Physics, 24.09.2019 22:30 campilk5

1. rank the following in terms of increasing density:
a. a100 g object with a volume of 25 cubic centimeters
b. a 200 g object with a volume of 100 cubic centimeters
c. a 100 g object with a volume of 100 cubic centimeters
d. a 400 g object with a volume of 50 cubic centimeters

Answers: 2

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Answers

Answer from: Abbymay17

C < B < A < D

Explanation:

Density is defined as the quotient of Mass divided by Volume, therefore we need to find that quotient for each case in order to rank them:

Case A: $D = \frac{100}{25} = 4 \frac{g}{cm^3}$

Case B: $D = \frac{200}{100} = 2 \frac{g}{cm^3}$

Case C: $D = \frac{100}{100} = 1 \frac{g}{cm^3}$

Case D: $D = \frac{400}{50} = 8 \frac{g}{cm^3}$

So the ranking in INCREASING density is:

C < B < A < D

Answer from: addisonrausch

The density of the objects in increasing order is given as C B A D.

Explanation:

Density is the property of mater which compares amount of mater of an object and volume of an object.

The density is given by the formula:

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

Where,

‘ρ’ is density

‘m’ is mass of an object

‘V’ is volume of an object

Object A:

Take object A of 100 g object with a volume of 25 cubic centimetres:

We know that, $m_{A}=100 \ \mathrm{g}$ and $V_{A}=25 \ \mathrm{cm}^{3}$

Density of A is:

$\Rightarrow \rho_{A}=\frac{m_{A}}{V_{A}}$

$\Rightarrow \rho_{A}=\frac{100}{25}$

$\therefore \rho_{A}=4 \ g / \mathrm{cm}^{3}$

Object B:

Take object A 200 g object with a volume of 100 cubic centimetres:

We know that, $m_{B}=200 \ \mathrm{g}$ and $V_{B}=100 \ \mathrm{cm}^{3}$

Density of B is:

$\Rightarrow \rho_{B}=\frac{m_{B}}{V_{B}}$

$\Rightarrow \rho_{B}=\frac{200}{100}$

$\therefore \rho_{B}=2 \ g / \mathrm{cm}^{3}$

Object C:

Take object A 100 g object with a volume of 100 cubic centimetres:

We know that, $m_{C}=100 \ \mathrm{g}$ and $V_{C}=100 \ \mathrm{cm}^{3}$

Density of C is:

$\Rightarrow \rho_{C}=\frac{m_{C}}{V_{C}}$

$\Rightarrow \rho_{C}=\frac{100}{100}$

$\therefore \rho_{C}=1 \ \mathrm{g} / \mathrm{cm}^{3}$

Object D:

Take object A 400 g object with a volume of 50 cubic centimetres:

We know that, $m_{D}=400 \ \mathrm{g}$ and $V_{D}=50 \ \mathrm{cm}^{3}$

Density of object D is:

$\Rightarrow \rho_{D}=\frac{m_{D}}{V_{D}}$

$\Rightarrow \rho_{D}=\frac{400}{50}$

$\therefore \rho_{D}=8 \ \mathrm{g} / \mathrm{cm}^{3}$

From the above calculation, on comparing,

Answer from: Quest

m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| = √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| = √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| = 14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s

Answer from: Quest

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