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Ierofanga [76]
3 years ago
8

An electron moves in the plane of this screen toward the top of the screen. A magnetic field is also in the plane of the screen

and directed toward the right. What is the direction of the magnetic force on the electron?
Physics
1 answer:
Ad libitum [116K]3 years ago
6 0

Answer: the direction of the magnetic force on the electron will be moving out of the screen, perpendicular to the magnetic field.

Explanation:

The magnetic force F on a moving electron at right angle to a magnetic field is given by the formula:

F = BqVSinØ

If an electron moves in the plane of this screen toward the top of the screen. A magnetic field is also in the plane of the screen and directed toward the right. Then, the direction of the magnetic force on the electron will be perpendicular to the magnetic field

According to the Fleming's left - hand rule, the direction of the magnetic force on the electron will be moving out of the plane of the screen.

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A ball is thrown straight up. What are the velocity and acceleration of the ball at the highest point in its path?
zubka84 [21]

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.

8 0
3 years ago
A ball has a mass of 1.5kg and is thrown straight up with a speed of 60m/s, what is the ball’s momentum:
madam [21]

Answer:

Assumption: the air resistance on this ball is negligible. Take g = 10\; \rm m \cdot s^{-2}.

a. The momentum of the ball would be approximately 60\;\rm kg \cdot m \cdot s^{-1} two seconds after it is tossed into the air.

b. The momentum of the ball would be approximately \rm \left(-45\; \rm kg \cdot m \cdot s^{-1}\right) three seconds after it reaches the highest point (assuming that it didn't hit the ground.) This momentum is smaller than zero because it points downwards.

Explanation:

The momentum p of an object is equal its mass m times its velocity v. That is: \vec{p} = m \cdot \vec{v}.

Assume that the air resistance on this ball is negligible. If that's the case, then the ball would accelerate downwards towards the ground at a constant g \approx -10\; \rm m \cdot s^{-2}. In other words, its velocity would become approximately 10\; \rm m \cdot s^{-1} more negative every second.

The initial velocity of the ball is 60\; \rm m \cdot s^{-1}. After two seconds, its velocity would have become 60\;\rm m \cdot s^{-1} + 2\; \rm s \times \left(-10\;\rm m \cdot s^{-1}\right) = 40\; \rm m \cdot s^{-1}. The momentum of the ball at that time would be around p = m \cdot v \approx 60\; \rm kg \cdot m \cdot s^{-1}.

When the ball is at the highest point of its trajectory, the velocity of the ball would be zero. However, the ball would continue to accelerate downwards towards the ground at a constant g \approx -10\; \rm m \cdot s^{-2}. That's how the ball's velocity becomes negative.

After three more seconds, the velocity of the ball would be 0\; \rm m \cdot s^{-1} + 3\; \rm s \times \left(-10\; \rm m \cdot s^{-2}\right) = -30 \; \rm m \cdot s^{-1}. Accordingly, the ball's momentum at that moment would be p = m \cdot v \approx \left(-45\; \rm kg \cdot m \cdot s^{-1}\right).

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Answer:

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Explanation:

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