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saw5 [17]
3 years ago
6

There is a refrigerator running in a room, the heat flowing into the refrigerator from the outside is 40 J/s, and the refrigerat

or must consume 100 W to maintain the low temperature set in it. (1) What is the heat discharged from the refrigerator at this time?
(2) If the doors and windows of the room are closed, the temperature of the room rises by a few degrees per hour? Assume that the air heat capacity of the entire room is 400 kJ/K.
Physics
1 answer:
Mamont248 [21]3 years ago
8 0

Answer:

(A) 140 j/sec (b) 1.26 K

Explanation:

We have given the heat heat flowing into the refrigerator = 40 J/sec

Work done = 40 W

(a) So the heat discharged from the refrigerator =heat\ flowing\ in\ refrigerator+work\ done=40+100=140j/sec

(b) Total heat absorbed =140 j/sec =140\times 3600=504000j/hour

Let the temperature be \Delta T

Heat absorbed per hour =504000 [tex]=400\times 10^3\times \Delta T

So  \Delta T=\frac{504000}{400000}=1.26K

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Answer:

protected under students first amendment rights

Explanation:

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4 0
3 years ago
a car is moving 8.80 m/s when it begins to accelerate at 2.45 m/s^2. how much time does it take to trav 138m. please help me (':
Ulleksa [173]

Answer:

7.6 s

Explanation:

Considering kinematics formula for final velocity as

v^{2}=u^{2}+2as

Where v and u are final and initial velocities, a is acceleration and s is distance moved.

Making v the subject then

v=\sqrt{u^{2}+2as}

Substituting 8.8 m/s for u, 138 m for s and 2.45 m/s2 for a then

v=\sqrt{8.8^{2}+2*2.45*138}\\v=27.45 m/s

Also, v=u+at and making t the subject of the formula

t=\frac {v-u}{a}

Substituting 27.45 m/s for v, 8.8 m/s for u and 2.45 m/s for a then

t=\frac {27.45-8.8}{2.45}=7.6122448979591\approx 7.6s

Therefore, it needs 7.6 seconds to travel

7 0
3 years ago
What are all the<br> invertebrates with a<br> large foot
lyudmila [28]

Answer:

Explanation:

Bobbitt worm ( Eunice aphroditois). This segmented polychaete marine worm can attain lengths of 10 feet. It bristles...

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5 0
3 years ago
A projectile is launched with an initial velocity of (40 m/s), at an angle of (30°) above
zlopas [31]

Answer:

330.5  m

Explanation:

In this case, the object is launched horizontally at 30° with an initial velocity of 40 m/s .

The maximum height will be calculated as;

h=\frac{v^2_isin^2\alpha }{2g}

where ∝ is the angle of launch = 30°

vi= initial launch velocity = 40 m/s

g= 10 m/s²

h= 40²*sin²40° / 2*10

h={1600*0.4132 }/ 20

h= 661.1/2 = 330.5  m

8 0
3 years ago
Can you label the parts of the eye?
photoshop1234 [79]

Answer:

  1. Iris
  2. Pupil
  3. Corona
  4. Anterior chamber
  5. lens
  6. Vitreous humor
  7. Blood vessels
  8. Optic nerve
  9. Hyaloid canal
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Hope it helps :)

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3 years ago
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