Answer:
down below
Explanation:
Since we aren't the given the time, lets say that an object to 25 seconds to fall 50 meters. We can use the formula [ s = d/t ] to solve.
s = 50/25
s = 2
Therefore, the object was falling at a rate of 2 meters per second.
Best of Luck!
Answer:
This can be the FBD of the bag.
(my bag looks more like a box tho ^^")
Answer:
F = 8.6 10⁻¹² N
Explanation:
For this exercise we use the law of conservation of energy
Initial. Field energy with the electron at rest
Em₀ = U = q ΔV
Final. Electron with velocity, just out of the electric field
Emf = K = ½ m v²
Em₀ = Emf
e ΔV = ½ m v²
v =√ 2 e ΔV / m
v = √(2 1.6 10⁻¹⁹ 51400 / 9.1 10⁻³¹)
v = √(1.8075 10¹⁶)
v = 1,344 10⁸ m / s
Now we can use the equation of the magnetic force
F = q v x B
Since the speed and the magnetic field are perpendicular the force that
F = e v B
F = 1.6 10⁻¹⁹ 1.344 10⁸ 0.4
For this exercise we use the law of conservation of energy
Initial. Field energy with the electron at rest
Emo = U = q DV
Final. Electron with velocity, just out of the electric field
Emf = K = ½ m v2
Emo = Emf
.e DV = ½ m v2
.v = RA 2 e DV / m
.v = RA (2 1.6 10-19 51400 / 9.1 10-31)
.v = RA (1.8075 10 16)
.v = 1,344 108 m / s
Now we can use the equation of the magnetic force
F = q v x B
Since the speed and the magnetic field are perpendicular the force that
F = e v B
F = 1.6 10-19 1,344 108 0.4
F = 8.6 10-12 N
The work W done by the electric field in moving the proton is equal to the difference in electric potential energy of the proton between its initial location and its final location, therefore:
where q is the charge of the proton,
, with
being the elementary charge, and
and
are the initial and final voltage.
Substituting, we get (in electronvolts):
and in Joule:
