Answer:
depth of well is 163.30 m
Explanation:
Given data
speed of sound = 343 m/s
timer = 6.25 s
to find out
depth of well
solution
let us consider depth d
so equation will be
depth = 1/2 ×g ×t² ..............1
and
depth = velocity of sound × time .................2
here we have given time 6.25 that is sum of 2 time
when stone reach at bottom that time
another is sound reach us after stone strike on bottom
so time 1 + time 2 = 6.25 s
so from equation 1 and 2 we get
1/2 ×g ×t² = velocity of sound × time
1/2 ×9.8 × t1² = 343 × (6.25 - t1 )
t1 = 5.77376 sec
so height = 1/2 ×g ×t²
height = 1/2 ×9.8 × (5.773)²
height = 163.30 m
Answer:
44.13015
Explanation:
use the 9.8067 newtons to 1 kg conversion
The final volume of the gas is 238.9 mL
Explanation:
We can solve this problem by using Charle's law, which states that for a gas kept at constant pressure, the volume of the gas (V) is proportional to its absolute temperature (T):
![\frac{V}{T}=const.](https://tex.z-dn.net/?f=%5Cfrac%7BV%7D%7BT%7D%3Dconst.)
Which can be also re-written as
![\frac{V_1}{T_1}=\frac{V_2}{T_2}](https://tex.z-dn.net/?f=%5Cfrac%7BV_1%7D%7BT_1%7D%3D%5Cfrac%7BV_2%7D%7BT_2%7D)
where
are the initial and final volumes of the gas
are the initial and final temperature of the gas
For the gas in the balloon in this problem, we have:
is the initial volume
is the initial absolute temperature
is the final volume
is the final temperature
Solving for
,
![V_2 = \frac{V_1 T_2}{T_1}=\frac{(700)(100)}{293}=238.9 mL](https://tex.z-dn.net/?f=V_2%20%3D%20%5Cfrac%7BV_1%20T_2%7D%7BT_1%7D%3D%5Cfrac%7B%28700%29%28100%29%7D%7B293%7D%3D238.9%20mL)
Learn more about ideal gases:
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Answer:
Voltage in primary coil is 3.91 V
Explanation:
For transformer we know that the working principle is given as
![\frac{V_1}{V_2} = \frac{N_1}{N_2}](https://tex.z-dn.net/?f=%5Cfrac%7BV_1%7D%7BV_2%7D%20%3D%20%5Cfrac%7BN_1%7D%7BN_2%7D)
here we know that
![V_1 [tex] = voltage in primary coil[tex]V_2 = 25 V](https://tex.z-dn.net/?f=V_1%20%5Btex%5D%20%3D%20voltage%20in%20primary%20coil%3C%2Fp%3E%3Cp%3E%5Btex%5DV_2%20%3D%2025%20V)
![N_1 = 500](https://tex.z-dn.net/?f=N_1%20%3D%20500)
![N_2 = 3200](https://tex.z-dn.net/?f=N_2%20%3D%203200)
Now we have
![\frac{V_1}{25} = \frac{500}{3200}](https://tex.z-dn.net/?f=%5Cfrac%7BV_1%7D%7B25%7D%20%3D%20%5Cfrac%7B500%7D%7B3200%7D)
![V_1 = 3.91 V](https://tex.z-dn.net/?f=V_1%20%3D%203.91%20V)