The magnetic force exerted by a field E to a charge q is given by F=Eq. In this case, F=4.30*10^4*(6.80mu C). 1mu C=10^-6C, so F=4.30*6.80=10^-2=0.29N. The direction is in the x direction, the direction that the field is applied because the charge is positive.
Answer: (-4.3, 2.5)
Explanation:
In the second quadrant, magnitude of the vector is 5 and angle is 30° from the negative x-axis.
We can write this in terms of its components:

Thus, the components of vector in the second quadrant are (-4.3, 2.5)
The object has been moved by 5 m
Explanation:
The work done by a force when moving an object (which is equal to the energy used to move the object) is given by

where
F is the magnitude of the force
d is the displacement of the object
is the angle between the direction of the force and of the displacement
In this problem, we have
W = 35 J is the work done
F = 7.0 N is the magnitude of the force
, assuming the force is applied parallel to the direction of motion of the object
Therefore, we can solve the formula for d to find the displacement of the object:

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Answer:
I believe it's A frontal. Hope this can help
Answer:
The net force will be:

Explanation:
The net force is given by:


I hope it helps you!