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3 years ago

How does the vacuum between the inner and outer walls of a thermos bottle limit energy loss through conduction and convection?

1 answer:

3 0

By having no solid matter between the walls (like styrofoam) conduction is eliminated since conduction requires matter to be in contact with other matter. The vacuum state eliminates convection since that is the heating of gas that then carries the energy to another body by traveling there and exchanging heat with it.

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A helium nucleus (charge = 2e, mass = 6.63 10-27 kg) traveling at 6.20 105 m/s enters an electric field, traveling from point ci

MA_775_DIABLO [31]

Answer:

$v_B=3.78\times 10^5\ m/s$

Explanation:

It is given that,

Charge on helium nucleus is 2e and its mass is $6.63\times 10^{-27}\ kg$

Speed of nucleus at A is $v_A=6.2\times 10^5\ m/s$

Potential at point A, $V_A=1.5\times 10^3\ V$

Potential at point B, $V_B=4\times 10^3\ V$

We need to find the speed at point B on the circle. It is based on the concept of conservation of energy such that :

increase in kinetic energy = increase in potential×charge

$\dfrac{1}{2}m(v_A^2-v_B^2)=(V_B-V_A)q\\\\\dfrac{1}{2}m(v_A^2-v_B^2)={(4\times 10^3-1.5\times 10^3)}\times 2\times 1.6\times 10^{-19}=8\times 10^{-16}\\\\v_A^2-v_B^2=\dfrac{2\times 8\times 10^{-16}}{6.63\times 10^{-27}}\\\\v_A^2-v_B^2=2.41\times 10^{11}\\\\v_B^2=(6.2\times 10^5)^2-2.41\times 10^{11}\\\\v_B=3.78\times 10^5\ m/s$

So, the speed at point B is $3.78\times 10^5\ m/s$ .

7 0

3 years ago

How much work is done on a satellite in a circular orbit about earth?

alexgriva [62]

Answer:

0 J

Explanation:

$W$ = Work done on the satellite in circular orbit about earth by earth

$F$ = Force on gravity on satellite by earth

$d$ = displacement of the satellite

$\theta$ = Angle between the force on gravity and displacement = 90

We know that, Work done is given as

$W = F d Cos\theta\\W = F d Cos90\\W = F d (0)\\W = 0 J$

3 0

3 years ago

A lamp has a power of 120 W and is left on for 24 hours . A television has a power of 350 W and is left on for three hours . com

Paladinen [302]

This is a power problem which requires the rearranging of a formula. The lamps energy used is 5 N, and the TV’s usage is 116.7 N (rounded from 116.6666repeating). Here my work:

7 0

2 years ago

Read 2 more answers

What does the energy hill represent on an energy diagram?

creativ13 [48]

Answer:

In my opinion I think the answer is C you don't have to choose C

8 0

3 years ago

What are (a) the charge and (b) the charge density on the surface of a conducting sphere of radius 0.20 m whose potential is 240

Lady bird [3.3K]

Answer:

(a) charge q=5.33 nC

(b) charge density σ=10.62 nC/m²

Explanation:

Given data

radius r=0.20 m

potential V=240 V

coulombs constant k=9×10⁹Nm²/C²

To find

(a) charge q

(b) charge density σ

Solution

For (a) charge q

$V_{potential}=kq/r\\ q=rV_{potential}/k\\q=\frac{(0.20)(240)}{9*10^{9} }\\ q=5.333*10^{-9}C\\or\\ q=5.33nC$

For (b) charge density

As charge density σ is given as:

σ=q/(4πR²)

σ=(5.333×10⁻⁹) / (4π×(0.20)²)

σ=10.62 nC/m²

3 0

3 years ago