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4 years ago

PLEASE HELP WILL GIVE 10 POINTS!!!!

Physics

2 answers:

Nimfa-mama [501]4 years ago

7 0

freezing, deposition, and condensation

serg [7]4 years ago

4 0

D. freezing, deposition, and condensation

Explanation:

In order to understand, let's see the definition of these changes of state:

- freezing: freezing occurs when a substance changes from liquid state to solid state

- deposition: deposition occurs when a substance changes from gas state to solid state, without passing through the liquid state

- condensation: this process occurs when a substance changes from gas state to liquid state

In all the three processes listed above, energy is released from the substance, which in fact loses energy. In fact:

- in a liquid, the particles have more energy than in a solid (because in liquid they can move, while in solid they can only vibrate), so during freezing, since a liquid turns into a solid, it releases energy to the surrounding

- in a gas, the particles have more energy than in a liquid (because in a gas they can move faster than in a liquid), so during condensation, since a gas turns into a liquid, energy is released to the surrounding

- in deposition, the particles change from the gas state (where they move faster) to a solid state (where they are not free to move), so they lose energy as well.

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Your window thermometer shows that the temperature outdoors is 89.3°f. how should you report this temperature to your friend in

SVETLANKA909090 [29]

A) Conversion from Fahrenheit to Celsius:
The relationship between the two scales is the following
$T_f = \frac{9}{5}T_c + 32$
where
$T_f$ is the temperature in Fahrenheit
$T_c$ is the temperature in Celsius

Re-arranging the formula, we find the inverse relationship:
$T_c= \frac{5}{9}(T_f-32)$
and we can use this equation to convert $T_f = 89.3^{\circ}F$ into Celsius:
$T_c = \frac{5}{9}(89.3-32)=31.7^{\circ} C$

(b) Conversion to Kelvin scale:
The conversion between Celsius scale and Kelvin scale is easier:
$T_k = T_c + 273.1$
where $T_k$ is the temperature in Kelvin. Using $T_c = 31.7^{\circ} C$ , we find
$T_k = 31.7 + 273.1 = 304.8 K$

5 0

3 years ago

A flat coil of wire consisting of 20 turns, each with an area of 50 cm2, is positioned perpendicularly to a uniform magnetic fie

jenyasd209 [6]

Answer:

0.5 A

Explanation:

N = 20, A = 50 cm^2 = 50 x 10^-4 m^2, dB = 6 - 2 = 4 T, dt = 2 s, R = 0.4 ohm

The induced emf is given by

e = - N dФ/dt

Where, dФ/dt is the rate of change of magnetic flux.

Ф = B A

dФ/dt = A dB/dt

so,

e = 20 x 50 x 10^-4 x 4 / 2 = 0.2 V

negative sign shows the direction of magnetic field.

induced current, i = induced emf / resistance = 0.2 / 0.4 = 0.5 A

5 0

3 years ago

A 62 kg boy and a 37 kg girl use an elastic rope while engaged in a tug-of-war on a friction-less icy surface. If the accelerati

Paraphin [41]

Answer:

The magnitude of the acceleration of the boy toward the girl is $1.31\ m/s^2$

Explanation:

It is given that,

Mass of the boy, $m_1=62\ kg$

Mass of the girl, $m_2=37\ kg$

The acceleration of the girl toward the boy is, $a_2=2.2\ m/s^2$

To find,

The acceleration of the boy toward the girl.

Solution,

Let $a_1$ is the magnitude of the acceleration of the boy toward the girl. We know that force acting on one object to other are equal in magnitude but opposite in direction. So,

$F_1=-F_2$

$m_1a_1=-m_2a_2$

$a_1=-\dfrac{m_2a_2}{m_1}$

$a_1=-\dfrac{37\times 2.2}{62}$

$a_1=-1.31\ m/s^2$

$|a_1|=1.31\ m/s^2$

So, the magnitude of the acceleration of the boy toward the girl is $1.31\ m/s^2$

7 0

3 years ago

5. A cable is attached 32.0 m from the base of a flagpole that is about to

soldi70 [24.7K]

Answer:

The length of the flagpole is approximately 87.43 m

Explanation:

The given parameters of the cable attached to the flagpole are;

The point along the flagpole at which the cable is attached = 32.0 m

The angle with respect to the ground at which the raising of the flagpole is halted = 60.0°

The downward force exerted by the cable, $F_v$ = 1.233 × 10⁴ N

The force exerted by the cable to the left = 1.233 × 10⁴ N

Let 'W' represent the weight of the flagpole, at equilibrium, we have;

The sum of vertical forces = 0

Therefore;

$F_v$ + W - R = 0

W - R = -1.233 × 10⁴ N

Taking moment about the support at the base of the pole, we get;

1.233 × 10⁴ × d × cos(60.0°) - 1.233 × 10⁴ ×d× sin(60.0°) + W × d/2 ×cos(60.0°) = 0

∴ W × d/2 ×cos(60.0°) ≈ 4513.093·d

W = 2 × 4513.093/(cos(60.0°)) ≈ 18,052.373 N

R = 18,052.373 + 1.233 × 10⁴ ≈ 30,382.373

R ≈ 30,382.373 N

Taking moment about the point of attachment of the cable to the ground, we have;

W × ((d/2) × cos(60.0°) + 32) = R × 32

∴ (d/2) = ((30,382.373 × 32/18,052.373) - 32)/(cos(60.0°)) ≈ 43.71281

d = 2 × 43.71281 ≈ 87.43

The length of the flagpole, d ≈ 87.43 m

7 0

3 years ago

Prove law of conservation of energy for a stone moving vertically down ( explain energy at B & C )

STALIN [3.7K]

As per given condition of point B we can see that height at point B is "h/2" from the ground

So we know that potential energy is given as

U = mgh

so here we have to put height h = h/2

so potential energy is U = mgh/2

now for kinetic energy we need to find the speed of it after falling the distance h/2

now by kinematics we will have

$v_f^2 - 0^2 = 2(g)(h/2)$

now for kinetic energy

$KE = \frac{1}{2}mv^2 = \frac{1}{2}m(gh)$

$KE = 1/2mgh$

now total energy will be given as

$E = mgh/2 + mgh/2 = mgh$

now for point C we can say that it is the point near to ground

So here height is ZERO

now potential energy will also be zero

U = 0

now for kinetic energy we need to find speed

$v^2 - 0^2 = 2(g)(h)$

now kinetic energy

$KE = \frac{1}{2}mv^2 = \frac{1}{2}m(2gh)$

$KE = mgh$

now again we have total energy

$E = 0 + mgh = mgh$

6 0

4 years ago