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3 years ago

3. A Tylenol has 80 mg of acetaminophen. How many grams is that

Physics

1 answer:

denis23 [38]3 years ago

6 0

$\boxed{\sf 1g=1000mg}$

$\\ \sf\longmapsto 80g$

$\\ \sf\longmapsto \dfrac{80}{1000}$

$\\ \sf\longmapsto 0.08g$

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An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 95 kPa and 27°C, and 75

katen-ka-za [31]

Answer:

Part A) 3899 kPa

Part B) 392.33 kJ/kg

Part C) 0.523

Part D) 495 kPa

Explanation:

Part A

First from the temperature at state 1 the relative specific volume and the internal energy at that state are determined from:

$u_{1}$ = 214.07 kJ/kg

$\alpha$ $r_{1}$ = 621.2

The relative specific volume at state 2 is obtained from the compression ratio:

$\alpha$ $r_{2}$ = $\frac{\alpha r_{1} }{r}$

=621.2/ 8

= 77.65

From this the temperature and internal energy at state 2 can be determined using interpolation with data from A-17(table):

$T_{2}$ = 673 K

$u_{2}$ = 491.2 kJ/kg

The pressure at state 2 can be determined by manipulating the ideal gas relations at state 1 and 2:

$P_{2}$ = $P_{1} r\frac{T_{2} }{T_{1} }$

= 95*8*673/300

= 1705 kPa

Now from the energy balance for stage 2-3 the internal energy at state 3 can be obtained:

$deltau_{2-3} =q_{in}\\ u_{3} -u_{2} =q_{in}\\u_{3}=u_{2}+q_{in}$

= 1241.2 kJ/kg

From this the temperature and relative specific volume at state 3 can be determined by interpolation with data from A-17(table):

$T_{3}$ = 1539 K

$\alpha r_{3}$ = 6.588

The pressure at state 3 can be obtained by manipulating the ideal gas relations for state 2 and 3:

$P_{3} =P_{2} \frac{T_{3} }{T_{2} }$

= 3899 kPa

Part B

The relative specific volume at state 4 is obtained from the compression ratio:

$\alpha r_{4}= r\alpha r_{3}$

= 52.7

From this the temperature and internal energy at state 4 can be determined by interpolation with data from A-17:

$T_{4}$ =775 K

$u_{4}$ = 571.74 kJ/kg

The net work output is the difference of the heat input and heat rejection where the heat rejection is determined from the decrease in internal energy in stage 4-1:

$w=q_{in}-q_{out}\\q_{in}-(u_{4} -u_{1} )\\=392.33 kJ/kg$

Part C

The thermal efficiency is obtained from the work and the heat input:

η= $\frac{w}{q_{in} }$

=0.523

Part D

The mean effective pressure is determined from its standard relation:

MEP= $\frac{w}{\alpha_{1}- \alpha_{2} }$

= $\frac{w}{\alpha_{1}(1- \frac{1}{r} }$

= $\frac{rwP_{1} }{RT_{1} (r-1) }$

=495 kPa

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3 years ago

If vector C is added to vector D, the result is a third vector that is perpendicular to D and has a magnitude equal to 3D. What

Jet001 [13]

Answer:

(e) 3.2

Explanation:

We are given that vector C and D.

Let R be the magnitude of C+D.

According to question

R=3D

We have to find the ratio of the magnitude of C to that of D.

By using right triangle property

$C^2=R^2+D^2$

$C^2=(3D)^2+D^2$

$C^2=9D^2+D^2$

$C^2=10D^2$

$C=\sqrt{10D^2}=3.2D$

$\frac{C}{D}=3.2$

Hence, the ratio of the magnitude of C to that of D=3.2

(e) 3.2

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3 years ago

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aev [14]

Light speed is the speed at which light can travel in a "vacuum" (space for example).

The speed of light is constant and is exactly: 299,792,458 meters per second

ONLY IN A VACUUM!!

The speed of light changes when it goes through different mediums (such as from space to Earth, light travels slower in Earth).

Good luck!

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Q.1 Lorna built the circuit diagram below. All the bulbs are identical. (a) Complete the table below by writing on or off for ea

irga5000 [103]

Answer:

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Explanation:

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jekas [21]

There are two atoms of potassium bonded to one atom of sulfur.

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