A string is stretched by two equal and opposite forces F newton each then the tension in sting is?

Fiber optic (FO) cables are based upon the concept of total internal reflection (TIR), which is achieved when the FO core and cl

kozerog [31]

Answer:

False

Explanation:

Though fiber active cable is based on the concept of internal reflection but it is achieved by refractive index which transmit data through fast traveling pulses of light. It has a layer of glass and insulating casing called “cladding,”and this is is wrapped around the central fiber thereby causing light to continuously bounce back from the walls of the Cable.

7 0

2 years ago

At 1 atm pressure, the heat of sublimation of gallium is 277 kj/mol and the heat of vaporization is 271 kj/mol. to the correct n

Andreyy89

Below are the choices that can be found elsewhere:

a. 268 kJ
b. 271 kJ 
c. 9 kJ 
d. 6 kJ

So the key thing to realize here is what the information given to you actually means. Sublimation is going from a sold to a gas. Vaporization is going from a liquid to a gas. Hence you can create two equations from the information that you have:

Ga (s) --> Ga (g) delta H = 277 kJ/mol 

Ga (l) --> Ga (g) delta H = 271 kJ/mol 

From these two equations, you can then infer how to get the melting equation be simply finding the difference between the sublimation (two steps) and vaporization (one step). 

Ga (s) --> Ga (l) delta H = 6 kJ/mol 

At this point, all you need to do is a bit of stoichiometry. You start with 1.50 mol and multiply by the amount of energy per mole (6 kJ/mol). 

*ANSWER* 
9 kJ/mol (C)

7 0

2 years ago

Th e heat capacity of air is much smaller than that of water, and relatively modest amounts of heat are needed to change its tem

Darina [25.2K]

Answer:

Q=1005 J

t= 0.67 sec

Explanation:

Lets take condition of room is 1 atm and 25°C.

Heat capacity ,c = 21 J /K.mol

If we assume that air is ideal gas that

P V = n R T

$V= 5.5\times 6.5\times 3\ m^3$

$V=107.25\ m^3$

$V=107.25\times 1000 L$

V= 107250 L

At STP number of moles given as

$n=\dfrac{V}{V_{at\ S.T.P.}}$

V=22.4 L at S.T.P.

$n=\dfrac{107250}{22.4}\ moles$

n=4787.94 moles

n= 4.784 Kmoles

So heat required to raise 10°C temperature

Q = n x c x ΔT

Q = 4.78794 x 21 x 10

Q=1004.64 J

Time t

t= Q/P

P= 1.5 KW

t = 1.004.64 /1.5

t= 0.66 sec

4 0

2 years ago

Read 2 more answers

a cannonball with a mass of 1.0 kilogram is fired horizontally from a 500.- kilogram cannon, initially at rest, on a horizontal,

nirvana33 [79]

(a) According to Newton’s third law, for every action there is an equal yet opposite reaction. In this case, the cannonball is affected by 8.0x10^3 newtons in one direction. Now the cannon must also be affected by the same amount of net force on the opposite direction.

- 8.0x10^3 Newtons (the negative symbol only shows that it is opposite direction)

(b) We solve this using the formula:

F = m a

8.0x10^3 N = (1.0 kg) a

a = 8,000 m/s^2

8 0

2 years ago

Sebanyak 0,2 mol gas ideal berada dalam wadah yang volumenya 10 liter dan tekanan 1 atm . berapakah suhu gas tersebut

tamaranim1 [39]

The temperature of the gas is about 600 K

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<h3>Further explanation</h3>

The Ideal Gas Law that needs to be recalled is:

$\large {\boxed {PV = nRT} }$

P = Pressure (Pa)

V = Volume (m³)

n = number of moles (moles)

R = Gas Constant (8.314 J/mol K)

T = Absolute Temperature (K)

Let us now tackle the problem !

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Question (Translation):

A total of 0.2 moles of ideal gas are in containers with volume of 10 liters and a pressure of 1 atm. What is the temperature of the gas?

Given:

number of moles = n = 0.2 moles

volume of gas = V = 10 liters

pressure of gas = P = 1 atm

gas contant = R = 0.0821 L.atm/mol.K

Unknown:

temperature of gas = T = ?

Solution:

$PV = nRT$

$1 \times 10 = 0.2 \times 0.0821 \times T$

$10 = 0.01642 \times T$

$T = 10 \div 0.01642$

$T \approx 600 \texttt{ K}$

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<h3>Learn more</h3>

Minimum Coefficient of Static Friction : brainly.com/question/5884009
The Pressure In A Sealed Plastic Container : brainly.com/question/10209135
Effect of Earth’s Gravity on Objects : brainly.com/question/8844454

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<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Pressure

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Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant , Liquid , Pressure

8 0

2 years ago