The answers are B, C, G, I
Hope I helped
(Don’t know if they are right)
Answer:
See the answers below
Explanation:
We can solve this problem using the principle of energy conservation. That is, the energy is conserved before and after dropping the bag.
For this case we have mechanical energy, which is the sum of the kinetic and potential energies.
where:
Ek = kinetic energy [J] (units of Joules)
Ep = potential energy [J]
In the final Energy (2), there is only potential energy. since when the balloon reaches the maximum height its velocity is zero, that is, there is no kinetic energy.
A)
![m*g*h+\frac{1}{2}*m*v^{2} =m*g*h_{1} \\9.81*50+0.5*(15)^{2}=9.81*h_{1}\\h_{1} = 61.46 [m]](https://tex.z-dn.net/?f=m%2Ag%2Ah%2B%5Cfrac%7B1%7D%7B2%7D%2Am%2Av%5E%7B2%7D%20%3Dm%2Ag%2Ah_%7B1%7D%20%5C%5C9.81%2A50%2B0.5%2A%2815%29%5E%7B2%7D%3D9.81%2Ah_%7B1%7D%5C%5Ch_%7B1%7D%20%3D%2061.46%20%5Bm%5D)
B)
With the value calculated above we can find the acceleration of the balloon.
The distance traveled is the difference between the maximum height and 50 meters.
![x = 61.46-50\\x = 11.46[m]](https://tex.z-dn.net/?f=x%20%3D%2061.46-50%5C%5Cx%20%3D%2011.46%5Bm%5D)
With the following equation of kinematics.
![0 = 15^{2} +2*a*11.46\\a = - 9.816 [m/s^{2} ]](https://tex.z-dn.net/?f=0%20%3D%2015%5E%7B2%7D%20%2B2%2Aa%2A11.46%5C%5Ca%20%3D%20-%209.816%20%5Bm%2Fs%5E%7B2%7D%20%5D)
The negative sign indicates that the acceleration acts downward. That is, in the opposite direction to the movement.
We can use the following equation of kinematics to find the final velocity after 4 seconds.
![v_{f}=v_{o}-a*t\\v_{f}=15-9.816*(4)\\v_{f}=-24.24 [m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%3Dv_%7Bo%7D-a%2At%5C%5Cv_%7Bf%7D%3D15-9.816%2A%284%29%5C%5Cv_%7Bf%7D%3D-24.24%20%5Bm%2Fs%5D)
Now the distance:
![v_{f}^{2} =v_{o}^{2}-2*a*x\\(24.24)^{2} =(15)^{2} -2*9.81*x\\x = 18.48 [m]\\x_{f}=50+18.48 = 68.48 [m]](https://tex.z-dn.net/?f=v_%7Bf%7D%5E%7B2%7D%20%3Dv_%7Bo%7D%5E%7B2%7D-2%2Aa%2Ax%5C%5C%2824.24%29%5E%7B2%7D%20%3D%2815%29%5E%7B2%7D%20-2%2A9.81%2Ax%5C%5Cx%20%3D%2018.48%20%5Bm%5D%5C%5Cx_%7Bf%7D%3D50%2B18.48%20%3D%2068.48%20%5Bm%5D)
c) Using the following equation of kinematics.
![v_{f}=v_{o}-a*t\\0 = 15-9.81*t\\15=9.81*t\\t = 1.52 [s]](https://tex.z-dn.net/?f=v_%7Bf%7D%3Dv_%7Bo%7D-a%2At%5C%5C0%20%3D%2015-9.81%2At%5C%5C15%3D9.81%2At%5C%5Ct%20%3D%201.52%20%5Bs%5D)
Answer:
An atom is the smallest unit of ordinary matter that forms a chemical element.
Matter is composed of very tiny or microscopic particles called "Atom".
(According to Dalton atomic theory some of his principles were wrong and later was corrected but the definition of an atom was right in his theory).
Answer:
Change volume / volume = .5 / 30 = .01666
coefficient of volume expansion = 3 * 24E-6 = 72E-6
Change in temp = .0167 / 24E-6 = 232 deg C
Final temp = 232 + 30 = 262 deg C
Change in volume / volume = 3 * coeff linear expansion * change in deg C
Answer:
θ = 1.591 10⁻² rad
Explanation:
For this exercise we must suppose a criterion when two light sources are considered separated, we use the most common criterion the Rayleigh criterion that establishes that two light sources are separated census the central maximum of one of them coincides with the first minimum of the other source
Let's write the diffraction equation for a slit
a sin θ = m λ
The first minimum occurs for m = 1, also field in these we experience the angles are very small, we can approximate the sin θ = θ
θ = λ / a
In our case, the pupil is circular, so the system must be solved in polar coordinates, so a numerical constant is introduced.
θ = 1.22 λ / D
Where D is the diameter of the pupil
Let's apply this equation to our case
θ = 1.22 600 10⁻⁹ / 0.460 10⁻²
θ = 1.591 10⁻² rad
This is the angle separation to solve the two light sources
