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1 year ago

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Photoelectrons with a maximum speed of 8.00 • 106 m/sec are ejected froma surface in the presence of light with a frequency of 6

.32 · 1014Hz. If themass of an electron is 9.10 . 10-31 kg, calculate the maximum kineticenergy of a single electron, in joules.a. 3.64 x 10-24 Jb. 2.88 x 10-16 Jc. 5.82 x 10-17 Jd. 2.91 x 10-17 J

1 answer:

Andru [333]1 year ago

5 0

The kinetic energy is given by:

$K=\frac{1}{2}mv^2$

We know the mass and the maximum speed, plugging their values in the expression above we have:

$\begin{gathered} K=\frac{1}{2}(9.1\times10^{-31})(8\times10^6)^2 \\ K=2.91\times10^{-17}\text{ J} \end{gathered}$

Therefore, the answer is d.

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Please hurry Describe why electric currents can be dangerous

konstantin123 [22]

Answer:

Cardiac Arrest, burns, and nerve damage.

Explanation:

Basically, the main risk is cardiac arrest, caused by the electric current interfering with the normal operation of the heart muscle. Other possible damages are burns due to the electric energy vaporizing the water inside the cells, and nerve damage caused by excessive current through the nerves.

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2 years ago

The sensor in the torso of a crash test dummy records the magnitude and direction of the net force acting on the dummy.If the du

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Here in crash test the two forces are acting on the dummy in two different directions

As we know that force is a vector quantity so we need to use vector addition laws in order to find the resultant force on it.

So here two forces are given in perpendicular direction with each other so as per vector addition law we need to use Pythagoras theorem to find the resultant of two vectors

so we can say

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here given that

$F_1 = 130.0 N$

$F_2 = 4500.0 N$

now we will plug in all data in the above equation

$F_{net} = \sqrt{4500^2 + 130^}$

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3 years ago

Read 2 more answers

a mass on a spring vibrates in simple harmonic motion at an amplitude of 8.0 cm. if the mass of the object is 0.20kg and the spr

Reil [10]

Answer:

4.06 Hz

Explanation:

For simple harmonic motion, frequency is given by

$f=\frac {1}{2\pi}\times \sqrt{\frac {k}{m}}$ where k is spring constant and m is the mass of the object.

Substituting 0.2 Kg for mass and 130 N/m for k then

$f=\frac {1}{2\pi}\times \sqrt{\frac {130}{0.2}}=4.057670803\\f\approx 4.06 Hz$

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3 years ago

When the play button is pressed, a CD accelerates uniformly from rest to 450 rev/min in 3.0 revolutions. If the CD has a radius

Marina CMI [18]

To solve this problem it is necessary to apply the kinematic equations of angular motion.

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$\tau = I\alpha$

where

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$\alpha =$ Angular acceleration

The angular acceleration at the same time can be defined as function of angular velocity and angular displacement (Without considering time) through the expression:

$2 \alpha \theta = \omega_f^2-\omega_i^2$

Where

$\omega_{f,i} =$ Final and Initial Angular velocity

$\alpha =$ Angular acceleration

$\theta =$ Angular displacement

Our values are given as

$\omega_i = 0 rad/s$

$\omega_f = 450rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})$

$\omega_f = 47.12rad/s$

$\theta = 3 rev (\frac{2\pi rad}{1rev}) \rightarrow 6\pi rad$

$r = 7cm = 7*10^{-2}m$

$m = 17g = 17*10^{-3}kg$

Using the expression of angular acceleration we can find the to then find the torque, that is,

$2\alpha\theta=\omega_f^2-\omega_i^2$

$\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}$

$\alpha = \frac{47.12^2-0^2}{2*6\pi}$

$\alpha = 58.89rad/s^2$

With the expression of the acceleration found it is now necessary to replace it on the torque equation and the respective moment of inertia for the disk, so

$\tau = I\alpha$

$\tau = (\frac{1}{2}mr^2)\alpha$

$\tau = (\frac{1}{2}(17*10^{-3})(7*10^{-2})^2)(58.89)$

$\tau = 0.00245N\cdot m \approx 2.45*10^{-3}N\cdot m$

Therefore the torque exerted on it is $2.45*10^{-3}N\cdot m$

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3 years ago

A 1750-kilogram cars travels at a constant speed of 15.0 meters per second around a horizontal, circular track with a radius of

bulgar [2K]

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3 years ago