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5 days ago

Points A, B, and C lie along a line from left to right, respectively. Point B is at a lower electric potential than point A. Poi

nt C is at a lower electric potential than point B. Which one of the following statements best describes the subsequent motion, if any, of a positively-charged particle released from rest at point B?
A) The particle will accelerate in the direction of point C.
B) The particle will move at constant velocity in the direction of point C.
C) The particle will remain at rest.
D) The particle will move at constant velocity in the direction of point A.
E) The particle will accelerate in the direction of point A.

Physics

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 A bartender slides a beer mug at 1.50 m/s toward a customer at the end of a frictionless bar that is 1.20 m tall. The customer

serg [3582]

Response:

a) The mug makes contact with the ground 0.7425m from the bar's end. b) |V|=5.08m/s θ= -72.82°

Clarification:

To address this issue, we begin with a diagram depicting the situation. (refer to the attached illustration).

The illustration shows that the problem involves motion in two dimensions. To determine how far from the bar the mug lands, we need to find the time the mug remains airborne by examining its vertical motion.

To compute the time, we utilize the following formula with the known values:

$y_{f}=y_{0}+v_{y0}t+\frac{1}{2}at^{2}$

We have $y_{f}=0$ and $v_{y0}=0$ , allowing us to simplify the equation to:

$0=y_{0}+\frac{1}{2}at^{2}$

Now, we can calculate for t:

$-y_{0}=\frac{1}{2}at^{2}$

$-2y_{0}=at^{2}$

$\frac{-2y_{0}}{a}=t^{2}$

$t=\sqrt{\frac{-2y_{0}}{a}}$

We know $y_{0}=1.20m$ and $a=g=-9.8m/s^{2}$

The negative gravity indicates the downward motion of the mug. Hence, we substitute these values into the provided formula:

$t=\sqrt{\frac{-2(1.20m)}{(-9.8m/s^{2})}}$

Which results in:

t=0.495s

This time helps us evaluate the horizontal distance the mug traverses. Since:

$V_{x}=\frac{x}{t}$

Solving for x, we have:

$x=V_{x}t$

Substituting the known values yields:

x=(1.5m/s)(0.495s)

This calculates to:

x=0.7425m

b) With the time determining when the mug strikes the ground established, we can find the final velocity in the vertical direction using the formula:

$a=\frac{v_{f}-v_{0}}{t}$

The initial vertical velocity being zero simplifies our calculations:

$a=\frac{v_{f}}{t}$

Thus, we can determine the final velocity:

$V_{yf}=at$

Given that the acceleration equates to gravity (showing a downward effect), we substitute that alongside the previously found time:

$V_{yf}=(-9.8m/s^{2})(0.495s)$

This leads to:

$V_{yf}=-4.851m/s$

Now, we ascertain the velocity components:

$V_{xf}=1.5m/s$ and $V_{yf]=-4.851m/s$

Next, we find the speed by calculating the vector's magnitude:

$|V|=\sqrt{V_{x}^{2}+V_{y}^{2}}$

<pThus:

$|V|=\sqrt{(1.5m/s)^{2}+(-4.851m/s)^{2}$

Yielding:

|V|=5.08m/s

Lastly, to ascertain the impact direction, we apply the equation:

$\theta = tan^{-1} (\frac{V_{y}}{V_{x}})$

<pFulfilling this provides:

$\theta = tan^{-1} (\frac{-4.851m/s}{(1.5m/s)})$

<pLeading to:

$\theta = -72.82^{o}$

4 0

2 months ago

Ellen does an experiment by releasing a ball from a height of 1 m above each floor in a tall building. She records the time it t

serg [3582]

The question Ellen is likely exploring is "In what way does distance influence the gravitational force acting on objects?"

Explanation:

7 0

2 months ago

Read 2 more answers

Two racecars are driving at constant speeds around a circular track. Both cars are the same distance away from the center of the

Yuliya22 [3333]

Response:

The acceleration of car 2 is four times that of car 1.

Rationale:

Centripetal acceleration occurs when an object travels in a circular route. It can be expressed as:

$a=\dfrac{v^2}{r}$

In this scenario, two race cars are moving at consistent speeds around a circular course. Both automobiles are located at an equal distance from the center, but car 2 is operating at twice the speed of car 1.

Thus,

$\dfrac{a_1}{a_2}=\dfrac{v_1^2}{v_2^2}$

1 and 2 represent the first and second cars, respectively.

$v_2=2v_1$

Consequently,

$\dfrac{a_1}{a_2}=\dfrac{v_1^2}{(2v_1)^2}\\\\\dfrac{a_1}{a_2}=\dfrac{v_1^2}{4v_1^2}\\\\\dfrac{a_1}{a_2}=\dfrac{1}{4}\\\\a_2=4\times a_1$

Therefore, car 2's acceleration is four times that of car 1.

4 0

2 months ago

A 250 GeV beam of protons is fired over a distance of 1 km. If the initial size of the wave packet is 1 mm, find its final size

Maru [3345]

Answer:

The final size is nearly the same as the initial size because the increase in size $1.055\times 10^{- 7}$ is remarkably small

Solution:

According to the problem:

The proton beam energy is E = 250 GeV = $250\times 10^{9}\times 1.6\times 10^{- 19} = 4\times 10^{- 8} J$

Distance traveled by the photon, d = 1 km = 1000 m

Proton mass, $m_{p} = 1.67\times 10^{- 27} kg$

Initial size of the wave packet, $\Delta t_{o} = 1 mm = 1\times 10^{- 3} m$

Now,

This operates under relativistic principles

The rest mass energy for the proton is expressed as:

$E = m_{p}c^{2}$

$E = 1.67\times 10^{- 27}\times (3\times 10^{8})^{2} = 1.503\times 10^{- 10} J$

This proton energy is $\simeq 250 GeV$

Thus, the speed of the proton, v $\simeq c$

The time to cover 1 km = 1000 m of distance is calculated as:

T = $\frac{1000}{v}$

T = $\frac{1000}{c} = \frac{1000}{3\times 10^{8}} = 3.34\times 10^{- 6} s$

According to the dispersion factor;

$\frac{\delta t_{o}}{\Delta t_{o}} = \frac{ht_{o}}{2\pi m_{p}\Delta t_{o}^{2}}$

$\frac{\delta t_{o}}{\Delta t_{o}} = \frac{6.626\times 10^{- 34}\times 3.34\times 10^{- 6}}{2\pi 1.67\times 10^{- 27}\times (10^{- 3})^{2} = 1.055\times 10^{- 7}$

Thus, the widening of the wave packet is relatively minor.

Hence, we can conclude that:

$\Delta t_{o} = \Delta t$

where

$\Delta t$ = final width

3 0

2 months ago