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3 years ago

Calculate the density of a sample of gas with a mass of 30g and volume of 7500 cm3

Physics

1 answer:

rewona [7]3 years ago

4 0

<span>D= .004 g/ml

Hope this helps!</span>

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All of the following are types of symbiotic relationships except for

nevsk [136]

The answer is D. the way I remember it is they all end with -alism.

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2 years ago

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vlabodo [156]

Answer:

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2 years ago

Dylan has two cubes of iron. The larger cube has twice the mass of the smaller cube. He measures the smaller cube. Its mass is 2

liubo4ka [24]

Answer:

The volume of the larger cube is 5.08 g/cm³.

Explanation:

Given that,

Mass of smaller cube = 20 g

Density of smaller cube $\rho= 7.87 g/cm^2$

Dylan has two cubes of iron.

The larger cube has twice the mass of the smaller cube.

$M_{l}=2m_{s}$

Density is same for both cubes because both cubes are same material.

The density is equal to the mass divided by the volume.

$\rho=\dfrac{m}{V}$

$V=\dfrac{m}{\rho}$

Where, V = volume

m = mass

$\rho=density$

We need to calculate the volume of smaller mass

The volume of smaller mass

$V_{s}=\dfrac{m_{s}}{\rho_{s}}$

$V_{s}=\dfrac{20}{7.87}$

$V_{s}=2.54\ cm^3$

Now, We need to calculate the volume of large cube

$V_{l}=\dfrac{m_{l}}{\rho_{l}}$

$V_{l}=\dfrac{2\times20}{7.87}$

$V_{l}=5.08\ g/cm^3$

Hence, The volume of the larger cube is 5.08 g/cm³.

8 0

2 years ago

A 4.0-cm tall light bulb is placed at distance of 8.3 cm from a concave mirror having a focal length of 15.2 cm. Determine the i

scoundrel [369]

Answer: the image distance is -18, 28 cm this means behind of the concave mirror. The image size is 2.2 higher that the original so it has 8.8 cm with the same orientation as original and it is a virtual imagen.

Explanation: In order to sove the imagen formation for a concave mirror we have to use the following equation:

1/p+1/q=1/f where p and q represents the distance to the mirror for the object and imagen, respectively. f is the focal length for the concave mirror.

replacing the values we obtain:

1/8.3+1/q=1/15.2

so 1/q=(1/15.2)-(1/8.3)=-54.7*10^-3

then q=-18.28 cm

The magnification is given by M=-q/p=-(-18,28)/8.3= 2.2

We also add a picture to see the imagen formation for this case.