A wave has a wavelength of 15 mm and a frequency of 3 hertz. What is its speed?

A simple pendulum consisting of a bob of mass m attached to a string of length L swings with a period T. If the pendulum is take

Juli2301 [7.4K]

To solve this problem we will use the definition of the period in a simple pendulum, which warns that it is dependent on its length and gravity as follows:

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

Here,

L = Length

g = Acceleration due to gravity

We can realize that $2 \pi$ is a constant so it is proportional to the square root of its length over its gravity,

$T \propto \sqrt{\frac{L}{g}}$

Since the body is in constant free fall, that is, a point where gravity tends to be zero:

$g \rightarrow 0 \Rightarrow T \rightarrow \infty$

The value of the period will tend to infinity. This indicates that the pendulum will no longer oscillate because both the pendulum and the point to which it is attached are in free fall.

5 0

3 years ago

Globalization is the process of

Delvig [45]

Answer: Globalization is the process of interaction and integration among people, companies, and governments worldwide.

Explanation:

5 0

3 years ago

Read 2 more answers

Which was an idea use to distribute the theory of plate tectonics

xxMikexx [17]

Answer:

continental drift

Explanation:

Wegener proposed that the continents plowed through crust of ocean basins, which would explain why the outlines of many coastlines look like they fit together like a puzzle

8 0

4 years ago

Jasmine and her partner want to test to see if there is a relationship between how much of her solar panel is exposed to sunligh

katen-ka-za [31]

Any one trial might have been done incorrectly.

7 0

3 years ago

What is the shortest-wavelength x-ray radiation in m that can be generated in an x-ray tube with an applied voltage of 93.3 kV?

VikaD [51]

(a) $1.33\cdot 10^{-11} m$

The x-rays in the tube are emitted as a result of the collisions of electrons (accelerated through the potential difference applied) on the metal target. Therefore, all the energy of the accelerated electron is converted into energy of the emitted photon:

$e \Delta V = \frac{hc}{\lambda}$

where the term on the left is the electric potential energy given by the electron, and the term on the right is the energy of the emitted photon, and where:

$e=1.6\cdot 10^{-19}C$ is the electron's charge

$\Delta V = 93.3 kV = 93300 V$ is the potential difference

$h=6.63\cdot 10^{-34} Js$ is the Planck constant

$c=3.00\cdot 10^8 m/s$ is the speed of light

$\lambda$ is the wavelength of the emitted photon

Solving the formula for $\lambda$ , we find:

$\lambda=\frac{hc}{e\Delta V}=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{(1.6\cdot 10^{-19})(93300)}=1.33\cdot 10^{-11} m$

(b) 93300 eV (93.3 keV)

The energy of the emitted photon is given by:

$E=\frac{hc}{\lambda}$

where

h is Planck constant

c is the speed of light

$\lambda=1.33\cdot 10^{-11} m$ is the wavelength of the photon, calculated previously

Substituting,

$E=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{1.33\cdot 10^{-11}}=1.50\cdot 10^{-14} J$

Now if we want to convert into electronvolts, we have to divide by the charge of the electron:

$E=\frac{1.50\cdot 10^{-14} J}{1.6\cdot 10^{-19} J/eV}=93300 eV$

(c) The following statements are correct:

The maximum photon energy is just the applied voltage times the electron charge. (1)

The value of the voltage in volts equals the value of the maximum photon energy in electron volts.

In fact, we see that statement (1) corresponds to the equation that we wrote in part (a):

$e \Delta V = \frac{hc}{\lambda}$

While statement (2) is also true, since in part (b) we found that the photon energy is 93.3 keV, while the voltage was 93.3 kV.

3 0

3 years ago