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ivolga24 [154]
3 years ago
15

The hydrogen stored inside a large weather balloon has a mass of 13.558 g. What is the volume of this balloon if the density of

hydrogen is 0.089 g/L?
Physics
2 answers:
sergij07 [2.7K]3 years ago
8 0

Answer:

152.337 L

Explanation:

To calculate density using volume we use the common equation

D = \frac{m}{V} \\

where

D is the density

m is the mass

V is the volume

This equation can be manipulated to work out volume to:

V = \frac{m}{D} \\\\V = \frac{13.558g}{0.089g/L} \\\\V= 152.337 L

sergejj [24]3 years ago
6 0

       (13.558 gm) · (1 L / 0.089 gm)  =  152.34 L  (rounded)
                                        
 (fraction equal to ' 1 ')  ^
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A sound wave travels through a column of hydrogen at STP. Assuming a density of rho = 0.0900 kg/m3 and a bulk modulus of β = 1.4
STALIN [3.7K]

Answer:

speed of sound will be 1256 m/sec

Explanation:

We have given density \rho =0.09kg/m^3

And bulk modulus \beta =1.42\times 10^5Pa

We know that speed of sound in a medium is given by

v=\sqrt{\frac{\beta }{\rho }}, here \beta is bulk modulus and \rho is density

So speed will be v=\sqrt{\frac{1.42\times 10^5}{0.09 }}=12.56\times 10^2=1256m/sec

So speed of sound will be 1256 m/sec

4 0
3 years ago
In Fig. 21-25, particle 1 of charge &1.0 mC and particle 2 of charge $3.0 mC are held at separation L ! 10.0 cm on an x axis
Kaylis [27]

Answer:

Since the particle 1 and 2 are on the x-axis, the 3rd particle should also be on the x-axis in order the net force on it to be zero.

Let's denote the distance between particles 1 and 3 as x. Therefore the distance between particles 2 and 3 is (0.1 - x), since the distance between 1 and 2 is 0.1 m.

Coulomb's Law states the force between charges as

F_{1-3} = \frac{1}{4\pi \epsilon_0}\frac{q_1q_3}{x^2}

F_{2-3} = \frac{1}{4\pi \epsilon_0}\frac{q_2 q_3}{(0.1 -x)^2}

The question asks that F_{1-3} = F_{2-3}, so

\frac{1}{4\pi \epsilon_0}\frac{1\times 10^{-3} \times q_3}{x^2} = \frac{1}{4\pi \epsilon_0}\frac{3\times 10^{-3}\times q_3}{(0.1 - x)^2}\\\frac{1}{x^2} = \frac{3}{(0.1 - x)^2}\\3x^2 = 0.01 -0.2x + x^2\\2x^2 + 0.2x - 0.01 = 0\\x_1 = 0.036m\\x_2 = -0.136m

We will take the positive root:

x = 0.036~m away from the first particle.

Explanation:

Since Fig. 21-25 is not given in the question, the exact locations are not known. However, the location of the third particle is found to be 0.036 m away from the first particle and the third particle is located between the particles 1 and 2.

8 0
3 years ago
Consider the collision of a 1500-kg car traveling east at 20.0 m/s (44.7 mph) with a 2000-kg truck traveling north at 25 m/s (55
masha68 [24]

Answer:

1. X_{cm}=-8.57m

2. Y_{cm}=-14.29m

3. V_{cm} = 16.66m/s

4. α = 59.05°

5. V_{cm} = 16.66m/s

6. α = 59.05°

Explanation:

The position of the center of mass 1s before the collision is:

X_{cm}=\frac{X_c*mc+X_t*m_t}{m_c+m_t}

where

X_c=-20m;  m_c=1500kg;  

X_t=0m;   m_t=2000kg;

Replacing these values:

X_{cm}=-8.57m

Y_{cm}=\frac{Y_c*mc+Y_t*m_t}{m_c+m_t}

where

Y_c=0m;  m_c=1500kg;  

Y_t=-25m;   m_t=2000kg;

Replacing these values:

Y_{cm}=-14.29m

The velocity of their center of mass is:

V_{cm-x}=\frac{V_{c-x}*mc+V_{t-x}*m_t}{m_c+m_t}

where

V_{c-x}=20m/s;  m_c=1500kg;  

V_{t-x}=0m/s;   m_t=2000kg;

Replacing these values:

V_{cm-x}=8.57m/s

V_{cm-y}=\frac{V_{c-y}*mc+V_{t-y}*m_t}{m_c+m_t}

where

V_{c-y}=0m;  m_c=1500kg;  

V_{t-y}=-25m;   m_t=2000kg;

Replacing these values:

V_{cm-y}=-14.29m

So, the magnitude of the velocity is:

V_{cm}=\sqrt{V_{cm-x}^2+V_{cm-y}^2}

V_{cm}=16.66m/s

The angle of the velocity is:

\alpha =atan(V_{cm-y}/V_{cm-x})

\alpha=59.05\°

Since on any collision, the velocity of the center of mass is preserved, then the velocity after the collision is the same as the previously calculated value of 16.66m/s at 59.0° due north of east

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3 years ago
Hii! help asap. i’ll give brainliest thanks!
telo118 [61]
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7 0
2 years ago
A ball falling through the air has what?
AysviL [449]
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Kinetic Energy - The further the ball fall the more KE it has...until terminal velocity is reach, then KE would become constant. 
Potential Energy - Conversely to that of KE, the further the ball falls the less PE it will have. 

<em>Heat/Thermal Energy is technically also present due to the friction from the air resistance, but the transfer of energy between the air and ball is quite complex and not necessary important for basic physics.  
</em>
The question itself seem kind of vague and open ended, but I could just be viewing it the wrong way. 
Comment if you need more help! 
8 0
3 years ago
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