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

answered: 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

answered: 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,

answered: Guest

Answer: it will take 4.36 seconds to reach her average velocity 10.69 miles/hour.explanation: we have equation of motion $s=ut+\frac{1}{2} at^2$ ,where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.average velocity = displacement/ time taken.average velocity = 10.69 miles/hr = 10.69*1.6*1000/(60*60) = 4.75 m/s4.75 = s/ts = 4. 1we also have acceleration = 2.18 m/s², initial velocity = 0 m/s $s=0*t+\frac{1}{2} *2.18*t^2$ $s=1.09t^2[/ 2equating equation 1 and 2 [tex]4.75t = 1.09t^2\\ \\ t=4.36 seconds.$ so it will take 4.36 seconds to reach her average velocity 10.69 miles/hour.

answered: Guest

weight of the box is given as

$w = 112.1 n$

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

$w_x = w sin\theta$

here we have

$w = 112.1 n$

$w_x = 112.1 sin42$

$w_x = 75 n$

so it requires 75 n force to move it upwards

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1. rank the following in terms of increasing density:
a. a100 g object with a volume of 25 cu...