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Lesechka [4]
3 years ago
10

The difference between an experimental value and an accepted value is

Physics
1 answer:
likoan [24]3 years ago
3 0

Answer:

HERE'S MY UNDERSTANDING OF THE DIFFERENCE

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Greeley [361]
You should come off like would you like to hang out sometime 
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3 years ago
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The period of the earth around the sun is 1 year and its distance is 150 million km from the sun. An asteroid in a circular orbi
SOVA2 [1]

Answer:

5.024 years

Explanation:

T1 = 1 year

r1 = 150 million km

r2 = 440 million km

let the period of asteroid orbit is T2.

Use Kepler's third law

T² ∝ r³

So,

\left ( \frac{T_{2}}{T_{1}} \right )^2=\left ( \frac{r_{2}}{r_{1}} \right )^3

\left ( \frac{T_{2}}{1} \right )^2=\left ( \frac{440}{150} \right )^3

T2 = 5.024 years

Thus, the period of the asteroid's orbit is 5.024 years.

4 0
3 years ago
A certain moving electron has a kinetic energy of 0.991 × 10−19 J. Calculate the speed necessary for the electron to have this e
alisha [4.7K]

Answer: The speed necessary for the electron to have this energy is 466462 m/s

Explanation:

Kinetic energy is the energy posessed by an object by virtue of its motion.

K.E=\frac{1mv^2}{2}

K.E= kinetic energy = 0.991\times 10^{-19}J

m= mass of an electron = 9.109\times 10^{-31}kg

v= velocity of object = ?

Putting in the values in the equation:

0.991\times 10^{-19}J=\frac{1\times 9.109\times 10^{-31}kg\times v^2}{2}

v=466462m/s

The speed necessary for the electron to have this energy is 466462 m/s

4 0
3 years ago
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Humans can hear the squeaks and clicks that bats emit while using echolocation to see their surroundings
MA_775_DIABLO [31]

Answer:

False?

Explanation:

Hope this helps you!

If this is wrong next time I will be better!

5 0
3 years ago
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What is the frequency of a monochromatic light used in a diffraction experiment that has a wavelength of 6.38 ✕ 10e-07 m?
Lady_Fox [76]

Answer:

f=4.70\times 10^{14}\ Hz

Explanation:

Given that,

The wavelength of light, \lambda=6.38\times 10^{-7}\ m

We need to find the frequency of the light. We know that,

c=f\lambda\\\\f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8}{6.38\times 10^{-7}}\\\\f=4.70\times 10^{14}\ Hz

So, the required frequency of light is equal to 4.70\times 10^{14}\ Hz.

4 0
3 years ago
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