De Broglie's identity gives the relationship between the momentum and the wavelength of a particle:
![p=mv= \frac{h}{\lambda}](https://tex.z-dn.net/?f=p%3Dmv%3D%20%5Cfrac%7Bh%7D%7B%5Clambda%7D%20)
where
p is the particle momentum
m is its mass
v its velocity
h is the Planck constant
![\lambda](https://tex.z-dn.net/?f=%5Clambda)
is the wavelength
By re-arranging the equation, we get
![\lambda= \frac{h}{mv}](https://tex.z-dn.net/?f=%5Clambda%3D%20%20%5Cfrac%7Bh%7D%7Bmv%7D%20)
and by using the data about the proton, given in the text, we can find the proton's wavelength:
Answer:
Meter, Gram and Liter.
Explanation:
In the metric system, the standard units for the below are;
Length - Meter
Mass - Gram
Volume - Liter.
Answer:
Average force is F = mass times change in V/ change in time so..
1 366.07143 N
Explanation:
51 kg x 15 m/s / 0.56
1 366.07143 m kg / s
1 366.07143 N
1 kilogram 1 meter per second per second = 1 N
Answer:
![R = 24.3 m](https://tex.z-dn.net/?f=R%20%3D%2024.3%20m)
Explanation:
As we know that the orbital speed is given as
![v = \sqrt{\frac{GM}{R + h}}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B%5Cfrac%7BGM%7D%7BR%20%2B%20h%7D%7D)
here we know that
v = 5500 m/s
![R = 4.48 \times 10^6 m](https://tex.z-dn.net/?f=R%20%3D%204.48%20%5Ctimes%2010%5E6%20m)
![h = 630 km](https://tex.z-dn.net/?f=h%20%3D%20630%20km)
now we have
![5500 = \sqrt{\frac{(6.6 \times 10^{-11})M}{4.48 \times 10^6 + 6.30\times 10^5}}](https://tex.z-dn.net/?f=5500%20%3D%20%5Csqrt%7B%5Cfrac%7B%286.6%20%5Ctimes%2010%5E%7B-11%7D%29M%7D%7B4.48%20%5Ctimes%2010%5E6%20%2B%206.30%5Ctimes%2010%5E5%7D%7D)
![M = 2.34 \times 10^24 kg](https://tex.z-dn.net/?f=M%20%3D%202.34%20%5Ctimes%2010%5E24%20kg)
now acceleration due to gravity of planet is given as
![a = \frac{GM}{R^2}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7BGM%7D%7BR%5E2%7D)
![a = \frac{(6.6 \times 10^{-11})(2.34 \times 10^{24})}{(4.48\times 10^6)^2}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7B%286.6%20%5Ctimes%2010%5E%7B-11%7D%29%282.34%20%5Ctimes%2010%5E%7B24%7D%29%7D%7B%284.48%5Ctimes%2010%5E6%29%5E2%7D)
![a = 7.7 m/s^2](https://tex.z-dn.net/?f=a%20%3D%207.7%20m%2Fs%5E2)
now range of the projectile on the surface of planet is given as
![R = \frac{v^2 sin2\theta}{g}](https://tex.z-dn.net/?f=R%20%3D%20%5Cfrac%7Bv%5E2%20sin2%5Ctheta%7D%7Bg%7D)
![R = \frac{14.6^2 sin(2\times 30.8)}{7.7}](https://tex.z-dn.net/?f=R%20%3D%20%5Cfrac%7B14.6%5E2%20sin%282%5Ctimes%2030.8%29%7D%7B7.7%7D)
![R = 24.3 m](https://tex.z-dn.net/?f=R%20%3D%2024.3%20m)
Answer:
12 or 24
Explanation:
i think it is i hope it is right