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1 year ago

How does energy from the sun affect the motion of molecules in a gas compared to molecules in a liquid?

Physics

2 answers:

Nookie1986 [14]1 year ago

6 0

<h2>Answer:</h2>

Correct option is A.

a. Molecules in a gas move faster than in a liquid.

<h2>Explanation:</h2>

Energy from the sun affect the motion of molecules in a gas compared to molecules in a liquid such as that molecules in a gas move faster than in a liquid. So, first option is correct.

AnnZ [28]1 year ago

3 0

Molecules in a gas move faster than in a liquid.

hope it helps.

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A sailboat starts from rest and accelerates at a rate of 0.21 m/s^2 over a distance of 280 m. find the magnitude of the boat's f

sasho [114]

We use the kinematic equations,

$v=u+at$ (A)

$S= ut + \frac{1}{2} at^2$ (B)

Here, u is initial velocity, v is final velocity, a is acceleration and t is time.

Given, $u=0$ , $a=0.21 \ m/s^2$ and $s= 280 m$ .

Substituting these values in equation (B), we get

$280 \ m = 0 +\frac{1}{2} (0.21 m/s^2) t^2 \\\\ t^2 = \frac{280 \times 2}{0.21 } \\\\ t= 51.63 \ s$ .

Therefore from equation (A),

$v = 0 + (0.21) \times (51.63 s)= 10.84 \ m/s$

Thus, the magnitude of the boat's final velocity is 10.84 m/s and the time taken by boat to travel the distance 280 m is 51.63 s

8 0

2 years ago

calculate the time rate of change in air density during expiration. Assume that the lung has a total volume of 6000mL, the diame

kipiarov [429]

Answer:

The time rate of change in air density during expiration is 0.01003kg/m³-s

Explanation:

Given that,

Lung total capacity V = 6000mL = 6 × 10⁻³m³

Air density p = 1.225kg/m³

diameter of the trachea is 18mm = 0.018m

Velocity v = 20cm/s = 0.20m/s

dv /dt = -100mL/s (volume rate decrease)

= 10⁻⁴m³/s

Area for trachea =

$\frac{\pi }{4} d^2\\= 0.785\times 0.018^2\\= 2.5434 \times10^-^4m^2$

0 - p × Area for trachea =

$\frac{d}{dt} (pv)=v\frac{ds}{dt} + p\frac{dv}{dt}$

$-1.225\times2.5434\times10^-^4\times0.20=6\times10^-^3\frac{ds}{dt} +1.225(-1\times10^-^4)$

⇒ $-0.623133\times10^-^4+1.225\times10^-^4=6\times10^-^3\frac{ds}{dt}$

$\frac{ds}{dt} = \frac{0.6018\times10^-^4}{6\times10^-^3} \\\\= 0.01003kg/m^3-s$

ds/dt = 0.01003kg/m³-s

Thus, the time rate of change in air density during expiration is 0.01003kg/m³-s

3 0

2 years ago

Read 2 more answers

How much heat Q2Q2 is transferred to the skin by 25.0 gg of steam onto the skin? The heat of vaporization for steam is L=2.256×1

astraxan [27]

Answer:

56400Joules

Explanation:

The quantity of heat required is expressed as;

Q = mL

m is the mass = 25g = 0.025kg

L is the latent heat of vaporization for steam = 2.256×10^6J/kg

Substitute into the formula as shown;

Q = 0.025×2.256×10^6

Q = 56400Joules

Hence the quantity of hear required is 56400Joules

3 0

2 years ago

8. Find the momentum of a photon in eV/c and in Kg. m/s if the wavelength is (a) 400nm ; (b) 1 Å = 0.1 nm, (c) 3 cm ; and (d) 2

nataly862011 [7]

We use the formula: p = E/c where E = hc / λ. hence, p = h/ λ. where h is the Planck's constant: 6.62607004 × 10-34 m2 kg / s and <span>λ is the wavelenght.
</span>
a) p = <span>6.62607004 × 10-34 m2 kg / s / 0.1 x10^-9 m = 6.62607 x 10-24 m kg/s
</span>b) p = 6.62607004 × 10-34 m2 kg / s / 3 x10^-2 m = 2.20869 <span>x 10-32 m kg/s
</span>b) p = 6.62607004 × 10-34 m2 kg / s / 2 x10^-9 m = 3.3130 <span>x 10-25 m kg/s</span>

7 0

2 years ago

Suppose that a sound has initial intensity β1 measured in decibels. This sound now increases in intensity by a factor f. What is

topjm [15]

Answer:

β2= β1+10*f

Explanation:

comparing β2 and β1, it is said that β2 is increased by a factor of f.

for each factor of f, there is a 10*f dB increase.

therefore if the β1 is increases by an intensity of factor f

the new intensity would be β1+ 10*f

4 0

2 years ago