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Alexus [3.1K]
2 years ago
15

Which of the following statements is true? The square root of the molecular weight of a gas is equal to its rate of diffusion. T

he rate of diffusion of a gas in inversely proportional to the square root of its molecular weight. The rate of diffusion of a gas is directly proportional to the square root of its molecular weight. The square root of the molecular weight of a gas is not related to its rate of diffusion.
Physics
2 answers:
forsale [732]2 years ago
8 0
The diffusion coefficient of the gas is proportional to the average rate of thermal motion of the molecules.

the average velocity is inversely proportional to the square root of the molar mass

so
The gas diffusion rate is inversely proportional to the square root of its molecular weight.
padilas [110]2 years ago
4 0

The answer is: The rate of diffusion of a gas in inversely proportional to the square root of its molecular weight.

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At what height h above the ground does the projectile have a speed of 0.5v?
maw [93]

Answer:

h=\dfrac{3v^2}{8g}

Explanation:

It is given that,

Speed of the projectile is 0.5 v. Let h is the height above the ground. Using the first equation of motion to find it.

v=u+at

v=u-gt

Initial speed of the projectile is v and final speed is 0.5 v.

0.5v=v-gt

t=\dfrac{v}{2g}

g is the acceleration due to gravity

Let h is the height above the ground. Using the second equation of motion as :

h=vt-\dfrac{1}{2}gt^2

h=v\dfrac{v}{2g}-\dfrac{1}{2}g(\dfrac{v}{2g})^2

h=\dfrac{3v^2}{8g}

So, the height of the projectile above the ground is \dfrac{3v^2}{8g}. Hence, this is the required solution.

6 0
3 years ago
Suppose the ring rotates once every 4.30 ss . If a rider's mass is 58.0 kgkg , with how much force does the ring push on her at
Stells [14]

Answer:

422.36 N

Explanation:

given,

time of rotation = 4.30 s

T = 4.30 s

Assuming the diameter of the ring equal to 16 m

radius, R = 8 m

v = \dfrac{2\pi R}{T}

v = \dfrac{2\pi\times 8}{4.30}

  v = 11.69 m/s

now, Force does the ring push on her at the top

- N - m g = \dfrac{-mv^2}{R}

N + m g = \dfrac{mv^2}{R}

N = \dfrac{mv^2}{R}- m g

N = m(\dfrac{v^2}{R}- g)

N = 58\times (\dfrac{11.69^2}{8}- 9.8)

N = 422.36 N

The force exerted by the ring to push her is equal to 422.36 N.

6 0
2 years ago
A 23 kg body is moving through space in the positive direction of an x axis with a speed of 130 m/s when, due to an internal exp
babymother [125]

Answer:

a) Vx = 1088m/s

b) Vy = -162.93m/s

c) 5246745J

Explanation:

Mass of unbroken body = 23kg

Its velocity along +ve X-axis = 130m/s

Mass of first broken body, m1= 9.4kg

Its velocity along +ve X-axis = 130m/s

Nass of 2nd broken body, m2 = 6.1kg

Its velocity long-lived X - axis = -550m/s

Mass of 3rd broken body = ?

m3 = (23 - 9.4 - 6.1)kg

m3 = 7.5kg

Let velocity along the x-axis = Vx

Let the velocity along the x-axis = Vy

Applying law of conservation of momentum along x-axis

a) m1×0 + m2×(-550) + m3×(Vx) =M × 130

9.4 × 0 + 6.1× (-550) + 7.5(Vx) = 23 ×130

0 + (-5170) + 7.5Vx = 2990

2990 + 5170 = 7.5Vx

8160 = 7.5Vx

Vx = 8160/7.5

Vx = 1088m/s

b) Aplying conservation of momentum along the x-axis

(m1×130) + (m2 × 0) + (m3× Vy) = 0

(9.4 × 130) + (6.1 ×550) + 7.5Vy = 0

1222 + 0 + 7.5Vy = 0

1222 = -7.5Vy

Vy = 1222/(-7.5)

Vy = -262.93m/s

c) The energy released or change in KE is given by:

1/2[(m1v1^2) + (m2v2^2) +(m3Vx^2) ]= MV^2

Change in KE = 1/2[ 9.4× 130^2 + 6.1 × 550^2 + 7.5 × 1088^2 ] - 1/2(23 × 130^2)

Change in KE = 1/2[158860 + 1845250 + 8878080] - 1/2[388700]

Change in KE = 5441095 - 194350

Change in KE = 5246745J

4 0
2 years ago
For a fixed amount of gas at a fixed temperature, what will happen if the volume is doubled?.
mash [69]

Answer:

the pressure will decrease by 1/2

Explanation:

PV=nRT

P=(nRT)/(V)

nRT are all constant so they will equal 1

V is 2

P=1/2

4 0
2 years ago
Which objects are opaque? Check all that apply.
Vinvika [58]
<span>textbook
track shoes
</span><span>basketball</span>
7 0
2 years ago
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