The short answer is that the displacement is equal tothe area under the curve in the velocity-time graph. The region under the curve in the first 4.0 s is a triangle with height 10.0 m/s and length 4.0 s, so its area - and hence the displacement - is
1/2 • (10.0 m/s) • (4.0 s) = 20.00 m
Another way to derive this: since velocity is linear over the first 4.0 s, that means acceleration is constant. Recall that average velocity is defined as
<em>v</em> (ave) = ∆<em>x</em> / ∆<em>t</em>
and under constant acceleration,
<em>v</em> (ave) = (<em>v</em> (final) + <em>v</em> (initial)) / 2
According to the plot, with ∆<em>t</em> = 4.0 s, we have <em>v</em> (initial) = 0 and <em>v</em> (final) = 10.0 m/s, so
∆<em>x</em> / (4.0 s) = (10.0 m/s) / 2
∆<em>x</em> = ((4.0 s) • (10.0 m/s)) / 2
∆<em>x</em> = 20.00 m
Answer:The electric field is zero and the potential is positive.
Explanation:
Two identical positive charges are separated by a certain distance and midway between charges two identical positive charges are placed near each other.
So the Electric field at midway is zero because the electric field due to both charges add up to give zero electric field.(because they point in opposite direction)
Potential is scalar quantity and charges are positive so they add up to give potential.
Answer:
The velocity of a falling object
Explanation:
The positive X axis is towards right and positive Y axis is towards up, so North direction is positive
A vector with less than 1 magnitude is not negative, because its magnitude may be in between 0 and 1 which is positive vector.
Any vector whose magnitude is greater than 1 is never be a negative vector.
The velocity of a falling object is towards bottom, that is towards negative Y axis. So that vector is negative.
You haven't said what 'high' resistance or 'low' current means, so there's way not enough info to nail the statement as true or false. The most precise answer is "certainly could be but not necessarily". Anyway, the current in the circuit depends on BOTH the resistance AND the voltage. So without knowing the voltage too, you can't say anything about the current.
I believe the acceleration would be 5m/s
All you would need to do is divide the final speed by the time it took to get there. I am only about 80 sure this answer is correct, so take my advise only if you feel comfortable.
