Here we apply conservation of linear momentum. The momentum of the truck with cargo and without cargo remains constant. That is,
.
Here
are initial mass and velocity.
are final mass and velocity. Here
and
.
The velocity of the truck be after its cargo is taken off is
![v'=\frac{mv}{m'} \\ v'=\frac{1800*10}{1500} \\ v'=12 m/s](https://tex.z-dn.net/?f=%20v%27%3D%5Cfrac%7Bmv%7D%7Bm%27%7D%20%5C%5C%20%20v%27%3D%5Cfrac%7B1800%2A10%7D%7B1500%7D%20%5C%5C%20%20v%27%3D12%20m%2Fs%20)
Using the given equation you get:
E = 1.99x10^-25 / 9.0x10^-6
Divide 1.99 by 9.0: 1.99/9.0 = 0.22
For the scientific notation, when dividing subtract the two exponents:
25 -6 = 19
So you now have 0.22 x 10^-19
Now you need to change the 0.22 to be in scientific notation form:
2.2 x 10^-20
The answer is B.
Answer:
Explanation:
The equation for this is
where f is the frequency, v is the velocity, and lambda is the wavelength. Filling in:
and
which means that
the wavelength is 1.37 m, rounded to the correct number of significant digits.
Answer: ![6.408(10)^{-19} C](https://tex.z-dn.net/?f=6.408%2810%29%5E%7B-19%7D%20C)
Explanation:
This problem can be solved by the following equation:
![\Delta K=q V](https://tex.z-dn.net/?f=%5CDelta%20K%3Dq%20V)
Where:
is the change in kinetic energy
is the electric potential difference
is the electric charge
Finding
:
![q=\frac{\Delta K}{V}](https://tex.z-dn.net/?f=q%3D%5Cfrac%7B%5CDelta%20K%7D%7BV%7D)
![q=\frac{7.37(10)^{-17} J}{115 V}](https://tex.z-dn.net/?f=q%3D%5Cfrac%7B7.37%2810%29%5E%7B-17%7D%20J%7D%7B115%20V%7D)
Finally:
![q=6.408(10)^{-19} C](https://tex.z-dn.net/?f=q%3D6.408%2810%29%5E%7B-19%7D%20C)
Have everything in control and in order and discuss about different issues.