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An electron with a charge value of 1.6 x 10-19 C is moving in the presence of an electric field of 400 N/C. What force does the

Vsevolod [243]

Answer:

4.0×10⁻¹⁷ N

Explanation:

4 0

3 years ago

If 710- nm and 660- nm light passes through two slits 0.65 mm apart, how far apart are the second-order fringes for these two wa

alexira [117]

0.23 mm far apart are the second-order fringes for these two wavelengths on a screen 1.5 m away.

<h3>Given wavelengths 710nm and 660nm,0.65mm apart two slits, and a screen 1.5m away.</h3>

Position of n the order fringe = n λ D / d

for n = 2

position = 2 λ D / d

λ = 710 nm , D = 1.5m

d = .65 x 10⁻³

position 1 = 2 x 710 x 10⁻⁹ x 1.5 / .65 x 10⁻³

= 3276.92 x 10⁻⁶ m

= 3.276x 10⁻³ m

= 3.276mm .

For λ = 660 nm

position = 2 λ D / d

λ = 660 nm , D = 1.5 m

d = .65 x 10⁻³

position 2 = 2 x 660 x 10⁻⁹ x 1.5 / .65 x 10⁻³

= 3046.15 x 10⁻⁶ m

= 3.046 x 10⁻³ m

= 3.046 mm .

Difference between their position

= 3.276mm ₋ 3.046 mm

= 0.23 mm .

To know more about Fringes refer to: brainly.com/question/15649748

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7 0

2 years ago

How can you turn off a magnetic field produced by an electrical current?

aleksklad [387]

Answer:By turning the electrical current off

Explanation:Trust me I took the test

6 0

3 years ago

Read 2 more answers

Which of the following describes an action-reaction pair?

Vlad [161]

Answer:

Explanation: LETTER A

8 0

2 years ago

Read 2 more answers

One end of an insulated metal rod is maintained at 100 ∘C and the other end is maintained at 0.00 ∘C by an ice–water mixture. Th

lozanna [386]

Answer:

$k=105.0359\times 10^4\,W.m^{-1}.K^{-1}$

Explanation:

Given:

temperature at the hotter end, $T_H=100^{\circ}C$

temperature at the cooler end, $T_C=0^{\circ}C$

length of rod through which the heat travels, $dx=0.7\,m$

cross-sectional area of rod, $A=1.1\times 10^{-4}\,cm^2$

mass of ice melted at zero degree Celsius, $m=8.7\times 10^{-3}\,kg$

time taken for the melting of ice, $t=15\times60=900\,s$

thermal conductivity k=?

By Fourier's Law of conduction we have:

$\dot{Q}=k.A.\frac{dT}{dx}$ ......................................(1)

where:

$\dot{Q}$ =rate of heat transfer

dT= temperature difference across the length dx

Now, we need the total heat transfer according to the condition:

we know the latent heat of fusion of ice, $L = 334000\,J.kg^{-1}$

$Q=m.L$

$Q=8.7\times 10^{-3}\times 334000$

$Q=2905.8\,J$

Now the heat rate:

$\dot{Q}=\frac{Q}{t}$

$\dot{Q}=\frac{2905.8}{900}$

$\dot{Q}=3.2287\,W$

Now using eq,(1)

$3.2287=k\times 1.1\times 10^{-4} \times \frac{100}{0.7}$

$k=205.4606\,W.m^{-1}.K^{-1}$

8 0

3 years ago