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3 years ago

Substances which naturally attract each other called what

Physics

1 answer:

grandymaker [24]3 years ago

5 0

Answer:

Ferromagnetism, physical phenomenon in which certain electrically uncharged materials strongly attract others. Two materials found in nature, lodestone (or magnetite, an oxide of iron, Fe3O4) and iron, have the ability to acquire such attractive powers, and they are often called natural ferromagnets.

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A gas bottle contains 0.250 mol of gas at 730 mm hg pressure. if the final pressure is 1.15 atm, how many moles of gas were adde

Ludmilka [50]

Answer: 0.049 mol

Explanation:

1) Data:

n₁ = 0.250 mol

p₁ = 730 mmHg

p₂ = 1.15 atm

n₂ - n₁ = ?

2) Assumptions:

i) ideal gas equation: pV = nRT

ii) V and T constants.

3) Solution:

i) Since the temperature and the volume must be assumed constant, you can simplify the ideal gas equation into:

pV = nRT ⇒ p/n = RT/V ⇒ p/n = constant.

ii) Then p₁ / n₁ = p₂ / n₂

⇒ n₂ = p₂ n₁ / p₁

iii) n₂ = 1.15atm × 760 mmHg/atm × 0.250 mol / 730mmHg = 0.299 mol

iv) n₂ - n₁ = 0.299 mol - 0.250 mol = 0.049 mol

7 0

3 years ago

An electron moves in a straight path with a velocity of +500,000 m/s. It undergoes constant acceleration of +500,000,000,000 met

77julia77 [94]

Answer:

Distance travelled, d = 0.21 m

Explanation:

It is given that,

Initial velocity of electron, u = 500,000 m/s

Acceleration of the electron, a = 500,000,000,000 m/s²

Final velocity of the electron, v = 675,000 m/s

We need to find the distance travelled by the electron. Let distance travelled is s. Using third equation of motion as :

$v^2-u^2=2as$

$s=\dfrac{v^2-u^2}{2a}$

$s=\dfrac{(675000\ m/s)^2-(500000\ m/s^2)^2}{2\times 500000000000\ m\s^2}$

s = 0.205 m

s = 0.21 m

So, the electron will travel a distance of 0.21 meters. Hence, this is the required solution.

6 0

3 years ago

The diagram shows monochromatic light passing through two openings.

geniusboy [140]

Answer: constructive interference in which waves strengthen each other

Explanation:

4 0

3 years ago

. Suppose the mass of a fully loaded module in which astronauts take off from the Moon is 1.00 × 104 kg. The thrust of its engin

Viktor [21]

Answer:

a) The module's acceleration in a vertical takeoff from the Moon will be $1.377 \frac{m}{s^2}$

b) Then we can say that a thrust of $3*10^{4} N$ won't be able to lift off the module from the Earth because it's smaller than the module's weight ( $9.8 *10^{4} N$ ).

Explanation:

a) During a vertical takeoff, the sum of the forces in the vertical axis will be equal to mass times the module's acceleration. In this this case, the thrust of the module's engines and the total module's weight are the only vertical forces. (In the Moon, the module's weight will be equal to its mass times the Moon's gravity acceleration)

$T-(m*g)=m*a$

Where:

$T=$ thrust $=3 *10^{4} N$

$m=$ module's mass $=1 *10^{4} N$

$g=$ moon's gravity acceleration $=1.623 \frac{m}{s^2}$

$a=$ module's acceleration during takeoff

Then, we can find the acceleration like this:

$a=\frac{T}{m} -g=\frac{3*{10}^4 N}{1*{10}^4 kg}-1.623\frac{m}{s^2}$

$a=1.377 \frac{m}{s^2}$

The module's acceleration in a vertical takeoff from the Moon will be $1.377 \frac{m}{s^2}$

b) To takeoff, the module's engines must generate a thrust bigger than the module's weight, which will be its mass times the Earth's gravity acceleration.

$weight=m*g=(1*{10}^4 kg)*(9.8 \frac{m}{s^2})=9.8 *10^{4} N$

Then we can say that a thrust of $3*10^{4} N$ won't be able to lift off the module from the Earth because it's smaller than the module's weight ( $9.8 *10^{4} N$ ).

8 0

4 years ago

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