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barxatty [35]
3 years ago
14

) A hot air balloon along with its cargo has a mass of 4.0 x 105 kg, and it holds 7.0 x 10 5 m3 of hot air. It is floating at a

constant height in air with a density of 1.29 kg/m3. What is the density of the hot air in the balloon
Physics
1 answer:
vitfil [10]3 years ago
8 0

Answer:

0.72 kg/m^3

Explanation:FIRST FIND UPTHRUST WHICH SHOULD BE EQUAL TO WEIGHT OF HOT BALLOON. SUBTRACT WEIGHT OF PERSONS TO GET WEIGHT WEIGHT OF HOT AIR . DIVIDE WEIGHT BY 9.8 TO GET MASS OF HOT AIR . SINCE VOLUME IS GIVEN FIND DENSITY . SEE ATTACHMENT

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Read 2 more answers
Exercise 2.4.5: Suppose we add possible friction to Exercise 2.4.4. Further, suppose you do not know the spring constant, but yo
Maru [420]

Answer:

a). C = b/2   and C = b/4

b).  $ \therefore T = 2 \pi \sqrt{\frac{m_1 +m_2}{k (m_1 + m_2)}} = 2 \pi \sqrt{ \mu/k}$

c). m = 63.4 kg (approx.)

Explanation:

Ex. 2.4.4

The total force acting on mass m is $ F = F_{spring }= -kx $ , where x is the displacement from the equilibrium position.

The equation of motion is $ m {\overset{..}x} + kx = 0 $

or $ {\overset{..}x}+ \frac{k}{m}x=0 $     or   $ {\overset{..}x} + w_0^2 x = 0 $ ,    where $ w_0 = \sqrt{\frac{k}{m}} $  

The solution is $ x = A \cos (w_0t + \phi) $ ,  where A and Ф are constants.

A is amplitude of  motion

$ w_0$ is the angular frequency of motion

Ф is the phase angle.

Now, $ w_0 = 2 \pi f_0 = \sqrt{k/m} $

or $ m = \frac{k}{4\pi f_0^2} $

Given $ f_0 = 0.8 Hz , k = 4 N/m $

a).  $ m = \frac{4}{4(3.14)^2(0.8)^2} = 0.158\ kg$

b). $ w_0^2 = k/m $  

or  $ m = k/ w_0^2 = k / (2\pi f_0)^2 = k / 4 \pi^2 f_0^2 $

Ex. 2.4.5

a). Total force acting on the mass m is $F = F_{spring}+f $

                                                                    $ = -kx-bv $

    The equation of motion is $ m {\overset{..}x}= -kx-b{\overset{.}x} $

    or $ w_0 = \sqrt{\frac{k}{m}} $ ,  angular frequency of the undamped oscillation.

   γ = b/2m  is called the damping coefficient (γ=C)

   $ k = m w_0^2 = 4 \pi^2 m f_0^2 $

   for 1 kg weight (= 9.8 N), $ f_0$ = 1.1 Hz

    k = 4 x (3.14)² x (9.8) x 1.1²  = 4.6 x 10² N/m

  For 2 kg weight (= 19.6 N), $ f_0$ = 0.8 Hz

    k = 4 x 9.8596 x 2 x 9.8 x 0.8²  = 5 x $ 10^7$ N/m

   $ \gamma = \frac{b}{2m_1} = \frac{b}{2m_2} $

or $ \gamma = \frac{b}{2 \times 1} = \frac{b}{2 \times 2} $

γ = b/2 (for 1 kg)  and   γ = b/4 (for 2 kg)

C = b/2   and C = b/4

b). $ w_0^2 = \frac{k}{m}  \Rightarrow \frac{k}{w_0^2} = \frac{k}{(2 \pi f_0)^2} = \frac{k}{4 \pi^2 f_0^2} $

 For two particle problem,

   $ w'_0^2 = \sqrt{\frac{k(m_1+m_2)}{m_1 +m_2}} $

$ \therefore T = 2 \pi \sqrt{\frac{m_1 +m_2}{k (m_1 + m_2)}} = 2 \pi \sqrt{ \mu/k}$

where, μ is the reduced mass.

This time period is same for both the particles.

c). $ m =\frac{k}{4 \pi^2 f_0^2}$

        $ = \frac{5 \times 10^2}{4 \times 9.14^2 \times 0.2}  = 63.4\ kg $  ( approx.)

5 0
3 years ago
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