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3 years ago

How do the magnetic fields of Uranus and Neptune suggest that the mantles inside those planets are fluid?

Physics

1 answer:

givi [52]3 years ago

8 0

Answer:

The surface must be a good absorber of light. Spectra of the planets proved that methane in the atmospheres of these planets is what absorbs red light.

Explanation:

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A spherical capacitor contains a charge of 3.50 nC when connected to a potential difference of 210.0 V. Its plates are separated

Lelu [443]

Incomplete question as we have not told to find what quantity.The complete question is here

A spherical capacitor contains a charge of 3.50 nC when connected to a potential difference of 210.0 V. Its plates are separated by vacuum and the inner radius of the outer shell is 5.00 cm.calculate: (a) the capacitance; (b) the radius of the inner sphere; (c) the electric field just outside the surface of the inner sphere.

Answer:

(a) $C=16.7pF$

(b) $r_{a} =3.749cm$

Explanation:

Given data

$Q=3.50nC\\V=210V\\r_{b}=5.0cm$

For part (a)

The Capacitance given by:

$C=\frac{Q}{V}\\ C=\frac{3.50*10^{-9} C}{210V}\\C=1.6666*10^{-11}F\\or\\C=16.7pF$

For part (b)

The Capacitance of coordinates is given as

$C=\frac{4\pi e}{\frac{1}{r_{a} }-\frac{1}{r_{b} } }\\ So\\{\frac{1}{r_{a} }-\frac{1}{r_{b} } }=\frac{4\pi *8.85*10^{-12} }{1.666*10^{-11}}=6.672m^{-1} \\ \frac{1}{r_{a} }=6.672+(1 /0.05)\\\frac{1}{r_{a} }=26.672\\r_{a} =1/26.672\\r_{a} =0.0375m\\r_{a} =3.749cm$

For part (c)

The electric field according to Gauss Law is given by:

$EA=\frac{Q}{e}\\ E=\frac{Q}{4\pi er_{a}^{2} }=\frac{kQ}{r_{a}^{2}}\\ E=\frac{9*10^{9}*3.50*10^{-9} }{(0.0375m)^{2} }\\ E=2.24*10^{4} N/C$

7 0

3 years ago

A 2.3 kg block of copper is heated at atmospheric pressure such that its temperature increases from 6 oC to 90 oC. How much heat

Shtirlitz [24]

Answer:

the heat absorbed by the block of copper is 74368.476J

Explanation:

Hello!

To solve this problem use the first law of thermodynamics that states that the heat applied to a system is the difference between the initial and final energy considering that the mass and the specific heat do not change so we can infer the following equation

Q=mCp(T2-T1)

Where

Q=heat

m=mass=2.3kg

Cp=0.092 kcal/(kg C)=384.93J/kgK

T2=Final temperatura= 90C

T1= initial temperature=6 C

solving

$Q=(2.3kg)(384.93\frac{J}{kgC} )(90C-6C)=74368.476J$

the heat absorbed by the block of copper is 74368.476J

7 0

3 years ago

Identify the physical property of a material that is NOT a good conductor of heat.

EleoNora [17]

Answer:D.no of the above

Explanation:get right with Christ

4 0

3 years ago

One hazard of space travel is debris left by previous missions. There are several thousand objects orbiting Earth that are large

Eva8 [605]

Answer:

6666.67 Newtons

Explanation:

The formula F=ma (force is equal to mass multiplied by acceleration) can be used to calculate the answer to this question.

In this case:

mass= 0.1mg= 1*10^-7 kg
velocity= 4.00*10^3 m/s
time= 6.00*10^-8 s

Using velocity and time, acceleration can be calculated as:

a= 6.667*10^10 m/s²

Substituting these values into the formula F=ma, the answer is:

F= (1*10^-7)kg * (6.667*10^10) m/s²
F= 6666.67 Newtons of force

5 0

3 years ago

How should the distance, D, between two point charges, q1 and q2, be changed to double their electric potential energy

Licemer1 [7]

The electrostatic potential energy, U, of one point charge q at position d in the presence of an electric field E is defined as the negative of the work W done by the electrostatic force to bring it from the reference position d to that position

$u \: = \: \frac{kq1q2}{d}$

Thus, to double the electric potential energy U we need to reduce the distance of separation by half (1/2) because they are inversely proportion

6 0

4 years ago