Ask question

Ask question

All categories

3 months ago

8

An electron and a proton, starting from rest, are accelerated through an electric potential difference of the same magnitude. in

the process, the electron acquires a speed ve, while the proton acquires a speed vp. find the ratio ve/vp.

1 answer:

ValentinkaMS [3.4K]3 months ago

4 0

Since the absolute values of the charges are identical, the changes in potential energy remain equivalent. Consequently, the changes in kinetic energy will also match. We have:

1 = Ke/Kp = m_e * v_e^2 / m_p * v_p^2, which simplifies to:

v_e/v_p = sqrt(m_p/m_e),

indicating that the velocity of the electron is sqrt(m_p/m_e) times greater than that of the proton.

You might be interested in

In mammals, the weight of the heart is approximately 0.5% of the total body weight. Write a linear model that gives the heart we

Yuliya22 [3333]

Answer:

The typical weight of a human heart is approximately 0.93 lbs.

Explanation:

Based on this,

the heart's weight constitutes about 0.5% of total body mass.

Total human weight = 185 lbs

Let the entire body weight be represented as w and the heart's weight as $w_{h}$ .

We aim to determine the heart's weight for a human

Using the provided information

$w_{h}=0.5\times w$

Where, h = heart weight

w = human weight

$w_{h}=\dfrac{0.5}{100}\times 185$

$w_{h}=0.93\ lbs$

The final weight of a human heart is 0.93 lbs.

8 0

4 months ago

A negatively charged object is located in a region of space where the electric field is uniform and points due north. the object

serg [3582]

- 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:
$\Delta U = e d$

- 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:
$\Delta U = 0$

- 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:
$\Delta U = - e d$

5 0

2 months ago

1)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it r

Softa [3030]

Answer:

1)

$v_{oy}=11.29\ m/s$

2)

$y=7.39\ m$

Explanation:

Projectile Motion

When an object is projected near the surface of the Earth at an angle $\theta$ to the horizontal, it follows a trajectory known as a parabola. The only force acting on it (ignoring wind resistance) is gravity, affecting the vertical axis.

The height of a projectile can be calculated using

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

where $y_o$ represents the initial height from ground level, $v_{oy}$ is the vertical component of the initial velocity, and t is the elapsed time.

The vertical speed component is expressed as

$v_y=v_{oy}-gt$

1) To proceed, we will determine the initial vertical velocity component since we lack sufficient data to calculate the absolute value of $v_o$ .

The peak height is attained when $v_y=0$ , which allows us to compute the time to reach that height.

$v_{oy}-gt_m=0$

Solving for $t_m$

$\displaystyle t_m=\frac{v_{oy}}{g}$

Thus, the maximum height reached is

$\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}$

We know this value is equal to 8 meters

$\displaystyle y_o+\frac{v_{oy}^2}{2g}=8$

Continuing with the calculations for $v_{oy}$

$\displaystyle v_{oy}=\sqrt{2g(8-y_o)}$

Substituting known values yields

$\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}$

$\displaystyle v_{oy}=11.29\ m/s$

2) At t=1.505 seconds, the ball is positioned above Julie’s head; we can calculate

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}$

$\displaystyle y=1.5\ m+16,991\ m-11.098\ m$

$y=7.39\ m$

5 0

4 months ago

Astronaut A can cover 10 meters per minute walking with the heavy shovel. What does

Softa [3030]

The astronaut's speed is described in the sentence. The astronaut moves at a rate of 10 meters each minute. To clarify: speed is defined as distance divided by time and is characterized solely by its magnitude, not its direction. Hence, the 10 meters per minute reflects this. We lack information about the astronaut's directional movement. In contrast to speed, velocity incorporates direction as well; for instance, a velocity of 10m/s due west provides a directional context. Consequently, without specified direction, the value indicated is merely speed.

3 0

3 months ago