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3 years ago

6

Marie is puzzled by her findings she has done several meticulous calculations and has gotten the numbers .37 rad, .89 rad and 1.

65 rad. she was expecting .90 rad each time. she has been which of the following?
a. neither precise and accurate
b. both precise and accurate
c. precise
d. accurate

1 answer:

Irina-Kira [14]3 years ago

6 0

Answer:

I think the answer is a

Explanation:

for it to be accurate has be to exactly 0.9 rad

it is not precise because the answer she is getting is different everytime and not even close. For instance,

It would have been precise if she had gotten 0.37 rad in every attempt. or 0.89 every attempt...

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3 years ago

At a construction site a pipe wrench struck the ground with a speed of 31 m/s. (a) From what height was it inadvertently dropped

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v = 31 m/s

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3 years ago

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At what temperature (degrees Fahrenheit) is the Fahrenheit scale reading equal to (a) 2 times that of the Celsius and (b) 1/4 ti

castortr0y [4]

Answer:

(a) F = 320

(b) = F = -5.1625

Explanation:

The formula that converts degree Celsius (C) to degree Fahrenheit (F) is:

F = 1.8C + 32

Solving (a): F = 2C

Substitute 2C for F in the above equation

F = 1.8C + 32

2C = 1.8C + 32

Collect like terms

2C - 1.8C = 32

0.2C = 32

Multiply both sides by 5

5 * 0.2C = 32 * 5

C = 160

Recall that F = 2C

F = 2 * 160

F = 320

Solving (b): F = ¼C

Substitute ¼C for F in the above formula

F = 1.8C + 32

¼C = 1.8C + 32

Convert fraction to decimal

0.25C = 1.8C + 32

Collect like terms

0.25C - 1.8C = 32

-1.55C = 32

Divide both sides by -1.55

C = 32/(-1.55)

C = -32/1.55

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Recall that: F = ¼C

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5 0

3 years ago

A piston having 7.23 g of steam at 110°C increases its temperature by 35°C. At the same time it expands from a volume of 2.00 L

My name is Ann [436]

Answer : The value of $q,w,\Delta U\text{ and }\Delta U$ is 505 J, -599 J, -94 J and -693 J respectively.

Explanation : Given,

Mass of steam = 7.23 g

Initial temperature = $110^oC$

Final temperature = $(110+35)^oC=145^oC$

Initial volume = 2 L

Final volume = 8 L

External pressure = 0.985 bar

Heat capacity of steam = 1.996 J/g.K

First law of thermodynamic : It states that the energy can not be created or destroyed, it can only change or transfer from one state to another state.

As per first law of thermodynamic,

$\Delta U=q+w$

First we have to calculate the heat absorbed by the system.

Formula used :

$Q=m\times c\times \Delta T$

or,

$Q=m\times c\times (T_2-T_1)$

where,

Q = heat absorbed by the system = ?

m = mass of steam = 7.23 g

$C_p$ = heat capacity of steam = $1.966J/g.K$

$T_1$ = initial temperature = $110^oC=273+110=383K$

$T_2$ = final temperature = $145^oC=273+145=418K$

Now put all the given value in the above formula, we get:

$Q=7.23g\times 1.966J/g.K\times (418-383)K$

$Q=505J$

Now we have to calculate the work done.

Formula used :

$w=-p_{ext}dV\\\\w=-p_{ext}(V_2-V_1)$

where,

w = work done = ?

$p_{ext}$ = external pressure = 0.985 bar = 0.985 atm (1 bar = 1 atm)

$V_1$ = initial volume of gas = 2.00 L

$V_2$ = final volume of gas = 8.00 L

Now put all the given values in the above formula, we get :

$w=-p_{ext}(V_2-V_1)$

$w=-(0.985atm)\times (8.00-2.00)L$

$w=-5.91L.atm=-5.91\times 101.3J=-599J$

conversion used : (1 L.atm = 101.3 J)

Now we have to calculate the change in internal energy of the system.

$\Delta U=q+w$

$\Delta U=505J+(-599J)$

$\Delta U=-94J$

Now we have to calculate the change in enthalpy of the system.

Formula used :

$\Delta H=\Delta U+P\Delta V$

$\Delta H=\Delta U+w$

$\Delta H=(-94J)+(-599J)$

$\Delta H=-693J$

Therefore, the value of $q,w,\Delta U\text{ and }\Delta U$ is 505 J, -599 J, -94 J and -693 J respectively.

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4 years ago