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3 years ago

Heat generated by Friction

Chemistry

1 answer:

dimulka [17.4K]3 years ago

4 0

Answer:

The answer is C

Explanation:

because if you pull or push something, friction will always push or pull the opposite way

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A thermometer having first-order dynamics with a time constant of 1 min is placed in a temperature bath at 100oF. After the ther

sveticcg [70]

Answer:

(a) See below

(b) 103.935 °F; 102.235 °F

Explanation:

The equation relating the temperature to time is

$T = T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )$

1. Calculate the thermometer readings after 0.5 min and 1 min

(a) After 0.5 min

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-0.5/1} \right )\\ & = & 100 + 10\left (1 - e^{-0.5} \right )\\ & = & 100 + 10 (1 - 0.6065)\\ & = & 100 + 10(0.3935)\\ & = & 100 + 3.935\\ & = & 103.935\,^{\circ}F\\\end{array}$

(b) After 1 min

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-1/1} \right )\\ & = & 100 + 10\left (1 - e^{-1} \right )\\ & = & 100 + 10 (1 - 0.3679)\\ & = & 100 + 10(0.6321)\\ & = & 100 + 6.321\\ & = & 106.321\,^{\circ}F\\\end{array}$

2. Calculate the thermometer reading after 2.0 min

T₀ =106.321 °F

ΔT = 100 - 106.321 °F = -6.321 °F

t = t - 1, because the cooling starts 1 min late

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-(t - 1)/\tau} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-(2 - 1)/1} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-1} \right )\\ & = & 106.321 - 6.321 (1 - 0.3679)\\ & = & 106.321 - 6.321 (0.6321)\\ & = & 106.321 - 3.996\\ & = & 102.325\,^{\circ}F\\\end{array}$

3. Plot the temperature readings as a function of time.

The graphs are shown below.

6 0

3 years ago

Question 10

olga2289 [7]

Answer:

Here it would be A,B and C because all the examples are violating safety precautions. They are all examples of un safe signs.

Hope this helps

Explanation:

6 0

3 years ago

Helium gas diffuses 4 times as fast as an unknown gas. What is relative molecular mass of the gas

Inessa05 [86]

The relative molecular mass of the gas : 64 g/mol

<h3>Further explanation</h3>

Given

Helium rate = 4x an unknown gas

Required

The relative molecular mass of the gas

Solution

Graham's Law

$\tt \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }$

r₁=4 x r₂

r₁ = Helium rate

r₂ = unknown gas rate

M₁= relative molecular mass of Helium = 4 g/mol

M₂ = relative molecular mass of the gas

Input the value :

$\tt \dfrac{4r_2}{r_2}=\sqrt{\dfrac{M_2}{4} }\\\\16=\dfrac{M_2}{4}\\\\M_2=64~g/mol$

7 0

2 years ago

A bedroom has a volume of 118 m3 What is its volume in km3?

ruslelena [56]

I believe the problem is just simply asking for us to convert the value from one unit to the other. This case from m^3 to km^3. From the SI units, we know 1 km is equal to 1000 m. We do as follows:

118 m^3 ( 1 km / 1000 m )^3 = 1.18 x 10^-7 km^3

5 0

3 years ago

Read 2 more answers

List three properties of metal that nonmetals typically DO NOT have

finlep [7]

-They can conduct heat
-They can conduct electricity
-They are typically stronger than non metals
Hope this helps, have a nice day! :)

7 0

3 years ago

Read 2 more answers