All categories

3 years ago

How can you use the volume of some quantity of a pure substance to calculate its mass

1 answer:

3 0

Firstly, you must identify the pure substance first. Then, from a reliable source obtain its density. Density is calculated through the equation,

density = mass / volume

From this equation, the volume can be calculated.

volume = mass / density

You might be interested in

Why are weathering, erosion and deposition a NECESSARY process in the rock cycle?

Zarrin [17]

Answer:

YESS well it is partly nessary but it depends on the situation

Explanation:

4 0

3 years ago

Read 2 more answers

Four +2 µC point charges are at the corners of a square of side 2 m. Find the potential at the center of the square (relative to

Romashka [77]

Answer:

(a) 51428.59 J/C

(b) 25714.29 J/C

(c) 0 J/C

Explanation:

Parameters given:

Q1 = 2 * 10^-6 C

Q2 = 2 * 10^-6 C

Q3 = 2 * 10^-6 C

Q4 = 2 * 10^-6 C

=> Q1 = Q2 = Q3 = Q4 = Q

Side of the square = 2m

The center of the square is the midpoint of the diagonals, i.e. Using Pythagoras theorem:

BD² = 2² + 2²

BD² = 8

BD = √(8) = 2.8m

OD = 1.4m

(The attached diagram explains better)

Hence, the distance between the center and each point charge, r, is 1.4m.

Electric Potential, V = kQ/r

k = Coulombs constant

(a) If all charges are positive:

V(Total) = V1 + V2 + V3 + V4

V1 = Potential due to Q1

V2 = Potential due to Q2

V3 = Potential due to Q3

V4 =Potential due to Q4

Since Q1 = Q2 = Q3 = Q4 = Q

=> V1 = V2 = V3 = V4

=> V(Total) = 4V1

V = (4 * 9 * 10^9 * 2 * 10^-6)/1.4

V = 51428.59J/C

(b) If 3 charges are positive and 1 is negative:

Since Q1 = Q2 = Q3 = Q

and Q4 = -Q

The total potential becomes:

V(Total) = V1 + V2 + V3 - V4

Since V1, V2, V3 and V4 have the same value,

V(Total) = V1 + V2

V(Total) = 2V1

V(Total) = (2 * 9 * 10^9 * 2 * 10^-6)/1.4

V(Total) = 25174.29 J/C

Since Q1 = Q2 = Q

and Q3 = Q4 = -Q

The total potential becomes:

V(Total) = V1 + V2 - V3 - V4

Since V1, V2, V3 and V4 have the same value,

V(Total) = 0 J/C

8 0

3 years ago

Which configuration is common to the outer shells of the elements neon, argon, and krypton?

Alexandra [31]

Answer: S2P6

The electronic configuraiton of given elements are as follows:

Ne =[He]2s2 2p6
Ar = [Ne]3s2 3p6
Kr = [Ar]4s2 3d10 4p6

As it can be seen that for all three elements, their outermost orbital are completely filled, that is it has both s orbital, p orbital and d orbital fulfilled. Noble or inert gas atoms like Neon, Argon, Krypton have fulfilled valence shell. Fulfilled outermost orbital is the most stable electronic state, hence all elements tends to achieve such stability. These noble gas elements are called inert gas because of their fulfilled outermost shell. This means they don't react easily or take part in eletron donating, receiving or sharing. This is because, for all other elements except inert gas atoms, their valence shell is incomplete and they tend to react by other atoms so as to complete their outermost shell , which we call as duplet (in case of Helium like) or Octet state. Such elements either donate some electrons or receive some to acheive such stable state..

5 0

3 years ago

Vector question,university physics zemanski

just olya [345]

Answer:

yes This is correct Answer

3 0

3 years ago

An electric drill

Lera25 [3.4K]

Answer:

Incomplete question. Complete question is: An electric drill starts from rest and rotates with a constant angular acceleration. After the drill has rotated through a certain angle, the magnitude of the centripetal acceleration of a point on the drill is twice the magnitude of the tangential acceleration. Determine the angle through which the drill rotates by this point.

The answer is : Δ θ = 1 rad

Explanation:

Ok, so the condition involves the centripetal acceleration and the tangential acceleration, so let’s start by writing expressions for each:

Ac= centripetal acceleration At= tangential acceleration

Ac = V² / r At = r α

Because we have to determine the angle ultimately, therefore we should convert the linear velocity into angular velocity in the expression for centripetal acceleration

V = r ω

Ac = (r ω)² / r = r² ω² / r

Ac = r ω²

now that we have expressions for the centripetal and tangential acceleration, we can write an equation that expresses the condition given: The magnitude of the centripetal acceleration is twice the magnitude of the tangential acceleration.

Ac = 2 At

That is,

r ω² = 2 r α

it is equivalent to;

ω² = 2 α

now we have the relation between angular speed and angular acceleration, but we also need to determine the angular displacement as well. Therefore choose a kinematics equation that doesn’t involve time because time is not mentioned in the question. Thus,

ω² – ω°² = 2 α Δ θ

such that ω° = 0

and ω² = 2 α

therefore;

2 α - 0 = 2 α Δ θ

2 α = 2 α Δ θ

So the angle will be : Δ θ = 1 rad

7 0

3 years ago