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pickupchik [31]

2 years ago

6

a lead block drops its temperature by 5.90 degrees celsius when 427 J of heat are removed from it. what is the mass of the block

?(unit=kg) IM GIVING 30 POINTS FOR THE CORRECT ANSWER

1 answer:

Airida [17]2 years ago

7 0

Answer:

577g

Explanation:

Given parameters:

Temperature change = 5.9°C

Amount of heat lost = 427J

Unknown:

Mass of the block = ?

Solution:

The heat capacity of a body is the amount of heat required to change the temperature of that body by 1°C.

H = m c Ф

H is the heat capacity

m is the mass of the block

c is the specific heat capacity

Ф is the temperature change

Specific heat capacity of lead is 0.126J/g°C

m = H / m Ф

m = $\frac{427}{0.126 x 5.9}$ = 577g

Mass of the lead block is 577g

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The density of gasoline is 730 kg/m3 at 0°C. Its average coefficient of volume expansion is 9.60 10-4(°C)−1. Assume 1.00 gal of

kipiarov [429]

Answer: 0.4911 kg

Explanation:

We have the following data:

$\rho_{0\°C}= 730 kg/m^{3}$ is the density of gasoline at $0\°C$

$\beta=9.60(10)^{-4} \°C^{-1}$ is the average coefficient of volume expansion

We need to find the extra kilograms of gasoline.

So, firstly we need to transform the volume of gasoline from gallons to $m^{3}$ :

$V=8.50 gal \frac{0.00380 m^{3}}{1 gal}=0.0323 m^{3}$ (1)

Knowing density is given by: $\rho=\frac{m}{V}$ , we can find the mass $m_{1}$ of 8.50 gallons:

$m_{1}=\rho_{0\°C}V$

$m_{1}=(730 kg/m^{3})(0.0323 m^{3})=23.579 kg$ (2)

Now, we have to calculate the factor $f$ by which the volume of gasoline is increased with the temperature, which is given by:

$f=(1+\beta(T_{f}-T_{o}))$ (3)

Where $T_{o}=0\°C$ is the initial temperature and $T_{f}=21.7\°C$ is the final temperature.

$f=(1+9.60(10)^{-4} \°C^{-1}(21.7\°C-0\°C))$ (4)

$f=1.020832$ (5)

With this, we can calculate the density of gasoline at $21.7\°C$ :

$\rho_{21.7\°C}=730 kg/m^{3} f=(730 kg/m^{3})(1.020832)$

$\rho_{21.7\°C}=745.207 kg/m^{3}$ (6)

Now we can calculate the mass of gasoline at this temperature:

$m_{2}=\rho_{21.7\°C}V$ (7)

$m_{2}=(745.207 kg/m^{3})(0.0323 m^{3})$ (8)

$m_{2}=24.070 kg$ (9)

And finally calculate the mass difference $\Delta m$ :

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$\Delta m=0.4911 kg$ (11) This is the extra mass of gasoline

6 0

3 years ago

An insect 5.00 mm tall is placed 20.0 cm to the left of a thin planoconvex lens. The left surface of this lens is flat, the righ

jenyasd209 [6]

Answer:

a) i = -9.63 cm , h ’= .0.24075 cm erect

b) i = 259.74 cm ,

Explanation:

For this exercise let's start by finding the focal length of the lens

1 / f = (n-1) (1 / R₁ - 1 / R₂)

1 / f = (1.70 -1)) 1 / ∞ - 1/13)

1 / f = 0.0538

f = - 18.57 cm

Now we can use the constructor equation

1 / f = 1 / o + 1 / i

1 / i = 1 / f - 1 / o

1 / i = -1 / 18.57 -1/20

1 / i = -0.1038 cm

I = -9.63 cm

For the height of the

image let's use magnification

m = h '/ h = - i / o

h ’= -h i / o

h ’= - 0.5 (-9.63) / 20

h ’= .0.24075 cm

b) we invert the lens

The focal length is

1 / f = (1.70 -1) (1/13 - 1 / int)

1 / f = 0.0538

f = 18.57 cm

1 / i = 1 / f -1 / o

1 / I = 1 / 18.57 - 1/20

1 / I = 3.85 10-3

i = 259.74 cm

h ’= - 0.5 259.74 / 20

h ’= 6.4935 cm

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