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3 years ago

7

Think about the difference between thermal energy and temperature. How might the melting icicles gain energy without changing te

mperature.
Ok someone pls answer this BY TM.

1 answer:

ivann1987 [24]3 years ago

5 0

No temperature change occurs from heat transfer if ice melts and becomes liquid water (i.e., during a phase change). For example, consider water dripping from icicles melting on a roof warmed by the Sun. Conversely, water freezes in an ice tray cooled by lower-temperature surroundings.

Hope this helped you

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A car of mass 2400kg moving at 20 m/s slams into a cement wall and comes to a halt?

Brut [27]

Impulse = change in momentum

The car's momentum was (mass) x (speed)

Momentum = (2400 kg) x (20 m/s)

Momentum = 48,000 km-m/s

To completely stop the car, the impulse = -48,000 km-m/s .

7 0

3 years ago

Who did not have experimental evidence to support his theory of the atom? Dalton Thomson Rutherford Democritus

Aloiza [94]

Democritus was the one who did not have experimental evidence to support his theory of the atom.

Answer: Option 4

<u>Explanation: </u>

The discovery of atoms were first stated by Democritus but due to the absence of any experimental proof, his statement was not noted as significant at that time.

After this, Dalton made the specific assumptions formulating some postulates for the atomic theory with proof. Then the cathode rays tube experiments performed by Thomson lead to the formation of plum pudding models of atom.

This is followed by Rutherford’s gold foil experiment discovering the presence of nucleus inside the atoms. So, Democritus first stated but due to absence of experimental evidences, his theory of atoms were not supported at that time.

3 0

3 years ago

In one experiment, the students allow the block to oscillate after stretching the spring a distance A. If the potential energy s

atroni [7]

Answer:

K = m g (A - A2)

Explanation:

In a block spring system the total energy is the sum of the potential energy plus the kinetic energy, for maximum elongation all the energy is potential

Em = U₀ = m g A

For when the system is at an ele

Elongation A2 less than A, energy has two parts

Em = K + U₂

K = Em –U₂

We substitute

K = m g A - m gA2

K = m g (A - A2)

3 0

3 years ago

Read 2 more answers

Consider a well-insulated rigid container with two chambers separated by a membrane. The total volume is 5.0 cubic meters. The f

mamaluj [8]

Answer:

The Entropy generated by the steam = 2.821 kJ/K

Explanation:

Total volume of container = 5m³

Heat transfer does not exist between system and surrounding, dQ = 0

At the first chamber, temperature of water at saturated liquid is 300°C

From the steam table:

Specific enthalpy of saturated liquid at 300°C , $h_{f} = 1344.8 kJ/kg$

Specific internal energy of saturated liquid at 300°C, $U_{f1} =$ 1332.7 kJ/kg

For closed system, the first law of thermodynamics state that:

dQ = dw + dU..................(1)

work done for free expansion, dw =0

0 = 0 + dU

dU = 0 , i.e. U₁ = U₂

At the second chamber,

The final pressure, P₂ = 50 kPa

From the steam table, at P₂ = 50 kPa, $U_{f2}$ = 340.49 kJ/kg

$(U_{fg} )_{2} = 2142.7 kJ/kg$

Let the dryness fraction at the second chamber = x

$U_{2} = U_{f2} + U_{fg2}$

$U_{2} = 340.49 + x2140.7$ Since U₁ = U₂

1332.7 = 340.49 + x2140.7

Dryness fraction, x = 0.463

From steam table, the specific volume is, $u_{f2} = 0.00103 m^{3} /kg\\$

$u_{2} = u_{f2} + xu_{fg2}$

$u_{2} = 0.00103 + 0.463(3.2393)\\u_{2} = 1.5 m^{3} /kg\\$

$u_{2} = \frac{v_{2} }{m_{2} }$

V₂ = 5 m³

1.5 = 5/m₂

m₂ = 3.33 kg

At 300°C $S_{1} = S_{f} = 3.2548 kJ/kg-k\\$

$S_{2} = S_{f2} + xS_{fg2}$

From the steam table,

$S_{f2} = 1.0912 kJ/kg-k\\S_{fg2} = 6.5019 kJ/kg-k\\S_{2} = 1.0912 + 0.463(6.5019)\\S_{2} = 4.102 kJ/kg-k$

Therefore the entropy generated will be :

Entropy = mass* (S₂ - S₁)

Entropy = 3.33* (4.102 - 3.2548)

Entropy = 2.821 kJ/K

5 0

3 years ago

Read 2 more answers

A hollow cylinder of mass 2.00 kg, inner radius 0.100 m, and outer radius 0.200 m is free to rotate without friction around a ho

Aneli [31]

Answer:

h=2.86m

Explanation:

In order to give a quick response to this exercise we will use the equations of conservation of kinetic and potential energy, the equation is given by,

$\Delta PE_i + \Delta KE_i = \Delta PE_f +\Delta KE_f$

There is no kinetic energy in the initial state, nor potential energy in the end,

$mgh+0=0+KE_f$

In the final kinetic energy, the energy contributed by the Inertia must be considered, as well,

$mgh = (\frac{1}{2}mv^2+\frac{1}{2}I\omega^2)$

The inertia of the bodies is given by the equation,

$I=\frac{m(R_1^2+R^2_2)}{2}$

$I=\frac{2(0.2^2+0.1^2)}{2}$

$I=0.05Kgm^2$

On the other hand the angular velocity is given by

$\omega =\frac{v}{R_2}=\frac{4}{1/5} = 2rad/s$

Replacing these values in the equation,

$(0.5)(9.8)(h) =\frac{1}{2}*0.5*4^2+\frac{1}{2}*0.05*20^2$

Solving for h,

$h=2.86m$

5 0

3 years ago