An 800-kHz radio signal is detected at a point 4.5 km distant from a transmitter tower. The electric field amplitude of the sign
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Answer:
The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the field.
The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the current.
The magnetic force on the current-carrying wire is strongest when the current is perpendicular to the magnetic field lines.
Explanation:
The magnitude of the magnetic force exerted on a current-carrying wire due to a magnetic field is given by
(1)
where I is the current, L the length of the wire, B the strength of the magnetic field, the angle between the direction of the field and the direction of the current.
Also, B, I and F in the formula are all perpendicular to each other. (2)
According to eq.(1), we see that the statement:
<em>"The magnetic force on the current-carrying wire is strongest when the current is perpendicular to the magnetic field lines.</em>"
is correct, because when the current is perpendicular to the magnetic field, and the force is maximum.
Moreover, according to (2), we also see that the statements
<em>"The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the field. "</em>
<em>"The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the current. "</em>
because F (the force) is perpendicular to both the magnetic field and the current.
(a) 5.65 times the Earth's radius
The escape velocity for a projectile on Earth is
where
G is the gravitational constant
M is the Earth's mass
R is the Earth's radius
If the projectile has an initial speed of 0.421 escape speed,
So its initial kinetic energy will be
where m is the mass of the projectile
At the point of maximum altitude, all this energy is converted into gravitational potential energy:
where r is the distance from the Earth's centre reached by the projectile. We can write r as a multiple of R, the Earth's radius:
And solving the equation we find
So, the projectile reaches a radial distance of 5.65 times the Earth's radius.
b) 2.36 times the Earth's radius
The kinetic energy needed to escape is:
This time, the projectile has 0.421 times this energy:
Again, at the point of maximum altitude, all this energy will be converted into potential energy:
and by solving for n we find
So, the projectile reaches a radial distance of 2.36 times the Earth's radius.
c)
The least initial mechanical energy needed for the projectile to escape Earth is equal to the gravitational potential energy of the projectile at the Earth's surface:
Indeed, the kinetic energy of the projectile must be equal to this value. In fact, if we use the formula of the escape velocity inside the formula of the kinetic energy, we find
Answer:
The magnitud of the velocity is
and the direccion:
degrees from the horizontal.
Explanation:
Fist we define our variables:
The letters i and j represent the direction of the movement, i in this case is the horizontal direction, and j is perpendicular to i.
velocities with sub-index 1 are the speeds before the crash, and with sub-index 2 are the velocities after the crash.
Using conservation of momentum:
Clearing for the velocity of the stone after the crash:
Substituting known values:
The magnitud of the velocity is :
and the direction:
this is -28.3 degrees from the +i direction or the horizontal direcction.
Note: i and j can also be seen as x and y axis.