A force of 15 N is applied to a spring, causing it to stretch 0. 3 m. What is the spring constant for this particular spring? N/
1 answer:
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Answer:
(a) 7.315 x 10^(-14) N
(b) - 7.315 x 10^(-14) N
Explanation:
As you referred at the final remark, the electron and proton undergo a magnetic force of same magnitude but opposite direction. Using the definition of magnetic force, a cross product must be done. One technique is either calculate the magnitude of the velocity and magnetic field and multiplying by sin (90°), but it is necessary to assure both vectors are perpendicular between each other ( which is not the case) or do directly the cross product dealing with a determinant (which is the most convenient approach), thus,
(a) The electron has a velocity defined as:
In respect to the magnetic field;
The magnetic force can be written as;
Bear in mind
thus,
Note: The cross product is operated as a determinant. Likewise, the product of the unit vector k is squared and that is operated as dot product whose value is equal to one, i.e,
(b) Considering the proton charge has the same magnitude as electron does, but the sign is positive, thus
Note: The cross product is operated as a determinant. Likewise, the product of the unit vector k is squared and that is operated as dot product whose value is equal to one, i.e,
Final remarks: The cross product was performed in R3 due to the geometrical conditions of the problem.
Answer:
(a) FN = m (g - )
(b) vmin = 17.146 m/s
Explanation:
The radius of the arc is
r = 30m
The normal force acting on the car form the highest point is
FN = m (g - )
If the normal force become 0 we have
m (g - ) = 0
or
g - = 0
This way, when FN = 0, then v = vmin, so
g - = 0
vmin = =