Classify each of the following statements as a characteristic (a) of electric forces only, (b) of magnetic forces only, (c) of b
1 answer:
(b) Magnetic forces only
When an object is placed in an electric field it experiences an electric force given by :
= |q| × E
Similarly when it is placed in a magnetic field it experiences a magnetic force given by :
= |q|×v×Bsinθ
When a particle is placed in electric field the particle gets accelerated.
The electric field has a direction, positive to negative. This is the direction that the electric field will cause a positive charge to accelerate. For example If a positive charge is moving in the same direction as the electric field vector the particle's velocity will increase. If it is moving in the opposite direction it will decelerate. ( No change in direction of motion in this case)
When a particle is moved in magnetic field ,the magnetic force is perpendicular to velocity and magnetic field . Since force is perpendicular to velocity ,it only changes the direction of motion.
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Answer:
F = 520 N
Explanation:
For this exercise the rotational equilibrium equation should be used
Σ τ = 0
Let's set a reference system with the origin at the back of the refrigerator and the counterclockwise rotation as positive. On the x-axis it is horizontal directed outward, eg the horizontal y-axis directed to the side and the z-axis vertical
Torque is
τ = F x r
the bold indicate vectors, we analyze each force
the applied force is horizontal along the -x axis, the arm (perpendicular distance) is directed in the z axis,
The weight of the body is the vertical direction of the z-axis, so the arm is on the x-axis
-F z + W x = 0
F z = W x
F = W
The exercise indicates the point of application of the force z = 1.5 m and the weight is placed in the center of mass of the body x = 0.6 m, we are assuming that the force is applied in the wide center of the refrigerator
let's calculate
F = 1300 0.6 / 1.5
F = 520 N
Answer:
(a) W = 650J
(b) Wf = 529.2J
(c) W = 0J
(d) W = 0J
(e) ΔK = 120.8J
(f) v2 = 2.58 m/s
Explanation:
(a) In order to find the work done by the applied force you use the following formula:
(1)
F: applied force = 130N
d: distance = 5.0m
The work done by the applied force is 650J
(b) The increase in the internal energy of the box-floor system is given by the work done of the friction force, which is calculated as follow:
(2)
μ: coefficient of friction = 0.300
M: mass of the box = 36.0kg
g: gravitational constant = 9.8 m/s^2
The increase in the internal energy is 529.2J
(c) The normal force does not make work on the box because the normal force is perpendicular to the motion of the box.
(d) The same for the work done by the normal force. The work done by the gravitational force is zero because the motion of the box is perpendicular o the direction of the gravitational force.
(e) The change in the kinetic energy is given by the net work on the box. You use the following formula:
(3)
You calculate the total work:
(4)
F: applied force = 130N
Ff: friction force
d: distance = 5.00m
The friction force is:
Next, you replace the values of all parameters in the equation (4):
The change in the kinetic energy of the box is 120.8J
(e) The final speed of the box is calculated by using the equation (3):
(5)
v2: final speed of the box
v1: initial speed of the box = 0 m/s
You solve the equation (5) for v2:
The final speed of the box is 2.58m/s