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AysviL [449]
3 years ago
9

Heat is transferred at a rate of 2 kW from a hot reservoir at 800 K to a cold reservoir at 300 K. Calculate the rate at which th

e entropy of the two reservoirs changes and determine if the second law is satisfied
Physics
1 answer:
vekshin13 years ago
6 0

Answer:

0.00417 kW/K or 4.17 W/K

Second law is satisfied.

Explanation:

Parameters given:

Rate of heat transfer, Q = 2kW

Temperature of hot reservoir, Th = 800K

Temperature of cold reservoir, Tc = 300K

The rate of entropy change is given as:

ΔS = Q * [(1/Tc) - (1/Th)]

ΔS = 2 * (1/300 - 1/800)

ΔS = 2 * 0.002085

ΔS = 0.00417 kW/K or 4.17 W/K

Since ΔS is greater than 0, te the second law of thermodynamics is satisfied.

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the lever in the figure below accomplishes 5.0 newtons-meters of work raising a 25 newton weight. How far,in meters is the weigh
miv72 [106K]

The simple definition of work is the product of the force applied to the object and the displacement:

W = Fd

W is the work, F is the applied force, and d is the displacement.

Givens:

W = 5.0N-m

F = 25N (the applied force must be at least equal to the weight to lift it)

Plug in the given values and solve for d:

5.0 = 25d

d = 0.20m

3 0
3 years ago
What is the work function of cadmium metal if light with = 9.85×1014hz is necessary to eject electrons?
Alika [10]

We use the Planck’s formula:

E = hv

where,

E = energy, h = planck’s constant = 6.6x10^-34 J s, v = frequency in Hz (Hz = 1 / s)

 

Subsituting the values to find for E:

E = (6.6×10^-34 J s) * 9.85×10^14 / s

E = 6.5x10^-19 J

3 0
3 years ago
A 1.0-kg standard cart collides on a low-friction track with cart A. The standard cart has an initial x component of velocity of
Andru [333]

Answer:

Explanation:

Mathematically, linear momentum is expressed as the product of mass and velocity. Linear momentum conservation law states that a body or system of bodies retains its total momentum unless an external force is applied to the system.

In this case, the system consists of two carts.

At the start, the linear momentum (P) of the system is equal to:

P=1.0kg*0.4m/s=0.4kg*m/s

It's only composed of linear momentum of the standard cart because cart A doesn't have any linear momentum at that moment.

After the collision, linear momentum has to be the same

P=0.4kg*m/s=1.0kg*0.20m/s+m_{A} *0.70m/s

where m_A is the mass of the cart A.

Solving for m_A

m_{A} =0.28kg

After the cart A rebounds, the linea momentum of the system has changed (because of the force present in the rebound). The new linear momentum is:

P=1kg*0.2m/s+0.29kg*(-0.7m/s)=-0.003kg*m/s

Then, the lump of putty is added to the system, but the linear momentum has to be the same, because we added a mass, not a force. The mass of that putty (m_p) has to be added to the equation of the system

-0.003kg*m/s=1.0kg*(-0.2m/s)+(0.29kg+m_{p} )(0.4m/s)

Solving for m_p

m_{p}=0.20kg

6 0
3 years ago
The standard wave format for any wave is a (transverse a longitudinal an electromagnetic) wave. When depicting a (transverse a l
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<span>Very wild guess at this one, since I've no idea what "standard wave format" means. First gap would be TRANSVERSE. Second gap could be LONGITUDINAL, or SOUND. Third gap could be ELECTROMAGNETIC (the two possible waves being the ELECTRIC wave and the MAGNETIC wave). http://perendis.webs.com</span>
7 0
3 years ago
Read 2 more answers
A child pushes horizontally on a box of mass m which moves with constant speed v across a horizontal floor. The coefficient of f
dexar [7]

Answer:

Explanation:

Given

mass of box is m

speed of box is v

coefficient of friction \mu =\frac{1}{4}=0.25

Box is moving at a constant speed i.e. Net Force on box is zero

i.e. Frictional Force and Force Applied are Equal

F=f_r=\mu mg

The Rate at which power is generated is

P=F\cdot v

P=\mu mg\cdot v

P=0.25mgv

4 0
3 years ago
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