Answer:
Magnetic field, B = 0.042 T
Explanation:
It is given that,
Speed of charged particle,
Angle between velocity and the magnetic field,
Charge,
Magnetic force, F = 0.002 N
The magnetic force is given by :
B is the magnetic field
B = 0.042 T
So, the strength of the magnetic field is 0.042 Tesla. Hence, this is the required solution.
I feel like it could be A
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
1. The time for a radioactive sample to reduce to half of its original mass.
4. 10% of 232 is 23.2 times 6 equals 139.2
68.9 x6 equals 413.4
5. 0.37kg
6. 2000 years
7. 6.84 seconds
Answer: 100 suns
Explanation:
We can solve this with the following relation:
Where:
is the diameter of a dime
is the diameter of the Sun
is the distance between the Sun and the pinhole
is the amount of dimes that fit in a distance between the sunball and the pinhole
Finding :
This is roughly the diameter of the Sun
Now, the distance between the Earth and the Sun is one astronomical unit (1 AU), which is equal to:
So, we have to divide this distance between in order to find how many suns could it fit in this distance: