Determine the magnitude and direction of the force on an electron traveling 3.58 E 6 m/s horizontally to the west in a verticall
1 answer:
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The question is incomplete but still I answer to assume your thinking.
The picture is attached below!.
Here,
F is the force with which you pull up the incline.
N is the normal force.
w is the weight acting downward.
Axis are mentioned in the attached picture.
Concept:
You can see there is no movement of object in the y-direction that means acceleration is zero in y-direction, sum of all the forces in y-direction equal to zero.
According to newton second law,
<span>∑ F = ma
</span>As, acceleration is zero in y-direction, so right hand side is zero in the above equation.
<span>∑ F = 0</span>
N-wcosθ=0
N= m*g*cos25°
N= m*(9.8)*(0.9063)
N= 8.8817*m
By putting the value of mass(m)(not given in the question) you will get the answer.
Hopefully, this is the answer of your question.
The picture is attached below!.
Here,
F is the force with which you pull up the incline.
N is the normal force.
w is the weight acting downward.
Axis are mentioned in the attached picture.
Concept:
You can see there is no movement of object in the y-direction that means acceleration is zero in y-direction, sum of all the forces in y-direction equal to zero.
According to newton second law,
<span>∑ F = ma
</span>As, acceleration is zero in y-direction, so right hand side is zero in the above equation.
<span>∑ F = 0</span>
N-wcosθ=0
N= m*g*cos25°
N= m*(9.8)*(0.9063)
N= 8.8817*m
By putting the value of mass(m)(not given in the question) you will get the answer.
Hopefully, this is the answer of your question.
Answer: