Answer:
Se the explanation below
Explanation:
We do not feel these forces of these bodies, because they are very small compared to the force of Earth's attraction. Although its mass is greater than that of a human being, its mass is not compared to the Earth's mass. In order to understand this problem we will use numerical data and the universal gravitation formula, to give validity to the explanation.
<u>Force exerted by the Earth on a human being</u>
<u />
Where:
G = universal gravitation constant = 6.673*10^-11 [N*m^2/kg^2]
m1 = mass of the person = 80 [kg]
m2 = mass of the earth 5.97*10^24[kg]
r = distance from the center of the earth to the surface or earth radius = 6371 *10^3 [m]
<u />
Now replacing we have
![F = 6.673*10^{-11} *\frac{80*5.97*10^{24}}{(6371*10^{3})^{2} } \\F = 785[N]](https://tex.z-dn.net/?f=F%20%3D%206.673%2A10%5E%7B-11%7D%20%2A%5Cfrac%7B80%2A5.97%2A10%5E%7B24%7D%7D%7B%286371%2A10%5E%7B3%7D%29%5E%7B2%7D%20%20%7D%20%5C%5CF%20%3D%20785%5BN%5D)
<u>Force exerted by a building on a human being</u>
<u />
Where:
G = universal gravitation constant = 6.673*10^-11 [N*m^2/kg^2]
m1 = mass of the person = 80 [kg]
m2 = mass of the earth 300000 [ton] = 300 *10^6[kg]
r = distance from the building to the person = 2[m]
![F = 6.673*10^{-11}*\frac{80*300*10^6}{2^{2} } \\F= 0.4 [N]](https://tex.z-dn.net/?f=F%20%3D%206.673%2A10%5E%7B-11%7D%2A%5Cfrac%7B80%2A300%2A10%5E6%7D%7B2%5E%7B2%7D%20%7D%20%20%5C%5CF%3D%200.4%20%5BN%5D)
As we can see the force exerted by the Earth is 2000 times greater than that exerted by a building with the proposed data.
Answer:
The answer is Mechanical Energy
Answer:
It is exerted in all directions at sea level, and this is because air molecules move in all directions.
Answer:
a. 4662.7W/m^2
b. the damage threshold is 100W/m^2 , therefore the laser is not safe to view if the intensity is 4662.7W/m^2
c. Bmax=6.24*10^-6T
d average intensity=0.01W/cm^2
Explanation:
A He-Ne laser produces light of 633 nm wavelength, 1.5 mW power, with a cylindrical beam of 0.64 mm in diameter.
(a) What is the intensity of this laser beam?
(b) The damage threshold of the retina is 100 W/m2 . Is this laser safe to view head-on?
(c) What are the maximum values of the electric and magnetic fields?
(d) What is the average energy density in the laser beam?
firstly, we get the relation between power and intensity to be
P=IA
1.5*10^-3=I*
I=4662.74W/m^2
b. the damage threshold is 100W/m^2 , therefore the laser is not safe to view if the intensity is 4662.7W/m^2
c.e=
e=(3512391.71)^0.5
e=1874.13W/m^2
maximum values of electric and magnetic fields
Bmax=e/c
c=speed of light 3*10^8 m/s^2
Bmax=1874/3*10^8
Bmax=6.24*10^-6T
d.intensity=average intensity(1/10^2)^2
intensity=0.01W/cm^2
Answer:
1)) ΔU = -8.96 J, 2) k = 8.18 10⁴ N / m, 3) v = 8.47 m / s
Explanation:
For this exercise we will use conservation of energy.
Starting point. Point where the pineapple comes out
Em₀ = U = m g h
where the reference frame is placed on the ground
Final point. Point where pineapple stops
Em_f = K_e + U = ½ k y² + m g y
1) the change in gravitational potential energy is
ΔU = U_f - U₀
ΔU = m g y - m g h
ΔU = mg (y-h)
let's calculate
ΔU = 0.116 9.8 (0.0148 - 7.9)
ΔU = -8.96 J
The negative sign indicates that the energy decreases
2) let's use energy conservation
Em₀ = Em_f
mg h = ½ k y² + mg y
k = mg (h-y) 
let's calculate
k = 0.116 9.8 (7.9 - 0.0148) 
k = 8.18 10⁴ N / m
3) we use the same starting point and as the end point we use this height (y₂ = 4.24 m)
Em_{f2} = K + U = ½ m v² + mg y₂
energy is conserved
Em₀ = Em_{f2}
mgh = ½ m v² + m g y₂
v =
let's calculate
v = 
v = 8.47 m / s
