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8 days ago

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Charge q is accelerated starting from rest up to speed v through the potential difference V. What speed will charge q have after

accelerating through potential difference 4V?

1 answer:

Ostrovityanka [942]8 days ago

7 0

Answer: v = $1.19 * 10^{6} m/s$

Explanation: q signifies the amount of electric charge = $1.609 * 10^{-19} c$

mass of an electron = $9.10 * 10^{-31} kg$

V = potential difference = 4V

v = speed of the electron

Using the work-energy theorem, the kinetic energy acquired by the electron must equal the work done in propelling it.

The kinetic energy = $\frac{mv^{2} }{2}$ , and potential energy = qV

Thus, $\frac{mv^{2} }{2} = qV$

$\frac{9.10 *10^{-31} * v^{2} }{2} = 1.609 * 10^{-16} * 4\\\\\\\\9.10*10^{-31} * v^{2} = 2 * 1.609 *10^{-16} * 4\\\\\\9.10 *10^{-31} * v^{2} = 1.287 *10^{-15} \\\\v^{2} = \frac{1.287 *10^{-15} }{9.10 *10^{31} } \\\\v^{2} = 1.414*10^{15} \\\\v = \sqrt{1.414*10^{15} } \\\\v = 1.19 * 10^{6} m/s$

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Compare the time period of two simple pendulums of length 4m and 16m at a place.

Ostrovityanka [942]

Answer:

The period of the pendulum measuring 16 m is double that of the 4 m pendulum.

Explanation:

Recall that the period (T) of a pendulum with length (L) is defined by:

$T=2\,\pi\,\sqrt{ \frac{L}{g} }$

where "g" denotes the local gravitational acceleration.

Since both pendulums are positioned at the same location, the value of "g" will be consistent for both, and when we compare the periods, we find:

$T_1=2\,\pi\,\sqrt{\frac{4}{g} } \\T_2=2\,\pi\,\sqrt{\frac{16}{g} } \\ \\\frac{T_2}{T_1} =\sqrt{\frac{16}{4} } =2$

Thus, the duration of the 16 m pendulum is two times that of the 4 m one.

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3 days ago

Light-rail passenger trains that provide transportation within and between cities speed up and slow down with a nearly constant

Yuliya22 [1153]

Answer:

$v_f = 13m/s + 0.75 \frac{m}{s^2} * 16 s= 13 m/s +12m/s = 25 m/s$

Explanation:

In this scenario, we determine the initial velocity as follows:

$v_i = 7 \frac{m}{s}$

The final velocity in this instance can be expressed as:

$v_f = 13 \frac{m}{s}$

It is noted that transitioning from 7m/s to 13m/s takes 8 seconds. We can apply a specific kinematic equation to find the acceleration for the first part of the journey:

$v_f = v_i +at$

Solved for acceleration, we find:

$a = \frac{v_f -v_i}{t} = \frac{13 m/s -7 m/s}{8 s}= 0.75 \frac{m}{s^2}$

For the subsequent route, we assume constant acceleration and that the train continues for 16 seconds, beginning with an initial velocity of 13m/s from the previous segment, allowing us to calculate the final speed via the following formula:

$v_f = v_ i +a t$

Substituting into the equation yields:

$v_f = 13m/s + 0.75 \frac{m}{s^2} * 16 s= 13 m/s +12m/s = 25 m/s$

5 0

17 days ago

A 1000-kg car is moving along a straight road down a 30∘30∘ slope at a constant speed of 20.0m/s20.0m/s. What is the net force a

Softa [913]

The overall force acting on the vehicle is zero

Explanation:

Let's evaluate the situation separately for the vertical direction and the horizontal direction along the slope.

Considering the direction perpendicular to the slope, two forces are in effect:

The weight component acting perpendicular to the slope, $mgcos \theta$ , directed into the slope
The normal force N, directed outward from the slope

Equilibrium exists here, indicating the net force in this direction is zero.

Now let’s examine the parallel direction to the slope. We have two forces present:

The weight component aligned with the slope, $mgsin \theta$ , directed down the slope
The frictional force $F_f$ , acting up the slope

The car moves at a constant speed in this direction, indicating that its acceleration is zero.

$a=0$

Thus, according to Newton's second law,

$F=ma$

implying the net force is zero:

$F=0$

Learn more about slopes and friction:

5 0

13 hours ago

A man makes a 27.0 km trip in 16 minutes. (a.) How far was the trip in miles? (b.) If the speed limit was 55 miles per hour, wa

Maru [1053]

Answer:

(a) 16.777 miles

(b) Yes, he exceeded the speed limit

Explanation:

(a)

We need to perform the necessary calculations to convert kilometers to miles:

$27km*\frac{1000m}{1km} *\frac{1mi}{1609.34m} =16.77706389mi$

Thus, the distance of the trip in miles is:

$d=16.77706389mi$

(b)

Next, we will compute the man's speed during the journey:

$v=\frac{d}{t}$

Before that, we must convert minutes to hours:

$16min*\frac{1h}{60min} =2.666666667h$

The resulting speed is:

$v=\frac{16.77706389mi}{2.666666667h} =62.91398959\frac{mi}{h}$

Consequently:

$62.91398959\frac{mi}{h}>55\frac{mi}{h}$

Thus, it can be concluded that the driver was speeding

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16 hours ago

A girl and boy pull in opposite directions on a stuffed animal. The girl exerts a force of 3.5 N. The mass of the stuffed animal

Keith_Richards [1021]

I will assume the girl is on the right while the boy is on the left.
The net force represents the total of all forces acting on an object, factoring in negatives.
Let the force from the boy be denoted as b. We’ll apply the formula F = ma.

b + 3.5 = 0.2(2.5)

This reduces to a straightforward algebraic problem. By solving, we find that the boy is applying a force of -3N to the left.

5 0

1 day ago

Read 2 more answers