Work is done on a locked door that remains closed while you try to pull it open. True False.
1 answer:
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The weight of the meterstick is:
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
from which we find the value of d2:
So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
from which we find the value of d2:
So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
Answer:
DOUBLE CHECK BECUASE IM ONLY 68.030303039999999% SURE!!!
(ANSWER IS HERE) ( D) It lacked practical examples in supporting theory
Know it's not B becuase there was no scientific community back then.
Know it's not C becuase it actully had lots of evidence.
But I'm not sure about A
To solve this problem it is necessary to take into account the concepts related to frequency and period, and how they are related to each other.
The relationship that defines both agreements is given by the equation,
Then the frequency for the previous period given (2sec) is
The beat frequency of two frequencies is equal to the difference between the two frequencies, then
<em>Hence option A is incorrect.</em>
We can do this process for 254Hz as and 258 Hz for
, then
<em>Hence option B is incorrect. </em>
We can also do this process for 255Hz as and 257 Hz for
, then
<em>Hence option C is incorrect. </em>
We can also do this process for 255.5Hz as f_1 and 256.5 Hz for f_2, then
<em>Hence option D is incorrect. </em>
We can also do this process for 255.75Hz as and 256.25 Hz for
, then
<em>Hence option E is incorrect. </em>
Therefore the sum of the frequencies in the sound wave would be 256.25Hz and 255.75Hz