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8090 [49]
3 years ago
5

The room contains a refrigerator, operated by an electric motor. The motor does work at the rate of 438 W when it is running. Th

e refrigerator removes heat from the food storage space at a rate of 450 W when the motor is running. In an effort to cool the room, you open the refrigerator door and let the motor run continuously. At what net rate is heat added to (+) or subtracted from (−) the room and all of its contents?
Physics
1 answer:
yKpoI14uk [10]3 years ago
8 0

Answer: Heat is removed from the room at 12W

Explanation:

Energy (heat) is generated by the the rotor at a rate

∆Q/t = 438W

Work is done (heat removal) by the refrigerator at a rate

∆W/t = 450W

Net rate of heat removed =

∆Q/t - ∆W/t

= (438 - 450)W

= -12W

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In the problem below, what is the student showing?
Phantasy [73]

Answer:

S<em>tudent showing the volume of cube</em>

                      V = a³

                       if a = 3 cm

  Then volume is a³ = 3 ×3 × 3

                                 = 27 cm³

6 0
3 years ago
Hi guys i need truly help!
lutik1710 [3]

Answer:

1) 327

2) 3

3) 109

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4) 1074

5) 1074 + 109t

6) 2382

Explanation:

7 0
3 years ago
In the movie "gravity" - Why did the escape pod get stuck and pulled back?
victus00 [196]

Answer:

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Explanation:

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2 years ago
Read 2 more answers
The spectral distribution of the radiation emitted by a diffuse surface may be approximated as follows. HW 2 Q2 (a) What is the
Ronch [10]

Answer:

a)<em> 2000 W/m²  </em><em>; </em>b) 636.94 W/m<em>².sr ; </em><em>c) </em>0.5

Explanation:

a)

The formula for calculation of total emissive power is:

Total emissive power = E = \int\limits^\alpha_0 E'<em>λdλ</em>

<em>                                    </em>= \int\limits^a_0(0)d<em>λ + </em>\int\limits^b_a(100)d<em>λ + </em>\int\limits^c_b(200)d<em>λ + </em>\int\limits^d_c(100)d<em>λ </em>\int\limits^e_d(0)d<em>λ</em>

<em>where a = 5; b = 10; c = 15; d = 20; e = 25</em>

<em>                                    = 0 +100(10-5) + 200(15-10) +100(20-15) + 0</em>

<em>                                    = 2000 W/m²</em>

b)

The formula for total intensity of radiation is:

I_{e} = E/π = 200/3.14 = 636.94 W/m<em>².sr  </em>

<em>c)</em>

Fo submissive power leaving the surface in range π/4 ≤θ≤π/2

[E(π/4 ≤θ≤π/2)]/E = \int\limits^f_0\int\limits^g_0\int\limits^i_h Icosθsinθ dθdΦdλ

where f = infinity, g=2π, h=π/4, i=π/2

By simplifying, we get

                           = (-1/2)[cos(2π/2)-cos(2π/2)]

                           = -0.5(-1-0)

                           =0.5

8 0
3 years ago
In a Broadway performance, an 77.0-kg actor swings from a R = 3.65-m-long cable that is horizontal when he starts. At the bottom
krek1111 [17]

Answer: h =1.22 m

Explanation:

from the question we were given the following

mass of performer ( M1 ) = 77 kg

length of cable ( R ) = 3.65 m

mass of costar ( M2 ) = 55 kg

maximum height (h) = ?

acceleration due to gravity (g) = 9.8 m/s^2  (constant value)

We first have to find the velocity of the performer. From the work energy theorem work done = change in kinetic energy

work done = 1/2 x mass x ( (final velocity)^2 - (initial velocity)^2 )

initial velocity is zero in this case because the performer was at rest before swinging, therefore

work done = 1/2 x 77 x ( v^2 - 0)

work done = 38.5 x ( v^2 ) ......equation 1

work done is also equal to m x g x distance ( the distance in this case is the length of the rope), hence equating the two equations we have

m x g x R =  38.5 x ( v^2 )

77 x 9.8 x 3.65 =  38.5 x ( v^2 )

2754.29 = 38.5 x ( v^2 )

( v^2 ) =  71.54

v = 8.4 m/s  ( velocity of the performer)

After swinging, the performer picks up his costar and they move together, therefore we can apply the conservation of momentum formula which is

initial momentum of performer (P1) + initial momentum of costar (P2) = final momentum of costar and performer after pick up (Pf)  

momentum = mass x velocity therefore the equation above now becomes

(77 x 8.4) + (55 x 0) = (77 +55) x Vf  

take note the the initial velocity of the costar is 0 before pick up because he is at rest

651.3 = 132 x Vf

Vf = 4.9 m/s

the performer and his costar is 4.9 m/s after pickup

to finally get their height we can use the energy conservation equation for from after pickup to their maximum height. Take note that their velocity at maximum height is 0

initial Kinetic energy + Initial potential energy = Final potential energy + Final Kinetic energy

where

kinetic energy = 1/2 x m x v^2

potential energy  = m x g x h

after pickup they both will have kinetic energy and no potential energy, while at maximum height they will have potential energy and no kinetic energy. Therefore the equation now becomes

initial kinetic energy = final potential energy

(1/2 x (55 + 77) x 4.9^2) + 0 = ( (55 + 77) x 9.8 x h) + 0

1584.7 = 1293 x h

h =1.22 m

3 0
3 years ago
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