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marta [7]
3 years ago
11

How can you change the phases of matter using thermal energy

Physics
1 answer:
Leokris [45]3 years ago
8 0

Answer:

These energy exchanges are not changes in kinetic energy. They are changes in bonding energy between the molecules. If heat is coming into a substance during a phase change, then this energy is used to break the bonds between the molecules of the substance.

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When a substance undergoes a phase change is still has a definite ...
PIT_PIT [208]
The answer is A, mass.
7 0
3 years ago
A 78.5-kg man floats in freshwater with 3.2% of his volume above water when his lungs are empty, and 4.85% of his volume above w
Dima020 [189]

Answer:

A) V_air = 1.295 L

B) Volume is not reasonable

Explanation:

A) Let;

m be total mass of the man

m_p be the mass of the man that pulled out of the water because of the buoyant force that pulled out of the lung

m_3 be the mass above the water with the empty lung

m_5 be the mass above the water with full lung

F_b be the buoyant force due to the air in the lung

V_a be the volume of air inside man's lungs

w_p be the weight that the buoyant force opposes as a result of the air.

Now, we are given;

m = 78.5 kg

m_3 = 3.2% × 78.5 = 2.512 kg

m_5 = 4.85% × 78.5 = 3.80725 kg

Now, m_p = m_5 - m_3

m_p = 3.80725 - 2.512

m_p = 1.29525 kg

From archimedes principle, we have the formula for buoyant force as;

F_b = (m_displaced water)g = (ρ_water × V_air × g)

Where ρ_water is density of water = 1000 kg/m³

Thus;

F_b = w_p = 1.29525 × 9.81

F_b = 12.7064 N

As earlier said,

F_b = (ρ_water × V_air × g)

Thus;

V_air = F_b/(ρ_water × × g)

V_air = 12.7064/(1000 × 9.81)

V_air = 1.295 × 10^(-3) m³

We want to convert to litres;

1 m³ = 1000 L

Thus;

V_air = 1.295 × 10^(-3) × 1000

V_air = 1.295 L

B) From research, the average lung capacity of an adult human being is 6 litres of air.

Thus, the calculated lung volume is not reasonable

4 0
3 years ago
7. Un bloque de 700 N se encuentra sobre una viga uniforme de 200 N y 6 m de longitud. El bloque está a una distancia de 1 m del
GrogVix [38]

Answer:

 x =  0.176 m

Explanation:

For this exercise we will take the condition of rotational equilibrium, where the reference system is located on the far left and the wire on the far right. We assume that counterclockwise turns are positive.

Let's use trigonometry to decompose the tension

      sin 60 = T_{y} / T

      T_{y} = T sin 60

       cos 60 = Tₓ / T

      Tₓ = T cos 60

we apply the equation

       ∑ τ = 0

       -W L / 2 - w x + T_{y} L = 0

 

the length of the bar is L = 6m

           -Mg 6/2 - m g x + T sin 60 6 = 0

             x = (6 T sin 60 - 3 M g) / mg

let's calculate

let's use the maximum tension that resists the cable T = 900 N

             x = (6 900 sin 60 - 3 200 9.8) / (700 9.8)

             x = (4676 - 5880) / 6860

             x = - 0.176 m

Therefore the block can be up to 0.176m to keep the system in balance.

5 0
3 years ago
This force can either push the block upward at a constant velocity or allow it to slide downward at a constant velocity. The mag
Dmitry [639]

Answer:

Part a)

F = 135.7 N

Part b)

F = 62.5 N

Explanation:

Part a)

If block is sliding up then net force must be zero and friction will be in opposite to the direction of motion of the block

Fcos\theta = mg + F_f

Fsin\theta = F_n

so we have

Fcos\theta = mg + \mu(Fsin\theta)

F(cos\theta - \mu sin\theta) = mg

F = \frac{mg}{cos\theta - \mu sin\theta}

F = \frac{55}{cos50 - 0.310(sin50)}

F = 135.7 N

Part b)

If block is sliding down then net force must be zero and friction will be in opposite to the direction of motion of the block

Fcos\theta = mg - F_f

Fsin\theta = F_n

so we have

Fcos\theta = mg - \mu(Fsin\theta)

F(cos\theta + \mu sin\theta) = mg

F = \frac{mg}{cos\theta + \mu sin\theta}

F = \frac{55}{cos50 + 0.310(sin50)}

F = 62.5 N

6 0
3 years ago
What is the acceleration of a dog that runs from 3 m/s to 6 m/s over a distance of 90m?
KonstantinChe [14]

Answer:

solution given:

acceleration (a)=?

initial velocity (u)=3m/s

final velocity (v)=6m/s

distance (s)=90m

we have

v²=u²+2as

substituting value

6²=3²+2*a*90

36=9+180a

36-9=180a

a=25/180

<u>a=0.1388m/s²</u>

6 0
2 years ago
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