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3 years ago

Please answer this No links please

Physics

1 answer:

Sladkaya [172]3 years ago

6 0

Your answer is- amplitudes

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Light of wavelength 600 nm passes though two slits separated by 0.22mm and is observed on a screen 1.1m behind the slits. The lo

Ne4ueva [31]

Answer:

$y(m=1)=\frac{(1)(600*10^{-9}m)(1.1m)}{0.22m}=3*10^{-6}m\\\\y(m=1)=\frac{(-1)(600*10^{-9}m)(1.1m)}{0.22m}=-3*10^{-6}m$

Explanation:

We have to take into account the expression for the position of the fringes

$dsen\theta=m\lambda\\y=\frac{m\lambda D}{d}$

where m is the number of the maximum, d is the separation of the slits, D is the distance to the screen.

(a) By replacing we obtain

$y(m=1)=\frac{(1)(600*10^{-9}m)(1.1m)}{0.22m}=3*10^{-6}m\\\\y(m=1)=\frac{(-1)(600*10^{-9}m)(1.1m)}{0.22m}=-3*10^{-6}m$

(b) more information is required to solve this point. Please complete the information.

HOPE THIS HELPS!

4 0

3 years ago

Read 2 more answers

According to Boyle’s Law, increasing the pressure of a gas will have which effect on its volume? increase decrease random effect

umka21 [38]

Volume will decrease if the heat remains constant

3 0

4 years ago

Please help me the formula is E = P x t

Rainbow [258]

The battery doesn't 'use' power. The battery produces the power that all the other electrical devices use.

If the starter motor is using 2,520 watts, then the battery is producing energy at the rate of 2,520 watts. That means 2,520 Joules of energy every second.

Thanks for giving us the formula.

E = P x t

Energy = Power x Time

Energy = (2,520 watts) x (1 second)

Energy = 2,520 Joules

6 0

3 years ago

Estimate (a) the maximum, and (b) the minimum thermal conductivity values (in W/m-K) for a cermet that contains 83 vol% carbide

velikii [3]

Answer:

k_max = 31.82 w/mk

k_min = 17.70 w/mk

Explanation:

a) the maximum thermal conductivity is given as

$k_{max} = k_m*v_m + k_p*v_p$

where k_m is thermal conductvitiy of metal

k_p is thermal conductvitiy of carbide

v_m = proportion of metal in the cement = 0.17

v_p = proportion of carbide in the cement = 0.83

= 66*0.17 + 28*0.83

k_max = 31.82 w/mk

b) the minimum thermal conductivity is given as

$k_{min} = \frac{k_{carbide}* k_{metal}}{k_{metal} v_{carbide} +k_{metal} v_{carbide}}$

= \frac{28+66}{28*0.17 +66*0.83}

k_min = 17.70 w/mk

7 0

3 years ago

Can someone pls help! I’m stuck on c and d. I’ll rate your answer the brainliest

Mkey [24]

Answer:

i think is c

Explanation:

7 0

3 years ago

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