To a stationary observer, a man jogs east at 2.5 m/s and a woman jogs west at 1.5 m/s. from the woman's frame of reference, what
2 answers:
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Answer:
C. The left wire attracts the right wire and exerts as much force as the right wire does.
Explanation:
To know what is the answer you first take into account the magnetic field generated by each current, for a distance of d:
Next, you use the formula for the magnetic force produced by the wires:
if the direction of the L vector is in +k direction, the first wire produced a magnetic field with direction +y, that is, +j and the second wire produced magnetic field with direction -y, that is, -j (this because the direction of the magnetic field is obtained by suing the right hand rule). Hence, the direction of the magnetic force on each wire, produced by the other one is:
Hence, due to this result you have that:
C. The left wire attracts the right wire and exerts as much force as the right wire does.
Answer:
The angular velocity is 15.37 rad/s
Solution:
As per the question:
Horizontal distance, x = 30.1 m
Distance of the ball from the rotation axis is its radius, R = 1.15 m
Now,
To calculate the angular velocity:
Linear velocity, v =
v =
v =
v =
Now,
The angular velocity can be calculated as:
Thus
Answer:
A) 667 J
B) 381.4 J
C) 0 J
D) 245.4 J
E) 40.2J
F) 2 m/s
Explanation:
Let g = 9.81 m/s2
A) The work done on the suitcase is the product of the force applied and the distance travelled:
w = Fs = 145 * 4.6 = 667 J
B) The work done by gravitational force the dot product between the gravity vector and the distance vector
C) As the normal force vector is perpendicular to the distance vector, the work done by the normal force is 0
D) The work done on the suitcase by friction force is the product of the force applied and the distance travelled, whereas friction force is the product of normal force and coefficient
E) The total workdone on the suite case would be the pulling work subtracted by gravity work and friction work
F) As the suit case has 0 kinetic and potential energy at the bottom, and the total work done is converted to kinetic energy at 4.6 m along the ramp, we can conclude that: