Answer:
See the answers below.
Explanation:
In order to solve this problem we must use the principle of energy conservation. Which tells us that the energy of a body will always be the same regardless of where it is located. For this case we have two points, point A and point B. Point A is located at the top at 120 [m] and point B is in the middle of the cliff at 60 [m].
![E_{A}=E_{B}](https://tex.z-dn.net/?f=E_%7BA%7D%3DE_%7BB%7D)
The important thing about this problem is to identify the types of energy at each point. Let's take the reference level of potential energy at a height of zero meters. That is, at this point the potential energy is zero.
So at point A we have potential energy and since a velocity of 18 [m/s] is printed, we additionally have kinetic energy.
![E_{A}=E_{pot}+E_{kin}\\E_{A}=m*g*h+\frac{1}{2}*m*v^{2}](https://tex.z-dn.net/?f=E_%7BA%7D%3DE_%7Bpot%7D%2BE_%7Bkin%7D%5C%5CE_%7BA%7D%3Dm%2Ag%2Ah%2B%5Cfrac%7B1%7D%7B2%7D%2Am%2Av%5E%7B2%7D)
At Point B the rock is still moving downward, therefore we have kinetic energy and since it is 60 [m] with respect to the reference level we have potential energy.
![E_{B}=m*g*h+\frac{1}{2}*m*v^{2}](https://tex.z-dn.net/?f=E_%7BB%7D%3Dm%2Ag%2Ah%2B%5Cfrac%7B1%7D%7B2%7D%2Am%2Av%5E%7B2%7D)
Therefore we will have the following equation:
![(6.5*9.81*120)+(0.5*6.5*18^{2} )=(6.5*9.81*60)+(0.5*6.5*v_{B}^{2} )\\3.25*v_{B}^{2} =4878.9\\v_{B}=\sqrt{1501.2}\\v_{B}=38.75[m/s]](https://tex.z-dn.net/?f=%286.5%2A9.81%2A120%29%2B%280.5%2A6.5%2A18%5E%7B2%7D%20%29%3D%286.5%2A9.81%2A60%29%2B%280.5%2A6.5%2Av_%7BB%7D%5E%7B2%7D%20%29%5C%5C3.25%2Av_%7BB%7D%5E%7B2%7D%20%3D4878.9%5C%5Cv_%7BB%7D%3D%5Csqrt%7B1501.2%7D%5C%5Cv_%7BB%7D%3D38.75%5Bm%2Fs%5D)
The kinetic energy can be easily calculated by means of the kinetic energy equation.
![KE_{B}=\frac{1}{2} *m*v_{B}^{2}\\KE_{B}=0.5*6.5*(38.75)^{2}\\KE_{B}=4878.9[J]](https://tex.z-dn.net/?f=KE_%7BB%7D%3D%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Av_%7BB%7D%5E%7B2%7D%5C%5CKE_%7BB%7D%3D0.5%2A6.5%2A%2838.75%29%5E%7B2%7D%5C%5CKE_%7BB%7D%3D4878.9%5BJ%5D)
In order to calculate the velocity at the bottom of the cliff where the reference level of potential energy (potential energy equal to zero) is located, we must pose the same equation, with the exception that at the new point there is only kinetic energy.
![E_{A}=E_{C}\\6.5*9.81*120+(0.5*9.81*18^{2} )=0.5*6.5*v_{C}^{2} \\v_{c}^{2} =\sqrt{2843.39}\\v_{c}=53.32[m/s]](https://tex.z-dn.net/?f=E_%7BA%7D%3DE_%7BC%7D%5C%5C6.5%2A9.81%2A120%2B%280.5%2A9.81%2A18%5E%7B2%7D%20%29%3D0.5%2A6.5%2Av_%7BC%7D%5E%7B2%7D%20%5C%5Cv_%7Bc%7D%5E%7B2%7D%20%3D%5Csqrt%7B2843.39%7D%5C%5Cv_%7Bc%7D%3D53.32%5Bm%2Fs%5D)
Answer:
The net force is zero.
Explanation:
Two opposing and equal forces cancel each other out, giving you a net force of zero.
Answer: An equation is missing in your question below is the missing equation
a) ≈ 8396
b) 150 nm/k
Explanation:
<u>A) Determine the number of Oscillators in the black body</u>
number of oscillators = 8395
attached below is the detailed solution
<u>b) determine the peak wavelength of the black body </u>
Black body temperature = 20,000 K
applying Wien's law / formula
λmax = b / T ------ ( 1 )
T = 20,000 K
b = 3 * 10^6 nm
∴ λmax = 150 nm/k
F - False.
The nucleus of an atom is positively charge.
Answer:
Same direction: t=234s; d=6.175Km
Opposite direction: t=27.53s; d=0.73Km
Explanation:
If the automobile and the train are traveling in the same direction, then the automobile speed relative to the train will be
(<em>the train must see the car advancing at a lower speed</em>), where
is the speed of the automobile and
the speed of the train.
So we have
.
So the train (<em>anyone in fact</em>) will watch the automobile trying to cover the lenght of the train L at that relative speed. The time required to do this will be:
![t = \frac{L}{v_{AT}} = \frac{1.3Km}{20Km/h} = 0.065h=234s](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7BL%7D%7Bv_%7BAT%7D%7D%20%3D%20%5Cfrac%7B1.3Km%7D%7B20Km%2Fh%7D%20%3D%200.065h%3D234s)
And in that time the car would have traveled (<em>relative to the ground</em>):
![d=v_At=(95Km/h)(0.065h)=6.175Km](https://tex.z-dn.net/?f=d%3Dv_At%3D%2895Km%2Fh%29%280.065h%29%3D6.175Km)
If they are traveling in opposite directions, <u>we have to do all the same</u> but using
(<em>the train must see the car advancing at a faster speed</em>), so repeating the process:
![v_{AT}=(95km/h)+(75Km/h)=170Km/h](https://tex.z-dn.net/?f=v_%7BAT%7D%3D%2895km%2Fh%29%2B%2875Km%2Fh%29%3D170Km%2Fh)
![t = \frac{L}{v_{AT}} = \frac{1.3Km}{170Km/h} = 0.00765h=27.53s](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7BL%7D%7Bv_%7BAT%7D%7D%20%3D%20%5Cfrac%7B1.3Km%7D%7B170Km%2Fh%7D%20%3D%200.00765h%3D27.53s)
![d=v_At=(95Km/h)(0.00765h)=0.73Km](https://tex.z-dn.net/?f=d%3Dv_At%3D%2895Km%2Fh%29%280.00765h%29%3D0.73Km)