All categories

4 years ago

A pure substance is anything that cannot be broken down or separated by _______________ means?

Physics

2 answers:

maxonik [38]4 years ago

6 0

Answer:

Explanation:

An element is a pure substance which cannot be broken down into anything simpler by chemical methods. Most elements consist of a mixture of isotopes and so are not pure substances. Isotopes of an element can be separated by physical methods such as diffusion and mass spectrometry.

MatroZZZ [7]4 years ago

3 0

Answer: Physical is the blank space

Explanation: Your welcome

You might be interested in

define atmospheric pressure, prove that presence of atmospheric pressure with the help of an activity

Anton [14]

I don understand this?

8 0

3 years ago

A 65-cm segment of conducting wire carries a current of 0.35 A. The wire is placed in a uniform magnetic field that has a magnit

dimaraw [331]

Answer:

e) 67°

Explanation:

the force on the wire can be calculated using the expression below

F = BILsinФ

But we are looking for the angle between the wire segment and the magnetic field, then we can make Ф the subject of the formula from the above expresion, then we have,

Ф =sin⁻¹ (F/BIL)

The parameters is defined as

I =current that is been carried by the wire= 0.35 A

Ф = angle between the wire segment and the magnetic field, which is the unknown?

L = length of the wire=65 cm

B = magnetic field = 1.24

F= force on the wire = 0.26 N,

Ф =sin⁻¹ (F/BIL)

Ф =sin⁻¹ X .....................eqn(#)

Where X= (F/BIL)

We can calculate for X= (F/BIL), from eqn(#) by substituting value of Force, Lenght and

magnetic field

X=(F/BIL)= 0.26/(1.24×0.35×0.65)

= 0.26/0.2821

=0.922

Then substitute X into eqn (Ф =sin⁻¹ X)

Then

Ф =sin⁻¹ (0.922)

Ф=66.42°

Ф=67° approximately

Therefore, the angle between the wire segment and the magnetic field is 67°

7 0

3 years ago

How does the media influence our view of beauty and gender roles?

shepuryov [24]

<span>They try to tell us what to think is beautiful, like they never have fat idols</span>

4 0

3 years ago

While performing an experiment involving a reaction between two chemicals, the scientist observes that the reaction container ha

AVprozaik [17]

When an object is becomes warm its an example of exothermic reaction.

8 0

3 years ago

A closed, rigid tank fitted with a paddle wheel contains 2 kg of air, initially at 300 K. During an interval of 5 minutes, the p

anzhelika [568]

Answer:

The final temperature of the air is $T_2= 605 K$

Explanation:

We can start by doing an energy balance for the closed system

$\Delta KE+\Delta PE+ \Delta U = Q - W$

where

$\Delta KE$ = the change in kinetic energy.

$\Delta PE$ = the change in potential energy.

$\Delta U$ = the total internal energy change in a system.

Q = the heat transferred to the system.

W = the work done by the system.

We know that there are no changes in kinetic or potential energy, so $\Delta KE = 0$ and $\Delta PE=0$

and our energy balance equation is $\Delta U = Q - W$

We also know that the paddle-wheel transfers energy to the air at a rate of 1 kW and the system receives energy by heat transfer at a rate of 0.5 kW, for 5 minutes.

We use this information to calculate the total internal energy change $\Delta U=W+Q$ using the energy balance equation.

We convert the interval of time to seconds $t = 5 \:min = 300\:s$

$\Delta \dot{U}=\dot{W}+ \dot{Q}\\=\Delta U=(W+ Q)\cdot t$

$\Delta U=(1 \:kW+0.5\:kW)\cdot 300\:s\\\Delta U=450 \:kJ$

We can use the change in specific internal energy $\Delta U = m(u_2-u_1)$ to find the final temperature of the air.

We are given that $T_1=300 \:K$ and the air can be describe by ideal gas model, so we can use the ideal gas tables for air to determine the initial specific internal energy $u_1$

$u_1=214.07\:\frac{kJ}{kg}$

Next, we will calculate the final specific internal energy $u_2$

$\Delta U = m(u_2-u_1)\\\frac{\Delta U}{m} =u_2-u_1$

$\frac{\Delta U}{m} =u_2-u_1\\u_2=u_1+\frac{\Delta U}{m}$

$u_2=214.07 \:\frac{kJ}{kg} +\frac{450 \:kJ}{2 \:kg}\\u_2= 439.07 \:\frac{kJ}{kg}$

With the value $u_2=439.07 \:\frac{kJ}{kg}$ and the ideal gas tables for air we make a regression between the values $u = 434.78 \:\frac{kJ}{kg},T=600 \:K$ and $u = 442.42 \:\frac{kJ}{kg}, T=610 \:K$ and we find that the final temperature $T_2$ is 605 K.

3 0

3 years ago