A beam of protons is directed in a straight line along the +z direction through a region of space in which there are crossed ele
1 answer:
Answer:
B = 7.6 T direction of + x
Explanation:
For the proton beam to continue in the same direction the electric and magnetic forces must be equal
= 0
= F_{e}
Fm = q E
The electric force is in the direction of the electric field because it is the charge of the positive proton, the electric force goes in the direction of –y, therefore, the magnetic force cancels this force must go in the direction of + y
The magnetic force is
F_{m} = q v x B = q v B sin θ
θ = 90
B = q E / q v
B = E / v
B = 800/105
B = 7.6 T
To find the direction of the magnetic field we use the right hand rule, the thumb goes in the direction of the proton velocity, the fingers extended in the direction of the magnetic field and the palm is the direction of force, for a positive charge.
Thumb goes in the direction of the + z axis
Palm in the direction of +y
Fingers point in the direction of + x
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Answer:
Explanation:
The total energy of the satellite when it is still in orbit is given by the formula
where
G is the gravitational constant
m = 525 kg is the mass of the satellite
is the Earth's mass
r is the distance of the satellite from the Earth's center, so it is the sum of the Earth's radius and the altitude of the satellite:
So the initial total energy is
When the satellite hits the ground, it is now on Earth's surface, so
so its gravitational potential energy is
And since it hits the ground with speed
it also has kinetic energy:
So the total energy when the satellite hits the ground is
So the energy transformed into internal energy due to air friction is the difference between the total initial energy and the total final energy of the satellite:
Answer:
r2 = 1 m
therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m
Explanation:
For this exercise we must use conservation of energy
the electric potential energy is
U =
for the proton at x = -1 m
U₁ =
for the electron at x = 1 m
U₂ =
starting point.
Em₀ = K + U₁ + U₂
Em₀ =
final point
Em_f =
energy is conserved
Em₀ = Em_f
\frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e^2 (- \frac{1}{r_2 +1} + \frac{1}{r_2 -1})
\frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e²( )
we substitute the values
½ 9.1 10⁻³¹ 450 + 9 10⁹ (1.6 10⁻¹⁹)² [ ) = 9 109 (1.6 10-19) ²(
)
2.0475 10⁻²⁸ + 2.304 10⁻³⁷ (5.0125 10⁻³) = 4.608 10⁻³⁷ ( )
2.0475 10⁻²⁸ + 1.1549 10⁻³⁹ = 4.608 10⁻³⁷
r₂² -1 = (4.443 10⁸)⁻¹
r2 =
r2 = 1 m
therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m