁻¹⁵ N.

To learn more about the gravitational force refer to the link;

brainly.com/question/24783651

#SPJ1

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A sphere of radius r = 5cm carries a uniform volume charge density rho = 400 nC/m^3. Q. What is the total charge Q of the sphere
Tanzania [10]

Answer:

The total charge Q of the sphere is 2.094\times10^{-10}\ C.

Explanation:

Given that,

Radius = 5 cm

Charge density J= 400\ nC/m^3

We need to calculate the total charge Q of the sphere

Using formula of charge

q=\rho V

Where, \rho = charge density

V = volume

Put the value into the formula

q=\rho\times(\dfrac{4}{3}\pi r^3)

Put the value into the formula

q=\dfrac{4}{3}\times\pi\times400\times10^{-9}\times(5\times10^{-2})^3

q=2.094\times10^{-10}\ C

Hence, The total charge Q of the sphere is 2.094\times10^{-10}\ C.

6 0
3 years ago
With what tension must a rope with length 3.00 mm and mass 0.105 kgkg be stretched for transverse waves of frequency 40.0 HzHz t
VladimirAG [237]

Answer:

the tension of the rope is 34.95 N

Explanation:

Given;

length of the rope, L = 3 m

mass of the rope, m = 0.105 kg

frequency of the wave, f = 40 Hz

wavelength of the wave, λ = 0.79 m

Let the tension of the rope = T

The speed of the wave is given as;

v = f\lambda = \sqrt{\frac{T}{\mu} } \\\\where;\\\\\mu \ is \ mass \ per \ unit \ length\\\\\mu  = \frac{0.105}{3} = 0.035 \ kg/m\\\\v = f\lambda = 40 \times 0.79 = 31.6 \ m/s\\\\v =  \sqrt{\frac{T}{\mu} } \\\\v^2 = \frac{T}{\mu} \\\\T = v^2 \mu\\\\T = (31.6^2)(0.035)\\\\T = 34.95 \ N

Therefore, the tension of the rope is 34.95 N

4 0
3 years ago
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Espero que esto te ayude y no me haya confundido
7 0
2 years ago
A electric heater that draws 13.5 a of dc current has been left on for 10 min. how many electrons that have passed through the h
kicyunya [14]
By definition, Ampere is a unit of current which is a measure of the amount of charge passing through a point in a circuit per unit  of time, with an equivalent charge of 1.602 x 10^(-19) Coulomb per electron. To determine the number of electrons passing through the heater, we use the definition of the current. We calculate as follows:

13.5 A = 13.5 C per second
Charge = 13.5 C/s (10 min) ( 60 s / 1 min)
Charge = 8100 C 

Number of electrons = 8100 C / 1.602 x 10^(-19) C per electron
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Therefore, there are 5.1 x10^22 electrons that assed through the heater for 10 minutes.
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Answer:

if you only have to control your chakra and know how to get all your vibes to pass it to objects and it takes time to practice

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