Please Help lol. just trying to get up the character count lol

Shanai drove 40 miles to the west, then turned around and drove 70 miles to the east. Draw vectors that show each segment of her

Dmitry [639]

Well she drove 30 more miles to the east than the west but I don’t understand what u are asking

4 0

2 years ago

True or False. Centripetal force holds an object towards the center of a circle.

Flura [38]

False’ because it is a force that makes a body follow a curved path

7 0

3 years ago

I love you Please answer my question :-)

Novay_Z [31]

Answer:

A- Astronomical body

C- Galaxy

D- Comet

B- Moon

Hope this helps you! Have a great day!

6 0

3 years ago

Read 2 more answers

Of the following transitions in the Bohr hydrogen atom, the ________ transition results in the absorption of the highest-energy

8_murik_8 [283]

Answer:

The n = 2 → n = 3 transition results in the absorption of the highest-energy photon.

Explanation:

$E_n=-13.6\times \frac{Z^2}{n^2}ev$

Formula used for the radius of the $n^{th}$ orbit will be,

where,

$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number

Here: Z = 1 (hydrogen atom)

Energy of the first orbit in H atom .

$E_1=-13.6\times \frac{Z^2}{1^2} eV=-13.6 eV$

Energy of the second orbit in H atom .

$E_2=-13.6\times \frac{Z^2}{(2)^2} eV=-3.40 eV$

Energy of the third orbit in H atom .

$E_3=-13.6\times \frac{Z^2}{(9)^2} eV=-1.51 eV$

Energy of the fifth orbit in H atom .

$E_5=-13.6\times \frac{Z^2}{(2)^2} eV=-0.544 eV$

Energy of the sixth orbit in H atom .

$E_6=-13.6\times \frac{Z^2}{(2)^2} eV=-0.378 eV$

Energy of the seventh orbit in H atom .

$E_7=-13.6\times \frac{Z^2}{(2)^2} eV=-278 eV$

During an absorption of energy electron jumps from lower state to higher state.So, absorption will take place in :

1) n = 2 → n = 3

2) n= 5 → n = 6

Energy absorbed when: n = 2 → n = 3

$E=E_3-E_2$

$E=(-1.51 eV) -(-3.40 eV)=1.89 eV$

Energy absorbed when: n = 5 → n = 6

$E'=E_6-E_5$

$E'=(-0.378 eV)-(-0.544 eV) =0.166 eV$

1.89 eV > 0.166 eV

E> E'

So,the n = 2 → n = 3 transition results in the absorption of the highest-energy photon.

4 0

3 years ago

A diffraction pattern forms when light passes through a single slit. The wavelength of the light is 691 nm. Determine the angle

expeople1 [14]

Explanation:

Given that,

Wavelength of the light, $\lambda=691\ nm=691\times 10^{-9}\ m$

(a) Slit width, $a=3.8\times 10^{-4}\ m$

The angle that locates the first dark fringe is given by :

$sin\theta=\dfrac{\lambda}{a}$

$sin\theta=\dfrac{691\times 10^{-9}}{3.8\times 10^{-4}}$

$\theta=0.104^{\circ}$

(b) Slit width, $a=3.8\times 10^{-6}\ m$

The angle that locates the first dark fringe is given by :

$sin\theta=\dfrac{\lambda}{a}$

$sin\theta=\dfrac{691\times 10^{-9}}{3.8\times 10^{-6}}$

$\theta=10.47^{\circ}$

Hence, this is the required solution.

7 0

3 years ago