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3 years ago

What happens when a roller coaster car moves down from the top of a hill?

2 answers:

6 0

-- It accelerates.
-- Its speed increases.
-- It gains momentum.
-- It loses altitude.
-- It loses potential energy.
-- It gains kinetic energy.
-- Its wheels make a lot of noise.
-- Everybody screams.

Stella [2.4K]3 years ago

3 0

Potential energy is transformed into kinetic and thermal energy.

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Jail

kykrilka [37]

Answer:

c.o Kg • ms 08

Explanation:

put all together

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

The light intensity incident on a metallic surface with a work function of 1.88 eV produces photoelectrons with a maximum kineti

ivanzaharov [21]

Answer:

KE = KE (incidental) - KE of emitted photons

or KE = h * f - Wf

So h * f = KE + Wf = 1.2 + 1.88 = 3.08 incident energy

If you double the frequency then h * f = 6.16

KE = 6.16 - 1.2 = 4.96 eV

7 0

3 years ago

Light striking a mirror at a 50° angle will be reflected at an angle _____.

Stells [14]

Equal to 50

law of reflection: angle of incidence equals angle of reflection

5 0

3 years ago

Read 2 more answers

How high does the water rise in the bell after enough time has passed for the air to reach thermal equilibrium

Minchanka [31]

The height risen by water in the bell after enough time has passed for the air to reach thermal equilibrium is 3.8 m.

<h3>Pressure and temperature at equilibrium </h3>

The relationship between pressure and temperature can be used to determine the height risen by the water.

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

where;

V₁ = AL
V₂ = A(L - y)
P₁ = Pa
P₂ = Pa + ρgh
T₁ = 20⁰C = 293 K
T₂ = 10⁰ C = 283 k

$\frac{PaAL}{T_1} = \frac{(P_a + \rho gh)A(L-y)}{T_2} \\\\\frac{PaL}{T_1} = \frac{(P_a + \rho gh)(L-y)}{T_2} \\\\L-y = \frac{PaLT_2}{T_1(P_a + \rho gh)} \\\\y = L (1 - \frac{PaT_2}{T_1(P_a + \rho gh)})\\\\y = 4.2(1 - \frac{101325 \times 283}{293(101325\ +\ 1000 \times 9.8 \times 100)} )\\\\y = 3.8 \ m$

Thus, the height risen by water in the bell after enough time has passed for the air to reach thermal equilibrium is 3.8 m.

The complete question is below:

A diving bell is a 4.2 m -tall cylinder closed at the upper end but open at the lower end. The temperature of the air in the bell is 20 °C. The bell is lowered into the ocean until its lower end is 100 m deep. The temperature at that depth is 10°C. How high does the water rise in the bell after enough time has passed for the air to reach thermal equilibrium?

Learn more about thermal equilibrium here: brainly.com/question/9459470

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3 0

2 years ago