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dusya [7]
3 years ago
12

A large room in a house holds 947 kg of dry air at 33.2°C. A woman opens a window briefly and a cool breeze brings in an additio

nal 62.4 kg of dry air at 13.2°C. At what temperature (in degrees Celsius) will the two air masses come into thermal equilibrium, assuming they form a closed system? (The specific heat of dry air is 1006 J/ kg * °C, although that value will cancel out of the calorimetry equation.)
°C
Physics
1 answer:
BartSMP [9]3 years ago
7 0

Answer:

31.96362 °C

Explanation:

m_1 = Mass of air in the room = 947 kg

m_2 = Mass of air entering the room = 62.4 kg

T_1 = Temperature in the room = 33.2°C

T_2 = Temperature air entering the room = 13.2°C

T = Equilibrium temperature

c = Specific heat of air = 1006 J/kg °C

In the case of thermal equilibrium we have the relation

m_1c(T_1-T)=m_2c(T-T_2)\\\Rightarrow T=\frac{m_2T_2+m_1T_1}{m_1+m_2}\\\Rightarrow T=\frac{62.4(13.2+273.15)+947(33.2+273.15)}{947+62.4}\\\Rightarrow T=305.11362\ K=305.11362-273.15=31.96362^{\circ}C

The temperature of thermal equilibrium is 31.96362 °C

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3 years ago
g How many diffraction maxima are contained in a region of the Fraunhofer single-slit pattern, subtending an angle of 2.12°, for
lbvjy [14]

Answer:

2

Explanation:

We know that in the Fraunhofer single-slit pattern,

maxima is given by

a\text{sin}\theta=\frac{2N+1}{2}\lambda

Given values

θ=2.12°

slit width a= 0.110 mm.

wavelength λ= 582 nm

Now plugging values to calculate N we get

0.110\times10^{-3}\text{sin}2.12=(\frac{2N+1}{2})582\times10^{-9}

Solving the above equation we get

we N= 2.313≅ 2

4 0
3 years ago
As shown in the diagram, two forces act on an object. The forces have magnitudes F1 = 5.7 N and F2 = 1.9 N. What third force wil
galina1969 [7]

Answer:

Second option 6.3 N at 162° counterclockwise from  

F1->

Explanation:

Observe the attached image. We must calculate the sum of all the forces in the direction x and in the direction y and equal the sum of the forces to 0.

For the address x we have:

-F_3sin(b) + F_1 = 0

For the address and we have:

-F_3cos(b) + F_2 = 0

The forces F_1 and F_2 are known

F_1 = 5.7\ N\\\\F_2 = 1.9\ N

We have 2 unknowns (F_3 and b) and we have 2 equations.

Now we clear F_3 from the second equation and introduce it into the first equation.

F_3 = \frac{F_2}{cos (b)}

Then

-\frac{F_2}{cos (b)}sin(b)+F_1 = 0\\\\F_1 = \frac{F_2}{cos (b)}sin(b)\\\\F_1 = F_2tan(b)\\\\tan(b) = \frac{F_1}{F_2}\\\\tan(b) = \frac{5.7}{1.9}\\\\tan^{-1}(\frac{5.7}{1.9}) = b\\\\b= 72\°\\\\m = b +90\\\\\m= 162\°

Then we find the value of F_3

F_3 = \frac{F_1}{sin(b)}\\\\F_3 =\frac{5.7}{sin(72\°)}\\\\F_3 = 6.01 N

Finally the answer is 6.3 N at 162° counterclockwise from  

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7 0
3 years ago
A jet accelerates from rest down a runway at 1.75m/s² for a distance of 1500 m before takeoff.
babunello [35]
A. Using the third equation of motion:
v2 = u2 + 2as
from the question;
the jet was initially at rest
hence u = 0
a = 1.75m/s2
s = 1500m
v2 = 02 + 2(1.75)(1500)
v2 = 5250
v = √5250
v = 72.46m/s
hence it moves with a velocity of 72.46m/s.
b. s = ut + 1/2at2
1500 = 0(t) + 1/2(1.75)t2
1500 × 2 = 2× 1/2(1.75)t2
3000 = 1.75t2
1714.29 = t2
41.4 = t
hence the time taken for the plane to down the runway is 41.4s.


Read more on Brainly.com - brainly.com/question/18743384#readmore
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