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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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Find the magnitude of the resultant force and the angle it makes with the positive x-axis. (Let |a| = 22 lb and |b| = 16 lb. Rou
SVEN [57.7K]

Incomplete question as the angle between the force is not given I assumed angle of 55°.The complete question is here

Two forces, a vertical force of 22 lb and another of 16 lb, act on the same object. The angle between these forces is 55°. Find the magnitude and direction angle from the positive x-axis of the resultant force that acts on the object. (Round to one decimal places.)  

Answer:

Resultant Force=33.8 lb

Angle=67.2°

Explanation:

Given data

Fa=22 lb

Fb=16 lb

Θ=55⁰

To find

(i) Resultant Force F

(ii)Angle α

Solution

First we need to represent the forces in vector form

\sqrt{x} F_{1}=22j\\ F_{2}=u+v\\F_{2}=16sin(55)i+16cos(55)j\\F_{2}=16(0.82)i+16(0.5735)j\\F_{2}=13.12i+9.176j

Total Force

F=F_{1}+F_{2}\\ F_{2}=22j+13.12i+9.176j\\F_{2}=13.12i+31.176j

The Resultant Force is given as

|F|=\sqrt{x^{2} +y^{2} }\\|F|=\sqrt{(13.12)^{2} +(31.176)^{2} }\\ |F|=33.8lb

For(ii) angle

We can find the angle bu using tanα=y/x

So

tan\alpha =\frac{31.176}{13.12}\\ \alpha =tan^{-1} (\frac{31.176}{13.12})\\\alpha =67.2^{o}

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Jupiter's satellite Europa orbits Jupiter with a period of 3.55 d at an orbital radius of 6.71 108 m (assume the orbit to be cir
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Answer:

(a)

M = 1.898 x 10^27 kg

(b)

v = 13.74 km/s

(c) E = 0.28 N/kg

Explanation:

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G = 6.67 x 10^-11 Nm^2/kg^2

(a) T=2\pi \sqrt{\frac{r^{3}}{GM}}

M=\frac{4\pi ^{2}r^{3}}{GT^{2}}

M=\frac{4\times3.14^{2}\times 6.71^{3}\times 10^{24}}{6.67\times 10^{-11}\times 306720^{2}}

M = 1.898 x 10^27 kg

(b) Let v be the orbital velocity

v=\frac{2\pi r}{T}

v=\frac{2\times 3.14\times 6.71\times 10^{8}}{306720}

v = 13739.5 m/s

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(b) The gravitational field E is given by

E = \frac{GM}{r^{2}}

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Answer:

Passed into the power grid for others to use the electricity

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If a home uses a large supply of solar panels to generate electricity, but has no battery system, surplus electricity that is produced is usually passed into the power grid for others to use the electricity, generating a income to the homeowner

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I think that the answer is c. good luck
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