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2 years ago

If the mass of a material is 47 grams and the volume of the material is 15 cm^3, what would the density of the material be?

1 answer:

8 0

Density is calculated as mass per unit volume. In this case, since the material has a mass of 47 grams and we have the volume of 15 cm^3, we can simply divide the values:
Density = 47 grams / 15 cm^3 = 3.1 g/cm^3
Therefore, the material has a density of 3.1 g/cm^3

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A 1.2 L weather balloon on the ground has a temperature of 25°C and is at atmospheric pressure (1.0 atm). When it rises to an el

Irina-Kira [14]

Answer:

71.19 C

Explanation:

25C = 25 + 273 = 298 K

Applying the ideal gas equation we have

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

where P, V and T are the pressure, volume and temperature of the gas at 1st and 2nd stage, respectively. We can solve for the temperature and the 2nd stage:

$T_2 = T_1\frac{P_2V_2}{P_1V_1} = 298\frac{0.77*1.8}{1.2*1} = 298*1.155 = 344.19 K = 344.19 - 273 = 71.19 C$

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3 years ago

The standard deviation of Eric’s data is 0.8°C. Martha conducted the same experiment. Her average temperature was 35.1 with a st

REY [17]

Answer:

less precise than

Explanation:

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2 years ago

3. A certain wire, 3 m long, stretches by 1.2 mm when under tension of 200 N. By how much does

nikitadnepr [17]

Answer:

The extension of the second wire is $e_2 = 0.0024 \ m = 2.4 mm$

Explanation:

From the question we are told that

The length of the wire is $L = 3 \ m$

The elongation of the wire is $e = 1.2mm = \frac{1.2}{1000} = 0.0012 m$

The tension is $F = 200 \ N$

The length of the second wire is $L_2 = 6 \ m$

Generally the Young's modulus(Y) of this material is

$Y = \frac{stress}{strain }$

Where $stress = \frac{F}{A}$

Where A is the area which is evaluated as

$A = \pi r^2$

and $strain = \frac{extention}{length} = \frac{e}{L}$

$Y = \frac{\frac{F}{\pi r^2 } }{ \frac{e}{L} }$

Since the wire are of the same material Young's modulus(Y) is constant

So we have

$\frac{F * L }{r^2 e} = \pi * Y = constant$

$F * L = constant * r^2 e$

Now the ration between the first and the second wire is

$\frac{F_1}{F_2} * \frac{L_1}{L_2} = \frac{r*2_1}{r^2} * \frac{e_1}{e_2}$

Since tension , radius are constant

We have

$\frac{L_1}{L_2} = \frac{e_1}{e_2}$

substituting values

$\frac{3}{6} = \frac{0.0012}{e_2}$

$0.5 e_2 = 0.0012$

$e_2 = \frac{ 0.0012 }{0.5}$

$e_2 = 0.0024 \ m = 2.4 mm$

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3 years ago

If the nucleus of an atom contains 12 protons how many electrons are there in a neutral atom why

kicyunya [14]

For a neutral atom number of electrons equals number of protons, in other for the net charge of the atom to be zero...
no. of electrons = 12

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3 years ago

If an object is placed between the focal point and twice the focal length of a convex lens, which type of image will be produced

valentinak56 [21]

<span>The image produced is real and enlarged.

Check for various positions of objects and Images for convex lens.

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3 0

2 years ago

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