For each of the motions described below, determine the algebraic sign (+, -, or 0) of the velocity and acceleration of the objec
1 answer:
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- The greatest potential energy increase occurs when the charge travels north. This happens because the charge is negative, which means it gains potential energy when moving in the same direction as the field (in contrast, a positive charge moving along the field loses potential energy, converting it to kinetic energy). The potential energy gained is calculated as the charge multiplied by the distance moved:
- The next largest increase occurs as the charge moves east. Here, the change in potential energy is actually zero since the charge moves perpendicular to the field, traversing points with constant potential. Therefore, there is no variation in potential energy in this case:
- Finally, when the charge moves south, it experiences a reduction in potential energy. This is due to moving against the electric field, and since it is a negative charge, it loses potential energy in this direction, which transforms into kinetic energy. Thus, in this scenario:
- The next largest increase occurs as the charge moves east. Here, the change in potential energy is actually zero since the charge moves perpendicular to the field, traversing points with constant potential. Therefore, there is no variation in potential energy in this case:
- Finally, when the charge moves south, it experiences a reduction in potential energy. This is due to moving against the electric field, and since it is a negative charge, it loses potential energy in this direction, which transforms into kinetic energy. Thus, in this scenario:
Hypothesis: An increase in voltage should result in a corresponding rise in current because according to Ohm's Law,
Ohm's Law indicates that current is proportional to voltage when resistance remains constant. Hence, if resistance stays the same, elevating the voltage will lead to an increase in current. Conversely, if voltage remains unchanged and resistance increases, current will decrease.
Answer:
130 m/s (to two significant figures)
Explanation:
In projectile motion, the launching velocity and launch angle help to determine both the horizontal and vertical velocity components.
u represents the initial projectile velocity = 150 m/s
uₓ = u cos θ = 150 cos 30° = 129.9 m/s
uᵧ = u sin θ = 150 sin 30° = 75.0 m/s
A projectile's motion can be viewed as made up of independent vertical and horizontal elements.
The vertical motion is affected by gravitational acceleration (which pulls down on the projectile), altering the vertical velocity component due to this acting force.
Conversely, there is no acting force in the horizontal direction, which means the horizontal component maintains a steady velocity throughout the projectile's flight.
Thus, at t = 4 s, the horizontal component of the projectile's speed remains equal to the initial horizontal velocity component.
At t = 4 s, the horizontal component of velocity is uₓ = u cos θ = 150 cos 30° = 129.9 m/s ≈ 130 m/s
Answer:
a) ∆x∆v = 5.78*10^-5
∆v = 1157.08 m/s
b) 4.32*10^{-11}
Explanation:
This problem can be addressed using Heisenberg's uncertainty principle, which is expressed as:
Where h represents Planck’s constant (6.62*10^-34 J s).
Assuming that the electron's mass remains the same, we proceed as follows:
Utilizing the electron's mass (9.61*10^-31 kg) and the uncertainty in position (50 nm), we can compute ∆x∆v and ∆v:
If we treat the electron like a classic particle, the time required to cross the channel is determined using the upper limit of the uncertainty in velocity: