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salantis [7]
3 years ago
14

As a space shuttle moves through the dilute ionized gas of Earth's ionosphere, the shuttle's potential is typically changed by -

1.4 V during one revolution. Assuming the shuttle is a conducting sphere of radius 15 m, estimate the amount of charge it collects.
Physics
1 answer:
postnew [5]3 years ago
4 0

Answer:

-2.3 × 10^-9 Coulombs(C).

Explanation:

So, we are given the following data or information or parameters that is going to help us to solve the problem effectively and efficiently;

=> " the shuttle's potential is typically changed by -1.4 V during one revolution. "

=> " Assuming the shuttle is a conducting sphere of radius 15 m".

So, in order to estimate the value for the charge we will be making use of the equation below:

Charge, C =( radius × voltage or potential difference) ÷ Coulomb's law constant.

Note that the value of Coulomb's law constant = 9 x 10^9 Nm^2 / C^2.

So, charge = { 15 × (- 1.4)} / 9 x 10^9 Nm^2 / C^2.

= -2.3 × 10^-9 Coulombs(C).

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Suppose a scoentist was able to construct a barometer with a liquid being denser than mercury , then how high would the liquid r
ki77a [65]

Answer:

the liquid has less height than the mercury

      h_{ liquid} = \frac{\rho_{Hg} }{\rho_{liqid}}  \  h_{Hg}

Explanation:

The pressure as a function of the height is given by

        P = ρ g h

where ρ is the density of the liquid, g the acceleration of gravity and h the height reached by the column of the liquid

In that case they say that the pressure is the standard one that is P = 1.01 10⁵ Pa = 760 mmHg

The first way to give the pressure is in SI units and the second way is the height that the mercury column reaches

In the case of building a barometer with a liquid that has a density greater than that of mercury

            ρ_liquid > ρ_Hg

             

the pressure

              P =ρ_lquid g h_liquid

if we have the same pressure

            ρ_{Hg} g h_{Hg} = ρ_{liquid}  g h_{liquid}

            h_{ liquid} = \frac{\rho_{Hg} }{\rho_{liqid}}  \  h_{Hg}

therefore the liquid has less height than the mercury

7 0
3 years ago
A certain electromagnetic wave traveling in seawater was observed to have an amplitude of 98.02 (V/m) at a depth of 10 m, and an
maksim [4K]

Answer:

The  value is   \alpha =  0.002 Np/m

Explanation:

From the question we are told that

  The first amplitude of the wave is  E_{max}1 =  98.02 \  V/m

  The first  depth  is  D_1 =  10 \  m

   The second amplitude is  E_{max}2 =  81.87 \  (V/m)

   The second depth is D_2 = 100 \ m

Generally from the spatial wave equation we have

   v(x) =  Ae^{-\alpha d}cos(\beta x  + \phi_o)

=>       \frac{v(x)}{v(x)} =\frac{  Ae^{-\alpha d}cos(\beta x  + \phi_o)}{ Ae^{-\alpha d}cos(\beta x  + \phi_o)}

So considering the ratio of the equation for the  two depth

\frac{A}{A_S}  =  \frac{e^{-D_1 \alpha }}{e^{-D_2 \alpha }}

=>   \frac{98.02}{81.87}  =  \frac{e^{-10 \alpha }}{e^{-100 \alpha }}

=>   \alpha  =  \frac{0.18}{90}

=>    \alpha =  0.002 Np/m

       

4 0
3 years ago
If you can answer my last post ill give you 75 points pls its very important and please make sure it correct!!!!!!!
Hoochie [10]

Answer:

I can't see the post :/

Explanation:

5 0
2 years ago
Read 2 more answers
Maaf sebelumnya ,bisa minta tolong kerjakan tugas saya? saya harus mengerjakan tugas Fisika yang lain . tolong bantu 4 nomer saj
AlladinOne [14]
Sorry cant not answer this question
8 0
3 years ago
Warm air is more dense than cool air. Warm air is less dense than cool air Warm air and cool air have the same density
Wewaii [24]

Answer:

You are exactly right. The molecules in hot air are moving faster than the molecules in cold air. Because of this, the molecules in hot air tend to be further apart on average, giving hot air a lower density. That means, for the same volume of air, hot air has fewer molecules and so it weighs less

7 0
2 years ago
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