Abby throws a ball straight up and times it. she sees that the ball goes by the top of a flagpole after 0.50 s and reaches the l
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Time=50s
speed=25m/s
Distance = speed×time
=25×50
=1250m
DISTANCE TRAVELLED IS =1250m
speed=25m/s
Distance = speed×time
=25×50
=1250m
DISTANCE TRAVELLED IS =1250m
Missing question in the text:
"A.What are the magnitude and direction of the electric field at the point in question?
"A.What are the magnitude and direction of the electric field at the point in question?
B.<span>What would be the magnitude and direction of the force acting on a proton placed at this same point in the electric field?"</span>
<span>Solution:
A) A charge q </span>under an electric field of intensity E will experience a force F equal to:
In our problem we have and
, so we can find the magnitude of the electric field:
The charge is negative, therefore it moves against the direction of the field lines. If the force is pushing down the charge, then the electric field lines go upward.
B) The proton charge is equal to
Therefore, the magnitude of the force acting on the proton will be
And since the proton has positive charge, the verse of the force is the same as the verse of the field, so upward.
Answer:
Explanation:
<u>Net Forces and Acceleration</u>
The second Newton's Law relates the net force acting on an object of mass m with the acceleration a it gets. Both the net force and the acceleration are vector and have the same direction because they are proportional to each other.
According to the information given in the question, two forces are acting on the swimming student: One of 256 N pointing to the south and other to the west of 104 N. Since those forces are not aligned, we must add them like vectors as shown in the figure below.
The magnitude of the resulting force is computed as the hypotenuse of a right triangle
The acceleration can be obtained from the formula
Note we are using only magnitudes here
