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Kruka [31]
3 years ago
5

A turbo molecular pump has rapidly rotating blades which hit the molecules of gas in the coating chamber. This collision transfe

rs some momentum to the particles. This process of momentum transfer is more efficient if the average linear velocity of a molecule is less than the linear velocity of the blade tip. Assuming that a particular pump has a linear blade tip velocity of 400 m/s, (a) calculate the ratio of the average room temperature (say, 27 0 C) molecular velocity to blade tip velocity for the gases such as hydrogen, nitrogen and xenon having molar masses of 2, 28 and 131 g/mol, respectively. (b) Is the turbo molecular pump better for pumping heavy element or light element?
Physics
1 answer:
ycow [4]3 years ago
6 0

Answer:

heavy molecules are pumped much more efficiently than light molecules. Most gases are heavy enough to be well pumped but it is difficult to pump hydrogen and helium efficiently.

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Andreas93 [3]
The kinetic energy of the proton is 3.4 kev
1 kev = 1.602 × 10^-16 joules
therefore 3.4 kev is equivalent to;
3.4 ×  (1.602 ×10^-16)= 5.4468 × 10^-16 J
Kinetic energy is calculated by the formula 1/2mv² where m is the mass and v is the velocity.
Therefore V = √((2 × ( 5.4468×10^-16))/ (1.67 ×10^-27))
                    = 8.077 × 10^5 m/s

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4 years ago
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coldgirl [10]
16, 5 , 3 = 16+5+3= 24 + 3

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4 years ago
A block of mass m1 = 290 g is at rest on a plane that makes an angle θ = 30° above the horizontal. The coefficient of kinetic fr
GrogVix [38]

Answer:

The speed of the block when is fallen 30cm

v=0.726\frac{m}{s}

Explanation:

∑F= (m2)g - ƒ - (m1)g*sin(θ) = (m1)a

g = 9.81 m/s²

ƒ  = μN = μ(m1)g

(m2)*g - u*(m1)*g - (m1)*g*sin(\alpha ) = (m1)*a

(0.200)(9.81) - (0.1)(0.290)(9.81) - (0.290)(9.81)sin(30°) = (0.290)a

(0.200kg)(9.81\frac{m}{s^{2}}) - (0.1)(0.290kg)(9.81\frac{m}{s^{2}}) - (0.290kg)(9.81\frac{m}{s^{2}})sin(30°) = (0.290kg)*a

0.511101=0.29*a\\a=0.879\frac{m}{s^{2} }

v_{f}^{2}=v_{o}^{2}+2*a(x_{f}^{2}-v_{o}^{2})\\v_{o}=0\\v_{o}=0\\v_{f}^{2}=2*a(x_{f})\\v_{f}=\sqrt{2*a(x_{f})}\\v_{f}=\sqrt{2*0.879\frac{m}{s^{2}}*0.30m} \\v_{f}=0.726 \frac{m}{s}

5 0
4 years ago
A car enters a freeway with initial velocity of 15.0 m/s and with con stant rate of acceleration, reaches a velocity of 22.5 m/s
LiRa [457]

Answer:

a) The acceleration is 2.14 m/s^{2}

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c) The average velocity is 18.75 m/s

Explanation:

Using the equations that describe an uniformly accelerated motion:

a) a=\frac{v_f - v_o}{t} =\frac{22.5m/s - 15.0 m/s}{3.50s}

b) d= d_0 + v_0 t + \frac{1}{2} a t^{2} = 0 +15.0 x 3.5 + \frac{2.14x3.50^{2} }{2} = 65.61 m

c) v_m =\frac{d}{t}=\frac{65.61}{3.5}  =18.75 m/s

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