In a certain region of space, the electric potential is V(x,y,z)=Axy−Bx^2+Cy, where A, B, and C are positive constants. (a) Calc
1 answer:
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Explanation:
The given data is as follows.
F =
q =
v = 385 m/s
= 0.876
Now, we will calculate the magnitude of magnetic field as follows.
B =
=
= T
= 10.65 T
Thus, we can conclude that magnitude of the magnetic field is 10.65 T.
Answer:
W = (F1 - mg sin θ) L, W = -μ mg cos θ L
Explanation:
Let's use Newton's second law to find the friction force. In these problems the x axis is taken parallel to the plane and the y axis perpendicular to the plane
Y Axis
N - =
N = W_{y}
X axis
F1 - fr - Wₓ = 0
fr = F1 - Wₓ
Let's use trigonometry to find the components of the weight
sin θ = Wₓ / W
cos θ = W_{y} / W
Wₓ = W sin θ
W_{y} = W cos θ
We substitute
fr = F1 - W sin θ
Work is defined by
W = F .dx
W = F dx cos θ
The friction force is parallel to the plane in the negative direction and the displacement is positive along the plane, so the Angle is 180º and the cos θ= -1
W = -fr x
W = (F1 - mg sin θ) L
Another way to calculate is
fr = μ N
fr = μ W cos θ
the work is
W = -μ mg cos θ L