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solong [7]
3 years ago
14

Does anyone know this???

Physics
1 answer:
Gelneren [198K]3 years ago
8 0

Answer:

C

Explanation:

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Jfdojbfkkjds vdf;ifdfdi
Oliga [24]

Answer:

hsjishhdhjs jdjyshskksndhu

6 0
3 years ago
If your speedometer has an uncertainty of 2.5 km/h at a speed of 92 km/h, what is the percent uncertainty
Elis [28]

Answer:

2.7%

Explanation:

Given:

Uncertainty of the speedometer (u)= 2.5km/h

Speed measured at that uncertainty (v) = 92km/h

Percent uncertainty (p) is given as the ratio of the uncertainty to the speed measured then multiplied by 100%. i.e

p = \frac{u}{v} * 100%

p = \frac{2.5}{92} * 100%

p = 2.7%

Therefore, the percent uncertainty is 2.7%

6 0
3 years ago
What is the significance of a standard system of measurement?
dimulka [17.4K]

Because scientists all over the world are working together, looking for answers to the same questions, just as much as if they all worked in the same physical laboratory in the same building.  They need a way to share data and experimental results in a form that everyone can understand. ( D )

Let's say I perform an experiment and get very exciting results. I'm a good scientist, so the next thing I want to do is to publish a complete description of how I did my experiment, and include all of my results.  That way, scientists around the world can read about what I did, they can find any mistakes that I made, and they can even repeat my experiment for themselves and see if they get the same results.

Now let's say that my results looked like this:

Result #1). 

The reaction stabilized when it reached the rate of 1.26 briligs per tove.

Result #2).

After running at that constant rate for 35 toves, a pile of product was produced whose mass was exactly 61.284 wibbles.

Result #3).

When the pile of product was allowed to settle for another 20 toves, it had spread out, and covered an area of 14.907 square filks.

Do YOU understand my results ?

All those other scientists would have a tough time trying to decide whether my results made sense.  And if they repeated my experiment, they would have no way to tell whether their results matched mine or not.

Without a standard system of measurement, and units that mean the same thing to everybody, us scientists literally could not communicate.


3 0
3 years ago
Read 2 more answers
Stars within the halo move in random orbits that extend above and below the disk of the galaxy, while stars in the disk move in
Blizzard [7]
<span>Answer: Spherical Distribution Feedback: Correct The stars in the halo component have highly-inclined random orbits that orbit the center of our Galaxy. The stars within the halo would therefore make up a spherical distribution of stars surrounding the center of the Galaxy. In comparison, the disk stars move in elliptical orbits, which are nearly circular and are confined to the disk of the Galaxy. Disk stars therefore have very small inclinations and do not move above or below the plane of the Galactic disk.</span>
4 0
3 years ago
The following table lists the work functions of a few common metals, measured in electron volts. Metal Φ(eV) Cesium 1.9 Potassiu
Citrus2011 [14]

A. Lithium

The equation for the photoelectric effect is:

E=\phi + K

where

E=\frac{hc}{\lambda} is the energy of the incident light, with h being the Planck constant, c being the speed of light, and \lambda being the wavelength

\phi is the work function of the metal (the minimum energy needed to extract one photoelectron from the surface of the metal)

K is the maximum kinetic energy of the photoelectron

In this problem, we have

\lambda=190 nm=1.9\cdot 10^{-7}m, so the energy of the incident light is

E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1.9\cdot 10^{-7} m}=1.05\cdot 10^{-18}J

Converting in electronvolts,

E=\frac{1.05\cdot 10^{-18}J}{1.6\cdot 10^{-19} J/eV}=6.5 eV

Since the electrons are emitted from the surface with a maximum kinetic energy of

K = 4.0 eV

The work function of this metal is

\phi = E-K=6.5 eV-4.0 eV=2.5 eV

So, the metal is Lithium.

B. cesium, potassium, sodium

The wavelength of green light is

\lambda=510 nm=5.1\cdot 10^{-7} m

So its energy is

E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5.1\cdot 10^{-7} m}=3.9\cdot 10^{-19}J

Converting in electronvolts,

E=\frac{3.9\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.4 eV

So, all the metals that have work function smaller than this value will be able to emit photoelectrons, so:

Cesium

Potassium

Sodium

C. 4.9 eV

In this case, we have

- Copper work function: \phi = 4.5 eV

- Maximum kinetic energy of the emitted electrons: K = 2.7 eV

So, the energy of the incident light is

E=\phi+K=4.5 eV+2.7 eV=7.2 eV

Then the copper is replaced with sodium, which has work function of

\phi = 2.3 eV

So, if the same light shine on sodium, then the maximum kinetic energy of the emitted electrons will be

K=E-\phi = 7.2 eV-2.3 eV=4.9 eV

7 0
3 years ago
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