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3 years ago

6

why do this I'm trying to cheat and you put a question omgggggggggggggggggggggg!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

1 answer:

Dahasolnce [82]3 years ago

5 0

Answer:

yeah

Explanation:

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Solar cells are often coated with a transparent, thin film of silicon monoxide (n = 1.45) to minimize reflective losses from the

Alinara [238K]

Answer:

94.82 nm

Explanation:

We have given that wavelength $\lambda=550\ nm$

We have to find the minimum film thickness that produces the least reflection

minimum thickness is given by $t=\frac{\lambda }{4n}$

here n is given n=1.45

so minimum thickness $t=\frac{550 }{4\times 1.45}=94.82\ nm$

so the minimum film thickness will be 94.82 nm

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4 years ago

A vector quantity is the magnitude of a given quantity? True Or False

Anettt [7]

Answer:

false

Explanation:

A vector quantity is the magnitude of a given quantity? True Or False

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4 years ago

Ezra is pulling a sled, filled with snow, by pulling on a rope attached to the sled. The rope makes an angle θ with respect to t

svetoff [14.1K]

Answer:

Explanation:

Given

rope makes an angle of $\theta$

Mass of sled and snow is m

Normal Force $=F_N$

applied Force is F

as Force is pulling in nature therefore normal reaction is given by

$F_N=mg-F\sin \theta$

Also $F\cos \theta =f_r$

$f_r=\mu _k\cdot F_N$

$f_r=\mu _k\cdot (mg-F\sin \theta )$

$F\cos \theta =\mu _kF_N$ -------1

$F\sin \theta =mg-F_N$ ---------2

Squaring 1 & 2 and then adding

$F^2=(\mu _kF_N)^2+(mg-F_N)^2$

$F=\sqrt{(\mu _kF_N)^2+(mg-F_N)^2}$

Substitute value of F in 1

$cos\theta =\frac{\mu _KF_N}{\sqrt{(\mu _kF_N)^2+(mg-F_N)^2}}$

$\theta =cos^{-1}(\frac{\mu _KF_N}{\sqrt{(\mu _kF_N)^2+(mg-F_N)^2}})$

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3 years ago

What did the scientists deduce from the fact that the ants eyes of the desert have multiple lenses

Evgesh-ka [11]

That the pupl is smaller than the nulian hope this helped

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3 years ago

A 100 kg roller coaster comes over the first hill at 2 m/sec (vo). The height of the first hill (h) is 20 meters. See roller dia

aleksandr82 [10.1K]

For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:

1) The total energy for the roller coaster at the <u>initial point</u> is 19820 J

2) The potential energy at <u>point A</u> is 19620 J

3) The kinetic energy at <u>point B</u> is 10010 J

4) The potential energy at <u>point C</u> is zero

5) The kinetic energy at <u>point C</u> is 19820 J

6) The velocity of the roller coaster at <u>point C</u> is 19.91 m/s

1) The total energy for the roller coaster at the <u>initial point</u> can be found as follows:

$E_{t} = KE_{i} + PE_{i}$

Where:

KE: is the kinetic energy = (1/2)mv₀²

m: is the mass of the roller coaster = 100 kg

v₀: is the initial velocity = 2 m/s

PE: is the potential energy = mgh

g: is the acceleration due to gravity = 9.81 m/s²

h: is the height = 20 m

The<em> total energy</em> is:

$E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J$

Hence, the total energy for the roller coaster at the <u>initial point</u> is 19820 J.

2) The <em>potential energy</em> at point A is:

$PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J$

Then, the potential energy at <u>point A</u> is 19620 J.

3) The <em>kinetic energy</em> at point B is the following:

$KE_{A} + PE_{A} = KE_{B} + PE_{B}$

$KE_{B} = KE_{A} + PE_{A} - PE_{B}$

Since

$KE_{A} + PE_{A} = KE_{i} + PE_{i}$

we have:

$KE_{B} = KE_{i} + PE_{i} - PE_{B} = 19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J$

Hence, the kinetic energy at <u>point B</u> is 10010 J.

4) The <em>potential energy</em> at <u>point C</u> is zero because h = 0 meters.

$PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J$

5) The <em>kinetic energy</em> of the roller coaster at point C is:

$KE_{i} + PE_{i} = KE_{C} + PE_{C}$

$KE_{C} = KE_{i} + PE_{i} = 19820 J$

Therefore, the kinetic energy at <u>point C</u> is 19820 J.

6) The <em>velocity</em> of the roller coaster at point C is given by:

$KE_{C} = \frac{1}{2}mv_{C}^{2}$

$v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s$

Hence, the velocity of the roller coaster at <u>point C</u> is 19.91 m/s.

Read more here:

brainly.com/question/21288807?referrer=searchResults

I hope it helps you!

3 0

3 years ago