Answer:
Wavelength = 1.36 * 10^{-34} meters
Explanation:
Given the following data;
Mass = 0.113 kg
Velocity = 43 m/s
To find the wavelength, we would use the De Broglie's wave equation.
Mathematically, it is given by the formula;
![Wavelength = \frac {h}{mv}](https://tex.z-dn.net/?f=%20Wavelength%20%3D%20%5Cfrac%20%7Bh%7D%7Bmv%7D%20)
Where;
h represents Planck’s constant.
m represents the mass of the particle.
v represents the velocity of the particle.
We know that Planck’s constant = 6.6262 * 10^{-34} Js
Substituting into the formula, we have;
![Wavelength = \frac {6.6262 * 10^{-34}}{0.113*43}](https://tex.z-dn.net/?f=%20Wavelength%20%3D%20%5Cfrac%20%7B6.6262%20%2A%2010%5E%7B-34%7D%7D%7B0.113%2A43%7D%20)
![Wavelength = \frac {6.6262 * 10^{-34}}{4.859}](https://tex.z-dn.net/?f=%20Wavelength%20%3D%20%5Cfrac%20%7B6.6262%20%2A%2010%5E%7B-34%7D%7D%7B4.859%7D%20)
Wavelength = 1.36 * 10^{-34} meters
The first: alright, first: you draw the person in the elevator, then draw a red arrow, pointing downwards, beginning from his center of mass. This arrow is representing the gravitational force, Fg.
You can always calculate this right away, if you know his mass, by multiplying his weight in kg by the gravitational constant
![g = 9.81 \frac{m}{s {}^{2} }](https://tex.z-dn.net/?f=g%20%3D%209.81%20%5Cfrac%7Bm%7D%7Bs%20%7B%7D%5E%7B2%7D%20%7D%20)
let's do it for this case:
![f_{g} = m \times g \\ f _{g} = 65kg \times 9.81 \frac{m}{s {}^{2} } = 637.65](https://tex.z-dn.net/?f=f_%7Bg%7D%20%20%3D%20m%20%5Ctimes%20g%20%5C%5C%20f%20_%7Bg%7D%20%20%3D%2065kg%20%5Ctimes%209.81%20%5Cfrac%7Bm%7D%7Bs%20%7B%7D%5E%7B2%7D%20%7D%20%20%3D%20637.65)
the unit of your fg will be in Newton [N]
so, first step solved, Fg is 637.65N
Fg is a field force by the way, and at the same time, the elevator is pushing up on him with 637.65N, so you draw another arrow pointing upwards, ending at the tip of the downwards arrow.
now let's calculate the force of the elevator
![f = m \times a \\ f = 65 \times 5 \frac{m}{s {}^{2} } \\ f = 325n](https://tex.z-dn.net/?f=f%20%3D%20m%20%5Ctimes%20a%20%5C%5C%20f%20%3D%2065%20%5Ctimes%205%20%5Cfrac%7Bm%7D%7Bs%20%7B%7D%5E%7B2%7D%20%7D%20%20%5C%5C%20f%20%3D%20325n)
so you draw another arrow which is pointing downwards on him, because the elevator is accelating him upwards, making him heavier
the elevator force in this case is a contact force, because it only comes to existence while the two are touching, while Fg is the same everywhere
Decreases. Air resistance will slow a falling object to its terminal velocity, placing a limit on its acceleration.
Answer:
![v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}](https://tex.z-dn.net/?f=v_%7By_0%7D%20%3D%20%5Cfrac%7B%5Cfrac%7Bg%7D%7B2%7Dt%28t%20-%202%5Csqrt%7B%5Cfrac%7B2h%7D%7Bg%7D%7D%29%7D%7B%5Csqrt%7B%5Cfrac%7B2h%7D%7Bg%7D%7D%20-%20t%7D)
Explanation:
We will apply the equations of kinematics to both stones separately.
First stone:
Let us denote the time spent after the second stone is thrown as 'T'.
