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2 years ago

5

The host at the end of the video claims that ___________ is crucial to his success as a driver. A. Reaction time B. A safe space

C. His seat belt

1 answer:

Snezhnost [94]2 years ago

5 0

Answer:

answer is C. his seat belt

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Define the terms (a) thermal conductivity, (b) heat capacity and (c) thermal diffusivity

IceJOKER [234]

Explanation:

<u>(a)</u>

<u>The measure of material's ability to conduct thermal energy (heat) is known as thermal conductivity.</u> For examples, metals have high thermal conductivity, it means that they are very efficient at conducting heat.<u> The SI unit of heat capacity is W/m.K.</u>

The expression for thermal conductivity is:

$q=-\kappa \bigtriangledown T$

Where,

q is the heat flux

$\kappa$ is the thermal conductivity

$\bigtriangledown T$ is the temperature gradient.

<u>(b)</u>

<u>Heat capacity for a substance is defined as the ratio of the amount of energy required to change the temperature of the substance and the magnitude of temperature change. The SI unit of heat capacity is J/K.</u>

The expression for Heat capacity is:

$C=\frac{E}{\Delta T}$

Where,

C is the Heat capacity

E is the energy absorbed/released

$\Delta T$ is the change in temperature

<u>(c)</u>

<u>Thermal diffusivity is defined as the thermal conductivity divided by specific heat capacity at constant pressure and its density. The Si unit of thermal diffusivity is m²/s.</u>

The expression for thermal diffusivity is:

$\alpha=\frac{\kappa}{C_p \times \rho}$

Where,

$\alpha$ is thermal diffusivity

$\kappa$ is the thermal conductivity

$C_p$ is specific heat capacity at constant pressure

$\rho$ is density

6 0

2 years ago

Calculate the number of atoms per cubic meter in Metal B (units atoms/m^3). Write your answer with 4 significant figures metal:

Iteru [2.4K]

Answer:

7,217*10^28 atoms/m^3

Explanation:

Metal: Vanadium
Density: 6.1 g/cm^3
Molecuar weight: 50,9 g/mol

The Avogadro's Number, 6,022*10^23, is the number of atoms in one mole of any substance. To calculate the number of atoms in one cubic meter of vanadium we write:

1m^3*(100^3 cm^3/1 m^3)*(6,1 g/1 cm^3)*(1 mol/50,9g)*(6,022*10^23 atoms/1 mol)=7,217*10^28 atoms

Therefore, for vanadium we have 7,217*10^28 atoms/m^3

6 0

3 years ago

The period of a pendulum T is assumed to depend only on the mass m, the length of the pendulum `, the acceleration due to gravit

zzz [600]

Answer:

The expression is shown in the explanation below:

Explanation:

Thinking process:

Let the time period of a simple pendulum be given by the expression:

$T = \pi \sqrt{\frac{l}{g} }$

Let the fundamental units be mass= M, time = t, length = L

Then the equation will be in the form

$T = M^{a}l^{b}g^{c}$

$T = KM^{a}l^{b}g^{c}$

where k is the constant of proportionality.

Now putting the dimensional formula:

$T = KM^{a}L^{b} [LT^{-} ^{2}]^{c}$

$M^{0}L^{0}T^{1} = KM^{a}L^{b+c}$

Equating the powers gives:

a = 0

b + c = 0

2c = 1, c = -1/2

b = 1/2

so;

a = 0 , b = 1/2 , c = -1/2

Therefore:

$T = KM^{0}l^{\frac{1}{2} } g^{\frac{1}{2} }$

T = $2\pi \sqrt{\frac{l}{g} }$

where k = $2\pi$

8 0

3 years ago

Are front-end engineers starting to decline in China?

laiz [17]

Yes. They are declining in China. Very fast

7 0

3 years ago

An equal-tangent sag vertical curve (with a negative initial and a positive final grade) is designed for 55 mi/h. The PVI is at

Varvara68 [4.7K]

Answer:

The lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

Explanation:

Length of curve is given as

$L=2(PVT-PVI)\\L=2(242+30-240+00)\\L=2(230)\\L=460 ft$

$G_2$ is given as

$G_2=\frac{E_{PVT}-E_{PVI}}{0.5L}\\G_2=\frac{127.5-122}{0.5*460}\\G_2=0.025=2.5 \%$

The K value is given from the table 3.3 for 55 mi/hr is 115. So the value of A is given as

$A=\frac{L}{K}\\A=\frac{460}{115}\\A=4$

A is given as

$-G_1=A-G_2\\-G_1=4.0-2.5\\-G_1=1.5\\G_1=-1.5\%$

With initial grade, the elevation of PVC is

$E_{PVC}=E_{PVI}+G_1(L/2)\\E_{PVC}=122+1.5%(460/2)\\E_{PVC}=125.45 ft\\$

The station is given as

$St_{PVC}=St_{PVI}-(L/2)\\St_{PVC}=24000-(230)\\St_{PVC}=237+70\\$

Low point is given as

$x=K \times |G_1|\\x=115 \times 1.5\\x=172.5 ft$

The station of low point is given as

$St_{low}=St_{PVC}-(x)\\St_{low}=23770+(172.5)\\St_{low}=239+42.5 ft\\$

The elevation is given as

$E_{low}=\frac{G_2-G_1}{2L} x^2+G_1x+E_{PVC}\\E_{low}=\frac{2.5-(-1.5)}{2*460} (1.72)^2+(-1.5)*(1.72)+125.45\\E_{low}=124.16 ft$

So the lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

3 0

3 years ago