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yarga [219]
2 years ago
6

If a substance changes from a vapor to a liquid it ____ HELP FAST

Physics
1 answer:
alisha [4.7K]2 years ago
5 0

Answer:

Explanation:

Condensing is the word used to indicate the change of state of a substance from vapor to liquid, as in this case. During condensation, the substance releases thermal energy to the environment, therefore the kinetic energy of the molecules in the vapor decreases until they become closer to each other and they start to be affected by the intermolecular forces and so the substance becomes a liquid.

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a sydiver jumps out of a plane. She fall dowward at a very fast speed. Which force is acting upon it?
UNO [17]

Answer:

There will be two forces acting on her: Gravitational force and Air resisitence

6 0
3 years ago
A spring of spring constant 30.0 N/m is attached to a 2.3 kg mass and set in motion. What is the period and frequency of vibrati
coldgirl [10]

Answer:

1. The period is 1.74 s.

2. The frequency is 0.57 Hz

Explanation:

1. Determination of the the period.

Spring constant (K) = 30 N/m

Mass (m) = 2.3 Kg

Pi (π) = 3.14

Period (T) =?

The period of the vibration can be obtained as follow:

T = 2π√(m/K)

T = 2 × 3.14 × √(2.3 / 30)

T = 6.28 × √(2.3 / 30)

T = 1.74 s

Thus, the period of the vibration is 1.74 s.

2. Determination of the frequency.

Period (T) = 1.74 s

Frequency (f) =?

The frequency of the vibration can be obtained as follow:

f = 1/T

f = 1/1.74

f = 0.57 Hz

Thus, the frequency of the vibration is 0.57 Hz

4 0
3 years ago
True or false: (a) Maxwell's equations apply only to fields that are constant over time.(b) The wave equation can be derived fro
Andre45 [30]

Answer:

(a) False

(b) True

(c) True

(d) True

(e) True

(f) True

Explanation:

(a) Maxwell's equations not only applies to constant fields but it applies to both the fields, i.e., Time variant field as well as Time Invariant field.

(b) We make use of the Modified form of the Ampere's law and Faraday's Law to derive the wave equation.

(c) Electromagnetic waves contains both the electric and magnetic fields and these fields oscillates at an angle of 90^{\circ}C to the direction of wave propagation.

(d) In free space both the electric and magnetic fields are in phase while considering electromagnetic waves.

(e) In free space or vacuum, the expression for the speed of light in terms of electric and magnetic field is given as:

c = \frac{E}{B}

Thus the ratio of the magnitudes of the electric and magnetic field vectors are equal to the speed of light in free space.

(f) In free space or in vacuum the energy density of the electromagnetic wave is divided equally in both the fields and hence are equal.

3 0
3 years ago
The box of a well-known breakfast cereal states that one ounce of the cereal contains 113 Calories (1 food Calorie = 4186 J). If
mafiozo [28]

Answer:

m=454.73 kg

Explanation:

To convert the energy of the cereal to know the weight:

m*g*h=Q'

Cal=0.0179*113 Cal=2.0227Cal

Q'=2.0227Cal*\frac{4186J}{1Cal}

m*g*h=8467.02J

m*g=\frac{Q'}{h}=\frac{8467.02J}{1.90m}

m*g=4456.33 N

Supuse the gravity as a g=9.8m/s^2

m=\frac{4456.33N}{9.8m/s^2}

m=454.73 kg

3 0
3 years ago
Read 2 more answers
A point charge of +3 C is located at the origin of a coordinate system and a second point charge of -6 C is at x = 1.0 m. At w
Butoxors [25]

Answer:

The point at which the electrical potential is zero is x = +0.33 m.

Explanation:

By definition the electrical potential is:

V_{E} = \frac{K*q}{r}

Where:

K: is Coulomb's constant = 9x10⁹ N*m²/C²

q: is the charge

r: is the distance

The point at which the electrical potential is zero can be calculated as follows:

V_{1} + V_{2} = 0

K(\frac{q_{1}}{r_{1}} + \frac{q_{2}}{r_{2}}) = 0    (1)

q₁ is the first charge = +3 mC

r₁ is the distance from the point to the first charge  

q₂ is the first charge = -6 mC

r₂ is the distance from the point to the second charge    

By replacing r₁ = 1 - r₂ into equation (1) we have:

K(\frac{q_{1}}{1 - r_{2}} + \frac{q_{2}}{r_{2}}) = 0   (2)

By solving equation (2) for r₂:

r_{2} = \frac{q_{1}}{q_{1} - q_{2}} = \frac{3 mC}{3 mC - (-6 mC)} = +0.33 m

                 

Therefore, the point at which the electrical potential is zero is x = +0.33 m.

I hope it helps you!  

8 0
2 years ago
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