Answer:
b. v = 0, a = 9.8 m/s² down.
Explanation:
Hi there!
The acceleration of gravity is always directed to the ground (down) and, near the surface of the earth, has a constant value of 9.8 m/s². Since the answer "b" is the only option with an acceleration of 9.8 m/s² directed downwards, that would solve the exercise. But why is the velocity zero at the highest point?
Let´s take a look at the height function:
h(t) = h0 + v0 · t + 1/2 g · t²
Where
h0 = initial height
v0 = initial velocity
t = time
g = acceleration due to gravity
Notice that the function is a negative parabola if we consider downward as negative (in that case "g" would be negative). Then, the function has a maximum (the highest point) at the vertex of the parabola. At the maximum point, the slope of the tangent line to the function is zero, because the tangent line is horizontal at a maximum point. The slope of the tangent line to the function is the rate of change of height with respect to time, i.e, the velocity. Then, the velocity is zero at the maximum height.
Another way to see it (without calculus):
When the ball is going up, the velocity vector points up and the velocity is positive. After reaching the maximum height, the velocity vector points down and is negative (the ball starts to fall). At the maximum height, the velocity vector changed its direction from positive to negative, then at that point, the velocity vector has to be zero.
Answer:
B = 9.16 10⁻² T
Explanation:
The speed selector is a configuration where the electric and magnetic force has the opposite direction, which for a specific speed cancel
q v B = q E
v = E / B
B = E / v
Let's calculate
B = 4.4 10⁵ / 4.8 10⁶
B = 9.16 10⁻² T
Answer:
<h2>3000 N</h2>
Explanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × acceleration
From the question we have
force = 1000 × 3
We have the final answer as
<h3>3000 N</h3>
Hope this helps you
Answer:
1. False
2. True
3. True
Explanation:
1- False —> The relation between electric potential and electric field is given such that
Therefore, for a uniform E field, electric potential is linearly proportional to the distance.
2- True —> The electric field lines always cross the equipotential lines perpendicularly.
3- True —> In order to be a potential difference, one source of electric field is enough. The electric potential will decrease radially according to the following formula:
There is no test charge in the formula, only the source charge. Even when there is no test charge, the potential difference between points in space can exist.
