When boating in shallow areas or seagrass beds, you see a mud trail in your wake where your boat has churned up the bottom. If y

Two electrodes, separated by a distance d, in a vacuum are maintained at a constant potential difference. An electron, accelerat

Yuliya22 [3333]

Answer:

Explanation:

The distance between the electrodes is denoted as d.

The kinetic energy of the electron is represented as Ek when the electrodes are positioned at a distance of "d" apart.

Our goal is to determine the kinetic energy when they are separated by a distance of d/3.

K.E = ½mv²

It’s important to note that the mass remains constant; only velocity varies.

Additionally,

K.E = Work done by the electron

K.E = F × d

K.E = W = ma × d

Assuming constant acceleration

Hence, m and a are fixed,

therefore,

K.E is directly related to d

Thus, as d increases, K.E increases, and conversely, when d decreases, K.E decreases.

Consequently,

K.E_1 / d_1 = K.E_2 / d_2

With K.E_1 equating to E_k

and d_1 being d

while d_2 is represented as d/3

This leads to K.E_2 = K.E_1 / d_1 × d_2

Thus, K.E_2 = E_k × ⅓d / d

Finally,

K.E_2 = ⅓E_k

Therefore, the resultant kinetic energy is one third of the original E_k

7 0

1 month ago

A spinning wheel is slowed down by a brake, giving it a constant angular acceleration of 25.60 rad/s2. During a 4.20-s time inte

Yuliya22 [3333]

We will use the equations of rotational kinematics,

$\theta =\theta _{0} + \omega_{0} t+ \frac{1}{2}\alpha t^2$ (A)

$\omega^2= \omega^2_{0} +2\alpha\theta$ (B)

Here, $\theta$ and $\theta _{0}$ denote the final and initial angular displacements, respectively, whereas $\omega$ and $\omega_{0}$ represent final and initial angular velocities, and $\alpha$ is the angular acceleration.

We are provided with $\alpha = - 25.60 \ rad/s^2, \theta = 62.4 \ rad$ and $t = 4.20 \ s$ .

By substituting these values into equation (A), we have

$62.4 \ rad = 0 + \omega_{0} 4.20 \ s + \frac{1}{2} (- 25.60 \ rad/s^2) ( 4.20)^2 \\\\ \omega_{0} = \frac{220.5+ 62.4 }{4.20} =67.4 \ rad/s$

Now, using equation (B),

$\omega^2=(67.4 \ rad/s)^2 + 2 (- 25.60 \ rad/s^2)62.4 \ rad \\\\\ \omega = 36.7 \ rad/s$

This indicates that the wheel's angular speed at the 4.20-second mark is 36.7 rad/s.

4 0

2 months ago

Three pendulums with strings of the same length and bobs of the same mass are pulled out to angles θ1, θ2, and θ3, respectively,

Sav [3153]

Answer:

All three pendulums will have the same angular frequencies.

Explanation:

For a simple pendulum, the time period using the approximation $sin(\theta )\approx \theta$ is expressed as:

$T=2\pi\sqrt{\frac{l}{g}}$

The angular frequency $\omega$ is defined as

$\omega =\frac{2\pi }{T}\\\\\omega =\frac{2\pi}{2\pi \sqrt{\frac{l}{g}}}\\\\\therefore \omega =\sqrt{\frac{g}{l}}$

Since the angular frequency remains unaffected by the initial angle (valid strictly for small angle approximations), we deduce that the angular frequencies of the three pendulums are identical.

6 0

2 months ago

A supplier wants to make a profit by buying metal by weight at one altitude and selling it at the same price per pound weight at

Ostrovityanka [3204]

The appropriate choice is C.

In physics, the law of gravity helps us understand how gravity varies with height. As altitude increases, so too does the experience of gravity. Changes in altitude also result in variations in weight, though these differences are not particularly significant. Consequently, weighing metals at different heights shows negligible variance as the impact of gravity remains constant across them.

3 0

1 month ago

A 2.1 × 103 kg car starts from rest at the top of a driveway that is sloped at an angle of 20.0° with the horizontal. An average

Keith_Richards [3271]

Answer:

The driveway measures 4.98 m

Explanation:

We aim to find the length of the driveway, thus utilizing the following equations

W=ΔK.E where W represents work and ΔK.E indicates the change in kinetic energy

Moreover,

$K.E = \frac{ MV^2}{2}$ also