Please I need help on this
1 answer:
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Answer:
E = (-3.61^i+1.02^j) N/C
magnitude E = 3.75N/C
Explanation:
In order to calculate the electric field at the point P, you use the following formula, which takes into account the components of the electric field vector:
(1)
Where the minus sign means that the electric field point to the charge.
k: Coulomb's constant = 8.98*10^9Nm^2/C^2
q = -4.28 pC = -4.28*10^-12C
r: distance to the charge from the point P
The point P is at the point (0,9.83mm)
θ: angle between the electric field vector and the x-axis
The angle is calculated as follow:
The distance r is:
You replace the values of all parameters in the equation (1):
The electric field is E = (-3.61^i+1.02^j) N/C with a a magnitude of 3.75N/C
1) 9.4 m/s
First of all, we can calculate the work done by the horizontal force, given by
W = Fd
where
F = 34.6 N is the magnitude of the force
d = 12.9 m is the displacement of the cart
Solving ,
W = (34.6 N)(12.9 m) = 446.3 J
According to the work-energy theorem, this is also equal to the kinetic energy gained by the cart:
Since the cart was initially at rest, , so
(1)
where
m is the of the cart
v is the final speed
The mass of the cart can be found starting from its weight, :
So solving eq.(1) for v, we find the final speed of the cart:
2)
The work done on the train is given by
W = Fd
where
F is the magnitude of the force
d is the displacement of the train
In this problem,
So the work done is
3)
According to the work-energy theorem, the change in kinetic energy of the train is equal to the work done on it:
where
W is the work done
is the change in kinetic energy
Therefore, the change in kinetic energy is
4) 37.2 m/s
According to the work-energy theorem,
where
is the final kinetic energy of the train
is the initial kinetic energy of the train, which is zero since the train started from rest
Re-writing the equation,
where
m = 36300 kg is the mass of the train
v is the final speed of the train
Solving for v, we find