TRUE OR FALSE??

A van de Graaff generator accelerates electrons so that they have energies equivalent to that attained by falling through a pote

Marat540 [252]

Answer:

Hello your question is incomplete hence I will give you a general answer on how A van de Graaff generator works

answer :

If the electrons falls through a PD of 150mV the electron will gain energy of 150MeV

Explanation:

when a Van de Graff generator is used to accelerate an electron through a PD ( potential difference ) of any value the particle ( electron ) the electron will gain energy ( eV ) which is is equivalent in value of the PD it accelerated through

hence if the electrons falls through a PD of 150mV the electron will gain energy of 150MeV

4 0

3 years ago

A certain thin lens is made of glass with refraction index ????lens=1.500. In air, where the index of refraction is 1.000, the l

son4ous [18]

Answer:

The focal length of the lens in ethyl alcohol is 41.07 cm.

Explanation:

Given that,

Refractive index of glass= 1.500

Refractive index of air= 1.000

Refractive index of ethyl alcohol = 1.360

Focal length = 11.5 cm

We need to calculate the focal length of the lens in ethyl alcohol

Using formula of focal length for glass air system

$\dfrac{1}{f}=(n_{g}-n_{a})(\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}})$

Using formula of focal length for glass ethyl alcohol system

$\dfrac{1}{f'}=(n_{g}-n_{ethyl})(\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}})$

Divided equation (II) by (I)

$\dfrac{f'}{f}=\dfrac{n_{g}-n_{a}}{n_{g}-n_{ethyl}}$

Where, $n_{g}$ = refractive index of glass

$n_{a}$ = refractive index of air

$n_{ethyl}$ = refractive index of ethyl

Put the value into the formula

$\dfrac{f'}{11.5}=\dfrac{1.500-1.000}{1.500-1.360}$

$\dfrac{f'}{11.5}=\dfrac{25}{7}$

$f'=\dfrac{25}{7}\times11.5$

$f'=41.07\ cm$

Hence, The focal length of the lens in ethyl alcohol is 41.07 cm.

7 0

3 years ago

An 1100kg car reached 30m/s in 9s from rest. What force did the engine provide?

malfutka [58]

Answer:

≅3666.67 N

Explanation:

Use Newton's 2nd law, F = ma where F=force applied, m = mass of the object,

a = acceleration acquired by the object.

a= (v-u)/t where v = final velocity, u = initial velocity and t = time taken

calculate a = (30-0)/9 ≅ 3.33 m/s2

Then F = 1100×a = 3666.67 N

3 0

3 years ago

Could I get some help with this

nika2105 [10]

please give me brainlest!!

the answer is A.

4 0

3 years ago

A-10A twin-jet close-support airplane is approximately rectangular with a wingspan (the length perpendicular to the flow directi

Sidana [21]

Solution :

Given :

Rectangular wingspan

Length,L = 17.5 m

Chord, c = 3 m

Free stream velocity of flow, $V_{\infty}$ = 200 m/s

Given that the flow is laminar.

$Re_L=\frac{\rho V L}{\mu _{\infty}}$

$=\frac{1.225 \times 200 \times 3}{1.789 \times 10^{-5}}$

$= 4.10 \times 10^7$

So boundary layer thickness,

$\delta_{L} = \frac{5.2 L}{\sqrt{Re_L}}$

$\delta_{L} = \frac{5.2 \times 3}{\sqrt{4.1 \times 10^7}}$

= 0.0024 m

The dynamic pressure, $q_{\infty} =\frac{1}{2} \rho V^2_{\infty}$

$=\frac{1}{2} \times 1.225 \times 200^2$

$=2.45 \times 10^4 \ N/m^2$

The skin friction drag co-efficient is given by

$C_f = \frac{1.328}{\sqrt{Re_L}}$

$=\frac{1.328}{\sqrt{4.1 \times 10^7}}$

= 0.00021

$D_{skinfriction} = \frac{1}{2} \rho V^2_{\infty}S C_f$

$=\frac{1}{2} \times 1.225 \times 200^2 \times 17.5 \times 3 \times 0.00021$

= 270 N

Therefore the net drag = 270 x 2

= 540 N

7 0

3 years ago