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3 years ago

Besides the force due to gravity or g, what else determines the period of a pendulum?

2 answers:

5 0

For a simple pendulum, the mathematical equation that describes the period (T) is T=√(l/g) where l is the length of the string and g is the acceleration due to gravity (9.81 metres per second). Therefore it can be seen that both gravity and the length of the string within a pendulum effects its period.

Sidana [21]3 years ago

4 0

Length of the wire affects the period of a pendulem.

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How do i calculate acceleration? pls help! thanks :)

LenKa [72]

The acceleration is 2 mph/min

Explanation:

The acceleration of an object is the rate of change of velocity. It is defined as follows:

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time taken for the velocity to change from u to v

For the car in this problem, we have (taking North as positive direction):

u = 25 mph

v = 35 mph

t = 5 min

Substituting into the equation, we find the acceleration:

$a=\frac{35 mph-25mph}{5min}=2 mph/min$

Learn more about acceleration:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

4 0

3 years ago

What is the uncertainty of the position of the bacterium? express your answer with the appropriate units?

lbvjy [14]

For two un-related quantities, the Heisenberg uncertainty equations holds: the prduct of the two uncertainty quantities is greater than $\hbar/2$
Example of unrelated quantities are position and momentum, energy and time.
Thus
$\Delta x*\Delta p \ \textgreater \ \hbar/2$
Knowing the speed of the bacteria the uncertainty in its position is
$\Delta x \ \textgreater \ \hbar/(2 \Delta p) =\hbar/(2mv)$

4 0

3 years ago

What is the energy of a photon that has the same wavelength as an electron having a kinetic energy of 15 ev?

serg [7]

Answer: $6.268(10)^{-16}J$

Explanation:

The kinetic energy of an electron $K_{e}$ is given by the following equation:

$K_{e}=\frac{(p_{e})^{2} }{2m_{e}}$ (1)

Where:

$K_{e}=15eV=2.403^{-18}J=2.403^{-18}\frac{kgm^{2}}{s^{2}}$

$p_{e}$ is the momentum of the electron

$m_{e}=9.11(10)^{-31}kg$ is the mass of the electron

From (1) we can find $p_{e}$ :

$p_{e}=\sqrt{2K_{e}m_{e}}$ (2)

$p_{e}=\sqrt{2(2.403^{-18}J)(9.11(10)^{-31}kg)}$

$p_{e}=2.091(10)^{-24}\frac{kgm}{s}$ (3)

Now, in order to find the wavelength of the electron $\lambda_{e}$ with this given kinetic energy (hence momentum), we will use the De Broglie wavelength equation:

$\lambda_{e}=\frac{h}{p_{e}}$ (4)

Where:

$h=6.626(10)^{-34}J.s=6.626(10)^{-34}\frac{m^{2}kg}{s}$ is the Planck constant

So, we will use the value of $p_{e}$ found in (3) for equation (4):

$\lambda_{e}=\frac{6.626(10)^{-34}J.s}{2.091(10)^{-24}\frac{kgm}{s}}$

$\lambda_{e}=3.168(10)^{-10}m$ (5)

We are told the wavelength of the photon $\lambda_{p}$ is the same as the wavelength of the electron:

$\lambda_{e}=\lambda_{p}=3.168(10)^{-10}m$ (6)

Therefore we will use this wavelength to find the energy of the photon $E_{p}$ using the following equation:

$E_{p}=\frac{hc}{lambda_{p}}$ (7)

Where $c=3(10)^{8}m/s$ is the spped of light in vacuum

$E_{p}=\frac{(6.626(10)^{-34}J.s)(3(10)^{8}m/s)}{3.168(10)^{-10}m}$

Finally:

$E_{p}=6.268(10)^{-16}J$

4 0

3 years ago

Which statement best defines an electric field?

dalvyx [7]

A region around a charged partical or object. Let me know if this works. Hope I could help you.

7 0

3 years ago

Your friend claims that weather predictions are just guesses. What would you say to change your friend's mind?

VashaNatasha [74]

To change my friends mind, after they've already said weather predictions are just guesses, I would say they are correct, but not only to they use guesses to predict what will happen, but they also, before hand, use atmospheric tools. They then predict from the data collected. They do both.

4 0

3 years ago