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katrin [286]
3 years ago
5

An astronaut is said to be weightlessness when he\she travels in a satellite. Does it mean that the earth does not attract him\h

er?
Physics
1 answer:
GarryVolchara [31]3 years ago
8 0

Answer:

Explanation:

The sense of weightlessness in orbiting satellite is because of the lack of any contact-forces. ... Thus, both the astronaut and the satellite have the same value of g, which is again equal to the centripetal acceleration of the satellite. These two neutralize each other, causing everything to appear weightless.

MAKE ME BRAINLIEST

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Which best describes internet wikis as a source of scientific information
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Answer:

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What atmospheric conditions can cause sound wave refractions similar to those observed in the ocean?
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3 years ago
A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 11.0 cm ,
viktelen [127]
Part A)
First of all, let's convert the radii of the inner and the outer sphere:
r_A = 11.0 cm = 0.110 m
r_B = 16.5 cm=0.165 m
The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
C=4 \pi \epsilon _0  \frac{r_A r_B}{r_B- r_A}=4\pi(8.85 \cdot 10^{-12}C^2m^{-2}N^{-1}) \frac{(0.110m)(0.165m)}{0.165m-0.110m}=
=3.67\cdot 10^{-11}F

Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
Q=CV=(3.67\cdot 10^{-11}F)(100 V)=3.67\cdot 10^{-9}C

Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
E\cdot (4 \pi r^2) =  \frac{Q}{\epsilon _0}
from which
E(r) =  \frac{Q}{4 \pi \epsilon_0 r^2}
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
E(0.111m)=2680 N/C

And so, the energy density at r=0.111 m is
U= \frac{1}{2} \epsilon _0 E^2 =  \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(2680 N/C)^2=3.17 \cdot 10^{-5}J/m^3

Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor: Q=3.67 \cdot 10^{-9}C. We just need to calculate the electric field E at a different value of r: r=16.4 cm=0.164 m, so
E(0.164 m)= \frac{Q}{4 \pi \epsilon_0 r^2}=1228 N/C

And therefore, the energy density at this distance from the center is
U= \frac{1}{2}\epsilon_0 E^2 =  \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(1228 N/C)^2=6.68 \cdot 10^{-6}J/m^3
8 0
3 years ago
A flexible balloon contains 0.400 molmol of an unknown polyatomic gas. Initially the balloon containing the gas has a volume of
Nastasia [14]

Answer:

Mass released = 8.6 g

Given data:

Initial number of moles nitrogen= 0.950 mol

Initial volume = 25.5 L

Final mass of nitrogen released  = ?

Final volume = 17.3 L

Formula:

V₁/n₁  = V₂/n₂

25.5 L / 0.950 mol = 17.3 L/n₂

n₂ =  17.3 L× 0.950 mol/25.5 L

n₂ = 16.435 L.mol /25.5 L

n₂ = 0.644 mol

Initial mass of nitrogen:

Mass = number of moles × molar mass

Mass = 0.950 mol × 28 g/mol

Mass = 26.6 g

Final mass of nitrogen:

Mass = number of moles × molar mass

Mass = 0.644 mol × 28 g/mol

Mass = 18.0 g

Mass released = initial mass - final mass

Mass released = 26.6 g - 18.0 g

Mass released = 8.6 g

Read more on Brainly.com - brainly.com/question/15623698#readmore

7 0
3 years ago
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