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grandymaker [24]
3 years ago
8

Given that the atomic weight of hydrogen is approximately 1 gram per mole, use the method above to estimate the average molecula

r mass for water vapor.____ g/mol.
Physics
1 answer:
denpristay [2]3 years ago
6 0

Answer:

2

Explanation:

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For work to be accomplished we must have
Nat2105 [25]
The answer they are looking for is the last one. However the last two are technically correct but the third one would result in negative work.
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3 years ago
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(8c8p49) A 115g Frisbee is thrown from a point 1.00 m above the ground with a speed of 12.00 m/s. When it has reached a height o
IgorLugansk [536]

Answer:

The work done on the Frisbee is 1.36 J.

Explanation:

Given that,

Mass of Frisbee, m = 115 g = 0.115 kg

Initial speed of Frisbee, u = 12 m/s at a point 1 m above the ground

Final speed of Frisbee , v = 10.9674 m/s when it has reached a height of 2.00 m. Let W is the work done on the Frisbee by its weight. According to work energy theorem, the work done is equal to the change in its kinetic energy. So,

W=\dfrac{1}{2}m(v^2-u^2)\\\\W=\dfrac{1}{2}\times0.115\times\left((10.9674)^{2}-(12)^{2})\right)\\\\W=-1.36\ J

So, the work done on the Frisbee is 1.36 J. Hence, this is the required solution.

3 0
3 years ago
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Can science answer every question
Alla [95]
No because your opinion and beliefs answers many questions
5 0
3 years ago
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"A steel rotating-beam test specimen has an ultimate strength of 120 kpsi. Estimate the life of the specimen if it is tested at
MrRissso [65]

Answer:

life (N) of the specimen is 117000  cycles

Explanation:

given data

ultimate strength Su = 120 kpsi

stress amplitude σa = 70 kpsi

solution

we first calculate the endurance limit of specimen Se i.e

Se = 0.5× Su   .............1

Se = 0.5 × 120

Se = 60 kpsi

and we know strength of friction f  = 0.82

and we take endurance limit Se is = 60 kpsi

so here coefficient value (a) will be

a = \frac{(f\times Su)^2}{Se}     ......................1  

put here value and we get

a = \frac{(0.82\times 120)^2}{60}  

a = 161.4  kpsi

so coefficient value (b) will be

b = -\frac{1}{3}log\frac{(f\times Su)}{Se}  

b =  -\frac{1}{3}log\frac{(0.82\times 120)}{60}  

b = −0.0716

so here number of cycle N will be  

N =  (\frac{ \sigma a}{a})^{1/b}

put here value  and we get

N =  (\frac{ 70}{161.4})^{1/-0.0716}

N = 117000

so life (N) of the specimen is 117000  cycles

7 0
3 years ago
With what velocity Must a 0.53 kg softball be moving to be equal to the momentum of a 0.31 kg baseball moving at 21 m/s
vladimir2022 [97]

Answer:12.28m/s

Explanation:

momentum of baseball =mass of baseball x velocity of baseball

Momentum of baseball =0.31x21

Momentum of baseball =6.51kgm/s

For a softball to have same momentum with the baseball we can say :momentum of baseball =mass of softball x velocity of softball

6.51=0.53 x velocity of softball

Velocity of softball =6.51/0.53

Velocity of softball =12.28m/s

3 0
3 years ago
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