![y - y_0 = v_{y_0}(t+T) + \frac{1}{2}a(t+T)^2\\0 - h = 0 + \frac{1}{2}(-g)(t+T)^2\\(t+T)^2 = \frac{2h}{g}\\T = \sqrt{\frac{2h}{g}}-t](https://tex.z-dn.net/?f=y%20-%20y_0%20%3D%20v_%7By_0%7D%28t%2BT%29%20%2B%20%5Cfrac%7B1%7D%7B2%7Da%28t%2BT%29%5E2%5C%5C0%20-%20h%20%3D%200%20%2B%20%5Cfrac%7B1%7D%7B2%7D%28-g%29%28t%2BT%29%5E2%5C%5C%28t%2BT%29%5E2%20%3D%20%5Cfrac%7B2h%7D%7Bg%7D%5C%5CT%20%3D%20%5Csqrt%7B%5Cfrac%7B2h%7D%7Bg%7D%7D-t)
Second stone:
![y - y_0 = v_{y_0}T + \frac{1}{2}aT^2\\0 - h = v_{y_0}T -\frac{1}{2}gT^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\sqrt{\frac{2h}{g}} - t)^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\frac{2h}{g} + t^2 - 2t\sqrt{\frac{2h}{g}})\\-h = v_{y_0}\sqrt{\frac{2h}{g}} - v_{y_0}t - h -\frac{g}{2}t^2 + gt\sqrt{\frac{2h}{g}}\\v_{y_0}(\sqrt{\frac{2h}{g}} - t) = \frac{g}{2}t^2 - gt\sqrt{\frac{2h}{g}}\\v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}](https://tex.z-dn.net/?f=y%20-%20y_0%20%3D%20v_%7By_0%7DT%20%2B%20%5Cfrac%7B1%7D%7B2%7DaT%5E2%5C%5C0%20-%20h%20%3D%20v_%7By_0%7DT%20-%5Cfrac%7B1%7D%7B2%7DgT%5E2%5C%5C-h%20%3D%20v_%7By_0%7D%28%5Csqrt%7B%5Cfrac%7B2h%7D%7Bg%7D%7D%20-%20t%29%20-%20%5Cfrac%7Bg%7D%7B2%7D%28%5Csqrt%7B%5Cfrac%7B2h%7D%7Bg%7D%7D%20-%20t%29%5E2%5C%5C-h%20%3D%20v_%7By_0%7D%28%5Csqrt%7B%5Cfrac%7B2h%7D%7Bg%7D%7D%20-%20t%29%20-%20%5Cfrac%7Bg%7D%7B2%7D%28%5Cfrac%7B2h%7D%7Bg%7D%20%2B%20t%5E2%20-%202t%5Csqrt%7B%5Cfrac%7B2h%7D%7Bg%7D%7D%29%5C%5C-h%20%3D%20v_%7By_0%7D%5Csqrt%7B%5Cfrac%7B2h%7D%7Bg%7D%7D%20-%20v_%7By_0%7Dt%20-%20h%20-%5Cfrac%7Bg%7D%7B2%7Dt%5E2%20%2B%20gt%5Csqrt%7B%5Cfrac%7B2h%7D%7Bg%7D%7D%5C%5Cv_%7By_0%7D%28%5Csqrt%7B%5Cfrac%7B2h%7D%7Bg%7D%7D%20-%20t%29%20%3D%20%5Cfrac%7Bg%7D%7B2%7Dt%5E2%20-%20gt%5Csqrt%7B%5Cfrac%7B2h%7D%7Bg%7D%7D%5C%5Cv_%7By_0%7D%20%3D%20%5Cfrac%7B%5Cfrac%7Bg%7D%7B2%7Dt%28t%20-%202%5Csqrt%7B%5Cfrac%7B2h%7D%7Bg%7D%7D%29%7D%7B%5Csqrt%7B%5Cfrac%7B2h%7D%7Bg%7D%7D%20-%20t%7D)
Answer:
we got time and velocity over time.
so the distance is again the area underneath the graph
for a triangle with known base and height it's
4*10 / 2
distance traveled is 20
deceleration occurs when velocity decreases. that happens from t=2 till t=4
in 2 time-units we loose 10 units of velocity, so we decelerate by 5 units per 1 time
a (from t=2 to t=4) = -5v/t