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2 years ago

The velocity components expressed in m/s

1 answer:

3 0

The scalar absolute value (magnitude) of velocity is called speed, being a coherent derived unit whose quantity is measured in the SI (metric system) as metres per second (m/s or m⋅s−1). For example, "5 metres per second" is a scalar, whereas "5 metres per second east" is a vector.

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3.8 LAB - Select lesson schedule with multiple joins

dem82 [27]

Answer:

The database has three tables for tracking horse-riding lessons: Horse with columns: ID - primary key; RegisteredName; Breed; Height; BirthDate.

Explanation:

4 0

2 years ago

A piston-cylinder device contains an ideal gas mixture of 3 kmol of He gas and 7 kmol of Ar gas (both gases are monatomic) at 27

lidiya [134]

Answer:

Q = 62 ( since we are instructed not to include the units in the answer)

Explanation:

Given that:

$n_{HCl} = 3 \ kmol\\n_{Ar} = 7 \ k mol$

$T_1 = 27^0 \ C = ( 27+273)K = 300 K$

$P_1 = 200 \ kPa$

Q = ???

Now the gas expands at constant pressure until its volume doubles

i.e if $V_1 = x\\V_2 = 2V_1$

Using Charles Law; since pressure is constant

$V \alpha T$

$\frac{V_2}{V_1} =\frac{T_2}{T_1}$

$\frac{2V_1}{V_1} =\frac{T_2}{300}$

$T_2 = 300*2\\T_2 = 600$

mass of He =number of moles of He × molecular weight of He

mass of He = 3 kg × 4

mass of He = 12 kg

mass of Ar =number of moles of Ar × molecular weight of Ar

mass of He = 7 kg × 40

mass of He = 280 kg

Now; the amount of Heat Q transferred = $m_{He}Cp_{He} \delta T + m_{Ar}Cp_{Ar} \delta T$

From gas table

$Cp_{He} = 5.9 \ kJ/Kg/K\\Cp_{Ar} = 0.5203 \ kJ/Kg/K$

∴ Q = $12*5.19*10^3(600-300)+280*0.5203*10^3(600-300)$

Q = $62.389 *10^6$

Q = 62 MJ

Q = 62 ( since we are instructed not to include the units in the answer)

5 0

4 years ago

Read 2 more answers

Nearlyof all serious occupational injuries and illnesses stem from overexertion of repetitive motion.

Ostrovityanka [42]

Answer:

a) 1/2

Explanation:

Overexertion accounted for more than half of all events that resulted in a disabling condition.

Furthermore, 30% of all overexertion cases were reported in the services industry, on the other hand, 25% of injuries resulting from contact with objects and equipment occurred in the manufacturing industry.

The above piece of information is taken from the bureau of labor statistics, Survey of Occupational Injuries and Illnesses

"LOST-WORKTIME INJURIES AND ILLNESSES: CHARACTERISTICS AND RESULTING DAYS AWAY FROM WORK, 2002"

8 0

3 years ago

For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of t

saveliy_v [14]

Complete Question

For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of this material elongate when a true stress of 411 MPa (59610 psi) is applied if the original length is 470 mm (18.50 in.)?Assume a value of 0.22 for the strain-hardening exponent, n.

Answer:

The elongation is $=21.29mm$

Explanation:

In order to gain a good understanding of this solution let define some terms

True Stress

A true stress can be defined as the quotient obtained when instantaneous applied load is divided by instantaneous cross-sectional area of a material it can be denoted as $\sigma_T$ .

True Strain

A true strain can be defined as the value obtained when the natural logarithm quotient of instantaneous gauge length divided by original gauge length of a material is being bend out of shape by a uni-axial force. it can be denoted as $\epsilon_T$ .

The mathematical relation between stress to strain on the plastic region of deformation is

$\sigma _T =K\epsilon^n_T$

Where K is a constant

n is known as the strain hardening exponent

This constant K can be obtained as follows

$K = \frac{\sigma_T}{(\epsilon_T)^n}$

No substituting $345MPa \ for \ \sigma_T, \ 0.02 \ for \ \epsilon_T , \ and \ 0.22 \ for \ n$ from the question we have

$K = \frac{345}{(0.02)^{0.22}}$

$= 815.82MPa$

Making $\epsilon_T$ the subject from the equation above

$\epsilon_T = (\frac{\sigma_T}{K} )^{\frac{1}{n} }$

$Substituting \ 411MPa \ for \ \sigma_T \ 815.82MPa \ for \ K \ and \ 0.22 \ for \ n$

$\epsilon_T = (\frac{411MPa}{815.82MPa} )^{\frac{1}{0.22} }$

$=0.0443$

From the definition we mentioned instantaneous length and this can be obtained mathematically as follows

$l_i = l_o e^{\epsilon_T}$

Where

$l_i$ is the instantaneous length

$l_o$ is the original length

$Substituting \ 470mm \ for \ l_o \ and \ 0.0443 \ for \ \epsilon_T$

$l_i = 470 * e^{0.0443}$

$=491.28mm$

We can also obtain the elongated length mathematically as follows

$Elongated \ Length =l_i - l_o$

$Substituting \ 470mm \ for l_o and \ 491.28 \ for \ l_i$

$Elongated \ Length = 491.28 - 470$

$=21.29mm$

4 0

3 years ago

Liquid flows with a free surface around a bend. The liquid is inviscid and incompressible, and the flow is steady and irrotation

lions [1.4K]

Answer:

9 cm

Explanation:

The liquid on the bend will be affected by two accelerations: gravity and centripetal force.

Gravity will be of 9.81 m/s^2 pointing down at all points.

The centripetal acceleration will be of

ac = v^2/r

Pointing to the center of the bend (perpendicular to gravity).

The velocity will depend on the radius

v = (1 m^2/s) / r

Replacing:

ac = (1/r)^2 / r

ac = (1 m^4/s^2) / r^3

If we set up a cylindrical reference system with origin at the center of the bend, the total acceleration will be

a = (-1/r^3 * i - 9.81 * j)

The surface of the liquid will be an equipotential surface, this means all points on the surface have the same potential energy.

The potential energy of the gravity field is:

pg = g * h

The potential energy of the centripetal force is:

pc = ac * r

Then the potential field is:

p = -1/r^2 * - 9.81*h

Points on the surface at r = 1 m and r = 3 m have the same potential.

-1/1^2 * - 9.81*h1 = -1/3^2 * - 9.81*h2

-1 - 9.81*h1 = -1/9 - 9.81*h2

-1 + 1/9 = 9.81 * (h1 - h2)

h1 - h2 = (-8/9) / 9.81

h2 - h1 = 0.09 m

The outer part will be 9 cm higher than the inner part.

3 0

3 years ago

The velocity components expressed in m/s​

The velocity components expressed in m/s