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9966 [12]
3 years ago
5

Consider two ideal gases, A & B, at the same temperature. The rms speed of the molecules of gas A is twice that of gas B. Ho

w does the molecular mass of A compare to that of B? (a) It is twice that of B. (b) It is one half that of B (c) It is four times that of B (d) It is one fourth that of B. (e) It is 1.4 times that of B.
Physics
1 answer:
uysha [10]3 years ago
7 0

Answer:

option (d)

Explanation:

The relation between the rms velocity and the molecular mass is given by

v   proportional to  \frac{1}{\sqrt{M}} keeping the temperature constant

So for two gases

\frac{v_{A}}{v_{B}}=\sqrt{\frac{M_{B}}{M_{A}}}

\frac{2v_{B}}{v_{B}}=\sqrt{\frac{M_{B}}{M_{A}}}

{\frac{M_{B}}{M_{A}}} = 4

{\frac{M_{B}}{4}} = M_{A}

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there are 5 influencing factors that may affect participation rate in physical activity. can you name them?
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3 years ago
A wire is stretched between two posts. Another wire is stretched between two posts that are twice as far apart. The tension in t
Margarita [4]

Answer: 996m/s

Explanation:

Formula for calculating velocity of wave in a stretched string is

V = √T/M where;

V is the velocity of wave

T is tension

M is the mass per unit length of the wire(m/L)

Since the second wire is twice as far apart as the first, it will be L2 = 2L1

Let V1 and V2 be the speed of the shorter and longer wire respectively

V1 = √T/M1... 1

V2 = √T/M2... 2

Since V1 = 249m/s, M1 = m/L1 M2 = m/L2 = m/2L1

The equations will now become

249 = √T/(m/L1) ... 3

V2 = √T/(m/2L1)... 4

From 3,

249² = TL1/m...5

From 4,

V2²= 2TL1/m... 6

Dividing equation 5 by 6 we have;

249²/V2² = TL1/m×m/2TL1

{249/V2}² = 1/2

249/V2 = (1/2)²

249/V2 = 1/4

V2 = 249×4

V2 = 996m/s

Therefore the speed of the wave on the longer wire is 996m/s

3 0
3 years ago
A tray of electronic components contains 15 components, 4 of which are defective. If 4 components are selected, what is the poss
dlinn [17]

Answer:

a) 0.0007326

b) 0.03223

c) 0.2418

d) 0.2418

Explanation:

To find different probabilities for the selection of components among eleven good and four defective components, we will use the Combination.

a) C(4,4) = 1; C(15,4) = 1365

P = \frac{C(4,4)}{C(15,4)} = \frac{1}{1365} = 0.0007326

b) C(4,3) = 4; C(11,1) = 11

P = \frac{C(4,3)*C(11,1)}{C(15,4)} = \frac{4*11}{1365} = 0.03223

c) C(4,2) = 6; C(11,2) = 55

P = \frac{C(4,2)*C(11,2)}{C(15,4)} = \frac{6*55}{1365} = 0.2418

d) C(11,4) = 330

P = \frac{C(11,4)}{C(15,4)} = \frac{330}{1365} = 0.2418

8 0
3 years ago
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