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polet [3.4K]
3 years ago
5

According to the second law of thermodynamics, energy of the universe must always increase.

Physics
1 answer:
irinina [24]3 years ago
3 0
The answer is false..................

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i'm begging ya'll i posted this about 3 times already and i'm really stuck. can someone please give me the answers to the cart?
Vedmedyk [2.9K]

Answer:

just put the molecules and atoms where they belong

Explanation:

4 0
2 years ago
Where a subducting plate slides beneath the lithosphere, melting takes place and a(n) ____ is created?
11111nata11111 [884]

under ground cave will be created


3 0
3 years ago
A bus is moving at a speed of 150km/hr. Begins to slow at a constant rate of 3.0m/s each second. Find how far it goes before sto
ladessa [460]

Answer:

Distance = 13.9 meters

Explanation:

Given the following data;

Maximum speed = 150 km/hr to meters per seconds = 150 * 1000/3600 = 41.67 m/s

Decelerating speed = 3m/s

To find the distance travelled with this speed;

Distance = maximum speed/decelerating speed

Distance = 41.67/3

Distance = 13.9 meters

Therefore, the bus would travel a distance of 13.9 meters before stopping.

4 0
2 years ago
1.Calculate the energy transferred by a 12V hairdryer, running on a current of 0.50A, that is left on for 8.0 minutes.
CaHeK987 [17]

Answer:

1. Energy = 2880 Joules.

2. Energy = 60 Joules.

3. Quantity of charge = 120 Coulombs.

Explanation:

Given the following data;

1. Voltage = 12 Volts

Current = 0.5 Amps

Time, t = 8 mins to seconds = 8 * 60 = 480 seconds

To find the energy;

Power = current * voltage

Power = 12 * 0.5

Power = 6 Watts

Next, we find the energy transferred;

Energy = power * time

Energy = 6 * 480

Energy = 2880 Joules

2. Charge, Q = 4 coulombs

Potential difference, p.d = 15V

To find the total energy transferred;

Energy = Q * p.d

Energy = 4 * 15

Energy = 60 Joules

3. Voltage = 6 Volts

Current = 1 Amps

Time = 2 minutes to seconds = 2 * 60 = 120 seconds

To find the quantity of charge;

Quantity of charge = current * time

Quantity of charge = 1 * 120

Quantity of charge = 120 Coulombs

8 0
3 years ago
A rigid, insulated tank whose volume is 10 L is initially evacuated. A pinhole leak develops and air from the surroundings at 1
balandron [24]

Answer:

The answer is "143.74^{\circ} \ C , 8.36\ g, and \ 2.77\ \frac{K}{J}"

Explanation:

For point a:

Energy balance equation:

\frac{dU}{dt}= Q-Wm_ih_i-m_eh_e\\\\

W=0\\\\Q=0\\\\m_e=0

From the above equation:

\frac{dU}{dt}=0-0+m_ih_i-0\\\\\Delta U=\int^{2}_{1}m_ih_idt\\\\

because the rate of air entering the tank that is h_i constant.

\Delta U = h_i \int^{2}_{1} m_i dt \\\\= h_i(m_2 -m_1)\\\\m_2u_2-m_1u_2=h_i(M_2-m_1)\\\\

Since the tank was initially empty and the inlet is constant hence, m_2u-0=h_1(m_2-0)\\\\m_2u_2=h_1m_2\\\\u_2=h_1\\\\

Interpolate the enthalpy between T = 300 \ K \ and\ T=295\ K. The surrounding air  

temperature:

T_1= 25^{\circ}\ C\ (298.15 \ K)\\\\\frac{h_{300 \ K}-h_{295\ K}}{300-295}= \frac{h_{300 \ K}-h_{1}}{300-295.15}

Substituting the value from ideal gas:

\frac{300.19-295.17}{300-295}=\frac{300.19-h_{i}}{300-298.15}\\\\h_i= 298.332 \ \frac{kJ}{kg}\\\\Now,\\\\h_i=u_2\\\\u_2=h_i=298.33\ \frac{kJ}{kg}

Follow the ideal gas table.

The u_2= 298.33\ \frac{kJ}{kg} and between temperature T =410 \ K \ and\  T=240\ K.

Interpolate

\frac{420-410}{u_{240\ k} -u_{410\ k}}=\frac{420-T_2}{u_{420 k}-u_2}

Substitute values from the table.

 \frac{420-410}{300.69-293.43}=\frac{420-T_2}{{u_{420 k}-u_2}}\\\\T_2=416.74\ K\\\\=143.74^{\circ} \ C\\\\

For point b:

Consider the ideal gas equation.  therefore, p is pressure, V is the volume, m is mass of gas. \bar{R} \ is\  \frac{R}{M} (M is the molar mass of the  gas that is 28.97 \ \frac{kg}{mol} and R is gas constant), and T is the temperature.

n=\frac{pV}{TR}\\\\

=\frac{(1.01 \times 10^5 \ Pa) \times (10\ L) (\frac{10^{-3} \ m^3}{1\ L})}{(416.74 K) (\frac{8.314 \frac{J}{mol.k} }{2897\ \frac{kg}{mol})}}\\\\=8.36\ g\\\\

For point c:

 Entropy is given by the following formula:

\Delta S = mC_v \In \frac{T_2}{T_1}\\\\=0.00836 \ kg \times 1.005 \times 10^{3} \In (\frac{416.74\ K}{298.15\ K})\\\\=2.77 \ \frac{J}{K}

5 0
2 years ago
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