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KIM [24]
4 years ago
13

A pupil wants to find the density of an oil. She uses a chemical balance which measures to the nearest gram (g). She places an e

mpty measuring cylinder on to the balance.
She pours some oil into the cylinder. The level of oil in the measuring cylinder is shown below in an attachment (named Assignment Diagram).

Use this data to find the density of oil.
! Show your workings.
! Explain each stage in your calculation

THANK YOU TO EVERYONE WHO STOPPED AND TOOK THEIR TIME TO ANSWER MY QUESTION, THIS REALLY MEANS A LOT TO ME.
FINALLY, IF ANYONE USES THIS OPPORTUNITY TO EARN FREE POINTS, I WILL REPORT YOU.
Physics
1 answer:
scoray [572]4 years ago
5 0

Answer:

2.29g/cm3.

Explanation:

The following data were obtained from the question.

Mass of the oil = 172g

Volume of oil = 75cm3

Density =..?

Density of a substance is simply defined as the mass of the substance per unit volume of the substance. It is represented mathematically as:

Density = Mass /volume

With the above formula, we can calculate the density of the oil as shown below:

Density = Mass /volume

Density = 172g/75cm3

Density = 2.29g/cm3.

Therefore, the density of the oil is 2.29g/cm3.

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A commuter airplane starts from an airport and takes theroute. The plane first flies to city A, located 175 km away in a directi
bearhunter [10]

Answer:

245.45km in a direction 21.45° west of north from city A

Explanation:

Let's place the origin of a coordinate system at city A.

The final position of the airplane is given by:

rf = ra + rb + rc    where ra, rb and rc are the vectors of the relative displacements the airplane has made. If we separate this equation into its x and y coordinates:

rfX = raX+ rbX + rcX = 175*cos(30)-150*sin(20)-190 = -89.75km

rfY = raY + rbY + rcT = 175*sin(30)+150*cos(20) = 228.45km

The module of this position is:

rf = \sqrt{rfX^2+rfY^2} = 245.45km

And the angle measure from the y-axis is:

\alpha =atan(rfX/rfY) = 21.45\°

So the answer is 245.45km in a direction 21.45° west of north from city A

6 0
3 years ago
Calculate the heat energy needed to change the temperature of 2 kg of copper from 10°C to 110°C.
serious [3.7K]
<span>The specific heat (or the amount of heat required to raise the temperature of a unit mass of a substance by 1 degree Celsius) of copper is about 0.386 J/g/degree Celsius. This means that if we supply 0.386 J of energy to 1 gram of copper, its temperature will increase by 1 degree Celsius.</span>
7 0
4 years ago
Read 2 more answers
Suppose that the dipole moment associated with an iron atom of an iron bar is 2.8 × 10-23 J/T. Assume that all the atoms in the
masya89 [10]

To solve this exercise it is necessary to apply the equations related to the magnetic moment, that is, the amount of force that an image can exert on the electric currents and the torque that a magnetic field exerts on them.

The diple moment associated with an iron bar is given by,

\mu = \alpha *N

Where,

\alpha = Dipole momento associated with an Atom

N = Number of atoms

\alpha y previously given in the problem and its value is 2.8*10^{-23}J/T

L = 5.8cm = 5.8*10^{-2}m

A = 1.5cm^2 = 1.5*10^{-4}m^2

The number of the atoms N, can be calculated as,

N = \frac{\rho AL}{M_{mass}}*A_n

Where

\rho = Density

M_{mass} = Molar Mass

A = Area

L = Length

A_n =Avogadro number

N = \frac{(7.9g/cm^3)(1.5cm)(5.8cm^2)}{55.9g/mol}(6.022*10^{23}atoms/mol)

N = 7.4041*10^{23}atoms

Then applying the equation about the dipole moment associated with an iron bar we have,

\mu = \alpha *N

\mu = (2.8*10^{-23})*(7.4041*10^{23})

\mu = 20.72Am^2

PART B) With the dipole moment we can now calculate the Torque in the system, which is

\tau = \mu B sin(90)

\tau = (20.72)(2.2)

\tau = 45.584N.m

<em>Note: The angle generated is perpendicular, so it takes 90 ° for the calculation made.</em>

3 0
3 years ago
A runner of mass 80 kg is moving at 8.0 m/s. Calculate her kinetic energy. ​
lana [24]

Answer:

2560J

Explanation:

By definition the kinetic energy can be calculated in the following way:

K = (mv²)/2 = 80kg·(8.0m/s)²/2 = 2560 J

7 0
2 years ago
One litre of crude oil weighs 9.6N. Calculate its specific weight, density and specific gravity.​
Zepler [3.9K]

Answer:

The answer is "\bold{9600 \frac{N}{m^3}, 978.59 \frac{kg}{m^3}, and \ 0.978}"

Explanation:

Given:

\to v=1\ liter= 10^{-3} \ m^3\\\\\to  w= 9.6 \ N\\

calculation:

Specific \ weight =\frac{w}{v}=\frac{9.6}{10^{-3}}=9600 \frac{N}{m^3} \\\\w=mg\\\\m= \frac{w}{g}=\frac{9.6}{9.81}=0.9785\ kg\\\\\rho\ (density)=\frac{m}{v}=\frac{0.9785}{10^{-3}}=978.59 \frac{kg}{m^3}\\\\specific \ gravity = \frac{\prho \ obj}{\rho w}=\frac{978.54}{1000}=0.978

4 0
3 years ago
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