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3 years ago

6 A single sheet of aluminium foil is folded twice to produce a stack of four sheets. The total

1 answer:

3 0

The correct answer is option C. i.e. 4%

<span>The total thickness of the stack of sheets is measured to be (0.80 ± 0.02) mm.

The maximum true thickness will be </span>(0.80 + 0.02) mm. = 1.00 mm
Ans, the maximum uncertainity will be <span> (0.20 + 0.02) mm = +/-0.04 mm

Thus, the four sheets true thickness will be </span>1.00 mm +/-0.04 mm
Thus, single sheet thickness will be 0.250 mm +/-0.01 mm
Thus, percentage error = (0.01/0.25) * 100 = 4%

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efrigerant-134a is expanded isentropically from 600 kPa and 70°C at the inlet of a steady-flow turbine to 100 kPa at the outlet.

PolarNik [594]

Answer:

Inlet : $v_i=0.0646\frac{m}{s}$

Outlet: $v_o=0.171\frac{m}{s}$

Explanation:

1) Notation and important concepts

Flow of mass represent "the mass of a substance which passes per unit of time".

Flow rate represent "a measure of the volume of liquid that moves in a certain amount of time"

Specific volume is "the ratio of the substance's volume to its mass. It is the reciprocal of density."

Isentropic process is a "thermodynamic process, in which the entropy of the fluid or gas remains constant".

We know that the flow of mass is given by the following expression

$\dot{m}=\frac{\dot{V}}{\upsilon}$ , where $\dot{V}$ represent the flow rate and $\upsilon$ the specific volume at the pressure and temperature given.

$A_i=0.5m^2$ is the inlet area

$P_i=600Kpa$ pressure at the inlet area

$T_i=70C$ temperature at the inlet area

$A_o=1m^2$ is the outlet area

$P_o=100Kpa$ pressure at the outlet area

$T_o=C$ temperature at the outlet area

$\dot{m}=0.75\frac{kg}{s}$ represent the flow of mass

If we look at the first figure attached Table A-13 we see that the specific volume for the inlet condition is

$\upsilon_i =0.04304\frac{kg}{m^3}$ and the entropy is $h_i=1.0645\frac{KJ}{KgK}=h_o$

With the value of entropy and the outlet pressure of 100 Kpa we can find we specific volume at the outlet condition since w ehave the entropy $h_o=1.0645\frac{KJ}{KgK}$

Since on the table we don't have the exact value we need to interpolate between these two values (see the second figure attached)

$h_1=1.0531\frac{KJ}{KgK} , \upsilon_1=0.22473\frac{kg}{m^3}$

$h_2=1.0829\frac{KJ}{KgK} , \upsilon_2=0.23349\frac{kg}{m^3}$

Our interest value would be given using interpolation like this:

$\upsilon=0.22473+\frac{(0.23349-0.22473)}{(1.0829-1.0531)}(1.0645-1.0531)=0.228\frac{kg}{m^3}$

2) Solution to the problem

Now since we have all the info required to solve the problem we can find the velocities on this way.

We know from the definition of flow of mass that $\dot{m}=\frac{\dot{V}}{\upsilon}$ , but since $\dot{V}=Av$ we have this:

$\dot{m}=\frac{Av}{\upsilon}$

If we solve from the velocity v we have this:

$v=\frac{\upsilon \dot{m}}{A}$ (*)

And now we just need to replace the values into equation (*)

For the inlet case:

$v_i=\frac{\upsilon_i \dot{m}}{A_i}=\frac{0.043069\frac{kg}{m^3}(0.75\frac{kg}{s})}{0.5m^2}=0.0646\frac{m}{s}$

For the oulet case:

$v_o=\frac{\upsilon_o \dot{m}}{A_o}=\frac{0.228\frac{kg}{m^3}(0.75\frac{kg}{s})}{1m^2}=0.171\frac{m}{s}$

7 0

3 years ago

What gas law looks at the relationship between pressure and temperature of a gas? A) Boyle's Law B) Charles' Law C) Gay Lussac's

Mandarinka [93]

Answer: it should be Boyle’s law

Explanation:

6 0

3 years ago

Read 2 more answers

Consider a river flowing toward a lake at an average speed of 3 m/s at a rate of 550 m3/s at a location 58 m above the lake surf

Vladimir [108]

Answer:

1. 0.574 kJ/kg

2. 315.7 MW

Explanation:

1. The mechanical energy per unit mass of the river is given by:

$E_{m} = E_{k} + E_{p}$

$E_{m} = \frac{1}{2}v^{2} + gh$

Where:

Ek is the kinetic energy

Ep is the potential energy

v is the speed of the river = 3 m/s

g is the gravity = 9.81 m/s²

h is the height = 58 m

$E_{m} = \frac{1}{2}(3 m/s)^{2} + 9.81 m/s^{2}*58 m = 0.574 kJ/Kg$

Hence, the total mechanical energy of the river is 0.574 kJ/kg.

2. The power generation potential on the river is:

$P = m(t)E_{m} = \rho*V(t)*E_{m} = 1000 kg/m^{3}*550 m^{3}/s*0.574 kJ/kg = 315.7 MW$

Therefore, the power generation potential of the entire river is 315.7 MW.

I hope it helps you!

4 0

3 years ago

Can you help me with this paper please I will give you 20 points!

sweet [91]

1) they are attracting because if you look at the arrows they’re all pointing the same way.

2) if the magnet was turned around they would do the opposite and not attract ( this is called repulsion)

3) magnetic pole

4)magnet

5) magnetic force

6) magnetism

Hope this helps

6 0

3 years ago

Read 2 more answers

When conducting an investigation outside, you can assume it is safe to touch my plant or animal you see.

noname [10]

False, all scene are combed for clues and photographed.

4 0

3 years ago