Answer:
Write the following Quantitiesin scientific notation.
a. 10130 Pa to 2 decimal place
b. 978.15m * s-2 to one decimal place
c 0.000001256 A to3 decimal place
Add your answer and earn points.
Answer
5.0/5
2
kobenhavn
Expert
5.5K answers
43M people helped
Answer: a.
b.
c.
Explanation:
Scientific notation is defined as the representation of expressing the numbers that are too big or too small and are represented in the decimal form with one digit before the decimal point times 10 raise to the power.
For example : 5000 is written as
a. 10130 Pa to 2 decimal place is written as
b. to 1 decimal place is written as
c. to 3 decimal places is written as
Explanation:
Answer:
If the wishing well was in a vacuum, then s=ut + 0.5 a t^2 (s=distance, ... wishing well if you drop a coin into it and hear the splash 10 seconds
Explanation:
Answer:
Velocity = 4.33[m/s]
Explanation:
The total energy or mechanical energy is the sum of the potential energy plus the kinetic energy, as it is known the velocity and the height, we can determine the total energy.
![E_{M}=E_{p} + E_{k} \\E_{p} = potential energy [J]\\E_{k} = kinetic energy [J]\\where:\\E_{p} =m*g*h\\E_{p} = 4*9.81*0.5=19.62[J]\\E_{k}=\frac{1}{2} *m*v^{2} \\E_{k}=\frac{1}{2} *4*(3)^{2} \\E_{k}=18[J]\\Therefore\\E_{M} =18+19.62\\E_{M}=37.62[J]](https://tex.z-dn.net/?f=E_%7BM%7D%3DE_%7Bp%7D%20%20%2B%20E_%7Bk%7D%20%5C%5CE_%7Bp%7D%20%3D%20potential%20energy%20%5BJ%5D%5C%5CE_%7Bk%7D%20%3D%20kinetic%20energy%20%5BJ%5D%5C%5Cwhere%3A%5C%5CE_%7Bp%7D%20%3Dm%2Ag%2Ah%5C%5CE_%7Bp%7D%20%3D%204%2A9.81%2A0.5%3D19.62%5BJ%5D%5C%5CE_%7Bk%7D%3D%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Av%5E%7B2%7D%20%20%5C%5CE_%7Bk%7D%3D%5Cfrac%7B1%7D%7B2%7D%20%2A4%2A%283%29%5E%7B2%7D%20%5C%5CE_%7Bk%7D%3D18%5BJ%5D%5C%5CTherefore%5C%5CE_%7BM%7D%20%3D18%2B19.62%5C%5CE_%7BM%7D%3D37.62%5BJ%5D)
All this energy will become kinetic energy and we can find the velocity.
![37.62=\frac{1}{2} *m*v^{2} \\v=\sqrt{\frac{37.62*2}{4} } \\v=4.33[m/s]](https://tex.z-dn.net/?f=37.62%3D%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Av%5E%7B2%7D%20%5C%5Cv%3D%5Csqrt%7B%5Cfrac%7B37.62%2A2%7D%7B4%7D%20%7D%20%5C%5Cv%3D4.33%5Bm%2Fs%5D)
Answer:
f = 409 Hz
Explanation:
We have,
Length of the open organ pipe, l = 0.29 m
Frequency of vibration of second overtone, 
It is required to find the fundamental frequency of the pipe. For the open organ pipe, the frequency of second overtone is given by :

v is speed of sound
Let f is the fundamental frequency. It is given by :

The relation between f and f₂ can be written as :

So, the fundamental frequency of the pipe is 409 Hz.