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3 years ago

Which electric component provides energy to the circuit of a flashlight?

2 answers:

6 0

"Battery" <span> provides energy to the circuit of a flashlight

In short, Your Answer would be Option A

Hope this helps!</span>

riadik2000 [5.3K]3 years ago

6 0

<span>The flow of the electricity is activated when you push the switch into the "on" position, giving you light. The flow of electricity is broken when the switch is pushed into the "off" position, thus turning off the light.</span>

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Which element below has chemical properties similar to the chemical properties of fluorine.

SVETLANKA909090 [29]

For number one the answer is Iodine because it is in the same group as fluorine. For number two the answer is Germanium for the same reason. For number three the answer is Aluminum for the same reason.

4 0

3 years ago

1. Black Panther ran 567.5 ks to the east in 2.3 hours then tumed around and ran 2218 km to

mario62 [17]

Answer:Average Velocity of the trip=98.77km/h

Explanation:

Given that:

Black Panther started his journey with

Distance ran towards east = 567.5km

Time ran towards east = 2.3 hours

and then turned back,

Distance ran towards west, = 2218 km

Time taken towards west, = 1.2 hours

Let us treat the westward movement as negative movement of the east

So total displacement =567.5 km east + (-2218km east )= 1,650.5km

But it seems the value of the distance ran west is wrong , because i do not think he can run 2218km under 1.2hours considering he ran 567.5m due east at 2.3hours .

So Let we use 221.8 km as distance ran due west.

So that Total Displacement becomes = 567.5 km east + (-221.8km east )= 345.7km

Total time = 2.3 + 1.2 = 3.5hrs.

Average velocity = Total displacement / total time

.345.7km / 3.5hr = 98.77km/h

7 0

3 years ago

2 things that can cause power output to decrease

Ad libitum [116K]

Answer:

idk and idk

Explanation:

i hope i helped you

3 0

3 years ago

Read 2 more answers

X rays of wavelength 0.0169 nm are directed in the positive direction of an x axis onto a target containing loosely bound electr

mamaluj [8]

Answer:

a) 4.04*10^-12m

b) 0.0209nm

c) 0.253MeV

Explanation:

The formula for Compton's scattering is given by:

$\Delta \lambda=\lambda_f-\lambda_i=\frac{h}{m_oc}(1-cos\theta)$

where h is the Planck's constant, m is the mass of the electron and c is the speed of light.

a) by replacing in the formula you obtain the Compton shift:

$\Delta \lambda=\frac{6.62*10^{-34}Js}{(9.1*10^{-31}kg)(3*10^8m/s)}(1-cos132\°)=4.04*10^{-12}m$

b) The change in photon energy is given by:

$\Delta E=E_f-E_i=h\frac{c}{\lambda_f}-h\frac{c}{\lambda_i}=hc(\frac{1}{\lambda_f}-\frac{1}{\lambda_i})\\\\\lambda_f=4.04*10^{-12}m +\lambda_i=4.04*10^{-12}m+(0.0169*10^{-9}m)=2.09*10^{-11}m=0.0209nm$

c) The electron Compton wavelength is 2.43 × 10-12 m. Hence you can use the Broglie's relation to compute the momentum of the electron and then the kinetic energy.

$P=\frac{h}{\lambda_e}=\frac{6.62*10^{-34}Js}{2.43*10^{-12}m}=2.72*10^{-22}kgm\\$

$E_e=\frac{p^2}{2m_e}=\frac{(2.72*10^{-22}kgm)^2}{2(9.1*10^{-31}kg)}=4.06*10^{-14}J\\\\1J=6.242*10^{18}eV\\\\E_e=4.06*10^{-14}(6.242*10^{18}eV)=0.253MeV$