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VARVARA [1.3K]
3 years ago
7

3.113 A heat pump is under consideration for heating a research station located on Antarctica ice shelf. The interior of the sta

tion is to be kept at 15℃. Determine the maximum theoretical rate of heating provided by a heat pump, in ????W p???????? ????W of power input in each of two cases: The role of the cold reservoir is played by (a) the atmosphere at −20℃, (b) ocean water at 5℃.
Physics
1 answer:
Effectus [21]3 years ago
6 0

Answer:

a. β = 8.23 K

b. β = 28.815 K

Explanation:

Heat pump can be find using the equation

β = TH / TH - TC

a.

TH = 15 ° C + 273.15 K = 288.15 K

TC = - 20 ° C + 273.15 K = 253.15 K

β = 288.15 K / 288.15 k - ( 253.15 K )

β = 8.23 K

b.

TH = 15 ° C + 273.15 K = 288.15 K

TC = 5 ° C + 273.15 K = 278.15 K

β = 288.15 K / 288.15 k - ( 278.15 K )

β = 28.815 K

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irina [24]

Answer: 14. 49 m

Explanation:

We can solve this problem with the following equations:

x=V_{o} cos \theta t (1)

y-y_{o}=V_{o} sin \theta t-\frac{1}{2}gt^{2} (2)

Where:

x is the horizontal distance between the cannon and the ball

V_{o}=23 m/s is the cannonball initial velocity

\theta=0\° since the cannonball was shoot horizontally

t is the time

y=0 is the final height of the cannonball

y_{o}=1.96 m is the initial height of the cannonball

g=9.8 m/s^{2} is the acceleration due gravity

Isolating t from (2):

t=\sqrt{-\frac{2(y-y_{o})}{g}} (3)

t=\sqrt{-\frac{2(0 m-1.96 m)}{9.8 m/s^{2}}} (4)

t=0.63 s (5)

Substituting (5) in (1):

x=(23 m/s) cos(0\°) 0.63 s (6)

Finally:

x=14.49 m

5 0
2 years ago
What is the energy of a photon whose frequency is 6.0 x 10^20?
omeli [17]

Answer:

3.75 MeV

Explanation:

The energy of the photon can be given in terms of frequency as:

E = h * f

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The energy (in Joules) is:

E = 6.63 x10^(-34) * 6 * 10^(20)

E = 39.78 * 10^(-14) J = 3.978 * 10^(-13) J

We are given that:

1 eV = 1.06 * 10^(-19) Joules

This means that 1 Joule will be:

1 J = 1 / (1.06 * 10^(-19)

1 J = 9.434 * 10^(18) eV

=> 3.978 * 10^(-13) J = 3.978 * 10^(-13) * 9.434 * 10^(18) = 3.75 * 10^(6) eV

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