3 years ago

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

You might be interested in

How to ask a person out in a different way? LIke insted of do you want to go? Like i like this guy and i gave him a ig teddy bea

Greeley [361]

You should come off like would you like to hang out sometime

6 0

3 years ago

Read 2 more answers

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

Read 2 more answers

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

Read 2 more answers

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