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Alexxandr [17]
3 years ago
7

A house has well-insulated walls. It contains a volume of 105 m3 of air at 305 K.

Physics
1 answer:
REY [17]3 years ago
6 0

Answer: 85.46\ kJ

Explanation:

Given

Volume of air V=105\ m^3

Temperature of air T=305\ K

Increase in temperature \Delta T=0.7^{\circ}C

Specific heat for diatomic gas is C_p=\dfrac{7R}{2}

Energy required to increase the temperature is

\Rightarrow Q=nC_pdT\\\\\Rightarrow Q=n\times \dfrac{7R}{2}\times \Delta T\\\\\Rightarrow Q=\dfrac{7}{2}nR\Delta T\\\\\Rightarrow Q=\dfrac{7}{2}\times \dfrac{PV}{T}\times \Delta T\quad [\text{using PV=nRT}]

Insert the values

\Rightarrow Q=\dfrac{7}{2}\times \dfrac{1.01325\times 10^5\times 105}{305}\times 0.7\\ \text{Assuming air pressure to be atmospheric P=}1.01325\times 10^5\ N/m^2\\\\\Rightarrow Q=0.8546\times 10^5\\\Rightarrow Q=85.46\ kJ

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List down all the jovian planets in order of increasing distance from the sun
Olenka [21]

Answer:

There are total eight planets in the solar system and the average distance from the sun to each planet in increasing order is given below.

Explanation:

The average distance from the sun is listed below in increasing order.

1. Mercury  - It is the most closet planet to Sun, 57 million km

2. Venus  - 108 million km

3. Earth  - 150 million km

4. Mars  - 228 million km

5. Jupiter  - 779 million km

6. Saturn  - 1.43 billion km

7. Uranus  - 2.88 billion km

8. Neptun - It is the most farthest from the Sun, 4.50 billion km

8 0
3 years ago
A life preserver is thrown from an helicopter straight down to a person in distress. The initial velocity of the life preserver
Leno4ka [110]

Answer:29.627 m

Explanation:

Given

Initial velocity of life preserver(u) is 1.6 m/s

it takes 2.3 s to reach the water

using equation of motion

v=u+at

v=1.6+9.81\times 2.3

v=24.163 m/s

Let s be the height of life preserver

v^2-u^2=2gs

24.163^2-1.6^2=2\times 9.81\times s

s=\frac{581.29}{2\times 9.81}

s=29.627 m

6 0
3 years ago
Sphere A with mass 80 kg is located at the origin of an xy coordinate system; sphere B with mass 60 kg is located at coordinates
IRINA_888 [86]

Answer:

Fc = [ - 4.45 * 10^-8 j ] N  

Explanation:

Given:-

- The masses and the position coordinates from ( 0 , 0 ) are:

       Sphere A : ma = 80 kg , ( 0 , 0 )

       Sphere B : ma = 60 kg , ( 0.25 , 0 )

       Sphere C : ma = 0.2 kg , ra = 0.2 m , rb = 0.15

- The gravitational constant G = 6.674×10−11 m3⋅kg−1⋅s−2

Find:-

what is the gravitational force on C due to A and B?

Solution:-

- The gravitational force between spheres is given by:

                       F = G*m1*m2 / r^2

Where, r : The distance between two bodies (sphere).

- The vector (rac and rbc) denote the position of sphere C from spheres A and B:-

 Determine the angle (α) between vectors rac and rab using cosine rule:

                   cos ( \alpha ) = \frac{rab^2 + rac^2 - rbc^2}{2*rab*rac} \\\\cos ( \alpha ) = \frac{0.25^2 + 0.2^2 - 0.15^2}{2*0.25*0.2}\\\\cos ( \alpha ) = 0.8\\\\\alpha = 36.87^{\circ \:}

 Determine the angle (β) between vectors rbc and rab using cosine rule:

                   cos ( \beta  ) = \frac{rab^2 + rbc^2 - rac^2}{2*rab*rbc} \\\\cos ( \beta  ) = \frac{0.25^2 + 0.15^2 - 0.2^2}{2*0.25*0.15}\\\\cos ( \beta  ) = 0.6\\\\\beta  = 53.13^{\circ \:}

- Now determine the scalar gravitational forces due to sphere A and B on C:

       Between sphere A and C:

                  Fac = G*ma*mc / rac^2

                  Fac = (6.674×10−11)*80*0.2 / 0.2^2  

                  Fac = 2.67*10^-8 N

                  vector Fac = Fac* [ - cos (α) i + - sin (α) j ]

                  vector Fac = 2.67*10^-8* [ - cos (36.87°) i + -sin (36.87°) j ]

                  vector Fac = [ - 2.136 i - 1.602 j ]*10^-8 N

       Between sphere B and C:

                  Fbc = G*mb*mc / rbc^2

                  Fbc = (6.674×10−11)*60*0.2 / 0.15^2  

                  Fbc = 3.56*10^-8 N

                  vector Fbc = Fbc* [ cos (β) i - sin (β) j ]

                  vector Fbc = 3.56*10^-8* [ cos (53.13°) i - sin (53.13°) j ]

                  vector Fbc = [ 2.136 i - 2.848 j ]*10^-8 N

- The Net gravitational force can now be determined from vector additon of Fac and Fbc:

                  Fc = vector Fac + vector Fbc

                  Fc = [ - 2.136 i - 1.602 j ]*10^-8  + [ 2.136 i - 2.848 j ]*10^-8

                  Fc = [ - 4.45 * 10^-8 j ] N  

3 0
3 years ago
A ball is dropped from an upper floor, some unknown distance above your apartment. As you look out of your window, which is 1.50
laila [671]

Answer:

The ball is dropped at a height of 9.71 m above the top of the window.

Explanation:

<u>Given:</u>

  • Height of the window=1.5 m
  • Time taken by ball to cover the window height=0.15

Now using equation of motion in one dimension we have

s=ut+\dfrac{at^2}{2}

Let u be the velocity of the ball when it reaches the top of the window

then

1.5=0.15u+\dfrac{9.8\times0.15^2}{2}\\u=3.96\ \rm m/s\\

Now u is the final velocity of the ball with respect to the top of the building

so let t be the time taken for it to reach the top of the window with this velocity

3.96=gt\\t=0.4\ \rm s\\

Let h be the height above the top of the window

h=\dfrac{gt^2}{2}\\\\h=\dfrac{9.8\times 0.4^2}{2}\\h=9.71\ \rm m

3 0
3 years ago
Where are electrons located in an atom?
Scilla [17]

Answer:

B. Outside the nucleus.

Explanation:

Electrons orbit the nucleus of the atom.

7 0
2 years ago
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