If you were going to describe the relationship between current, voltage, and power, you could say: (1 point) Group of answer cho
1 answer:
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The potential energy of the rock is 30,000 J
Explanation:
The mechanical energy of an object is equal to the sum of its gravitational potential energy (PE) and its kinetic energy (KE):
where
PE is the gravitational potential energy, which is the energy possessed by the object due to its position in the gravitational field
KE is the kinetic energy, which is the energy possessed by the object due to its motion
In this problem, the rock is at rest, so its kinetic energy is zero:
KE = 0
Therefore, the energy of the rock is just equal to its potential energy, which is:
where
m = 10.2 kg is the mass of the rock
is the acceleration of gravity
h = 300 m is the height of the rock above the ground
Substituting and solving, we find
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The magnitude of the force on the charge by the influence of the magnetic field will be 6.6*10^-3 N
<h3>What is magnetic force?</h3>Magnetic force, attraction or repulsion that arises between electrically charged particles because of their motion.The magnitude of the magnetic force acting on the charge is given by:
where
The magnitude of the charge
The velocity of the charge
The magnitude of the magnetic field
The angle between the directions of v and B
By substituting the values we will get:
Hence the magnitude of the force on the charge by the influence of the magnetic field will be 6.6*10^-3 N
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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