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GREYUIT [131]
3 years ago
10

1)Which statement best describes energy in an open system?

Physics
2 answers:
zimovet [89]3 years ago
6 0
B an open system flow both 
Aleks04 [339]3 years ago
3 0

the first question is  b) Energy and matter flow into and out of the system. im not sure about the second one

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The largest and brightest galaxies are those that contain _____.
user100 [1]
They contain quasars
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3 years ago
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A hollow conducting sphere with an outer radius of 0.295 m and an inner radius of 0.200 m has a uniform surface charge density o
IrinaK [193]

Answer:

a. 6.032\times10^{-6}C/m^2

b.6.816\times10^5N/C

Explanation:

#Apply  surface charge density, electric field, and Gauss law to solve:

a. Surface charge density is defined as charge per area denoted as \sigma

\sigma=\frac{Q}{4\pi r_{out}^2}, and the strength of the electric field outside the sphere E=\frac{\sigma _{new}}{\epsilon _o}

Using Gauss Law, total electric flux out of a closed surface is equal to the total charge enclosed divided by the permittivity.

\phi=\frac{Q_{enclosed}}{\epsilon_o}\\\\\sigma=\frac{Q}{4\pi r_{out}^2}\\\\\sigma=\frac{0.370\times 10^{-6}}{4\pi \times (0.295m)^2}\\\\=3.383\times10^{-7}C/m^2  #surface charge outside sphere.

\sigma_{new}=\sigma_{s}-\sigma\\\\\sigma_{new}=6.37\times10^{-6}C/m^2-3.383\times10^{-7}C/m^2\\\\\sigma_{new}=6.032\times10^{-6}C/m^2

Hence, the new charge density on the outside of the sphere is 6.032\times10^{-6}C/m^2

b. The strength of the electric field just outside the sphere is calculated as:

From a above, we know the new surface charge to be 6.032\times10^{-6}C/m^2,

E=\frac{\sigma _{new}}{\epsilon _o}\\\\=\frac{6.032\times10^{-6}C/m^2}{\epsilon _o}\\\\\epsilon _o=8.85\times10^{-12}C^2/N.m^2\\\\E=\frac{6.032\times10^{-6}C/m^2}{8.85\times10^{-12}C^2/N.m^2}\\\\E=6.816\times10^5N/C

Hence, the strength of the electric field just outside the sphere is 6.816\times10^5N/C

5 0
3 years ago
Which of the following is the most accurate statement concerning the properties of matter?
DIA [1.3K]

Answer:

b. they can be observed and measured

Explanation:

Matter is anything that has weight and occupy space. There are three states of matter namely Solid, liquid and gas.

The properties of matter are both physical and chemical in nature. Both properties can be measured and observed. Phhysical properties are anything that can be measured without changing the state of the matter. Example of physical properties includes mass, volume, length, color etc.

Chemical properties is another properties of matter. This is the ability of the states of matters to combine with other substance to form a new product for example, rusting of iron, formation of salt etc.

All this as discussed are both measurable and can be observed.

8 0
4 years ago
Very Important!!
Murljashka [212]

Answer:

A. an alcohol - induced blackout

4 0
3 years ago
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An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 1.60 mm. If a
nata0808 [166]

Answer:

Explanation:

(a) For the calculation of the Electric field we use

E=\frac{V}{d}=\frac{15.0V}{1.6*10^{-3}m} =9375\frac{N}{C}

(b) The capacitance is calculate by using the expression

C=\frac{\epsilon_{0}A}{d}=\frac{8.85*10^{-12}C^{2}/(Nm^{2})*(7.6*10^{-4}m^{2})}{1.6*10^{-3}m}=4.2*10^{-12}C

(c) Finally, the charge on each plate is

Q=CV=(4.2*10^{-12}C)(15V)=6.3*10^{-11}C

I hope this is useful for you

Regards

4 0
3 years ago
Read 2 more answers
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