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4 years ago

PLEASE HELP ME Color corresponds to the ______________ of light waves. wave speed cycles wavelength

1 answer:

5 0

Wavelength. Each wavelength is a certain color. For instance, shorter wavelengths (like 470nm) will be blue or violet, while longer wavelengths (like 650nm) will be red. Hope this helps! :)

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The half-life of plutonium 239 is 24,200 years. Assume that the decay rate is proportional to the amount. Determine the amount o

kotykmax [81]

Answer:

time taken is equal to 14,156 years

Explanation:

we know,

$Y=Ae^{-kt}$

at t = 0

Y(0) = A

given that half life of plutonium 239 = 24,200

$\dfrac{A}{2}=Ae^{-kt}\\0.5=e^{-kt}\\k\times 24200 = ln(2)\\k = \dfrac{ ln(2)}{24200}$

$Y=Ae^{-kt}$

$\frac{3}{2} = e^{-kt}\\ln(1.5)=-\dfrac{ ln(2)}{24200}\times t\\t=-\dfrac{ln(1.5)\times 24200}{ ln(2)}\\t=14,156 \ years$

hence time taken is equal to 14,156 years

5 0

3 years ago

A MEMS-based accelerometer has a mass of m = 2 grams, an equivalent spring constant of k = 5 N/m, and an equivalent damping coef

pychu [463]

Answer:

The natural frequency = 50 rad/s = 7.96 Hz

Damping ratio = 0.5

Explanation:

The natural frequency is calculated in this manner

w = √(k/m)

k = spring constant = 5 N/m

m = mass = 2 g = 0.002 kg

w = √(5/0.002) = 50 rad/s

w = 2πf

50 = 2πf

f = 50/(2π) = 7.96 Hz

Damping ratio = c/[2√(mk)] = 0.1/(2 × √(5 × 0.002)) = 0.5

5 0

3 years ago

The phenomenon of vehicle "tripping" is investigated here. The sport-utility vehicle is sliding sideways with speed v1 and no an

Mrrafil [7]

Answer:

$v_1 = 3.5 \ m/s$

Explanation:

Given that :

mass of the SUV is = 2140 kg

moment of inertia about G , i.e $I_G$ = 875 kg.m²

We know from the conservation of angular momentum that:

$H_1= H_2$

$mv_1 *0.765 = [I+m(0.765^2+0.895^2)] \omega_2$

$2140v_1*0.765 = [875+2140(0.765^2+0.895^2)] \omega_2$

$1637.1 v_1$ $= 3841.575 \omega_2$

$\omega_2 = \frac{1637.1 v_1}{3841.575}$

$\omega _2 = 0.4626 \ v_1$

From the conservation of energy as well;we have :

$T_2 +V_{2 \to 3} = T_3 \\ \\ \\ \frac{1}{2} I_A \omega_2^2 - mgh =0$

$[\frac{1}{2} [875+2140(0.765^2+0.895^2)](0.4262 \ v_1)^2 -2140(9.81)[\sqrt{0.76^2+0.895^2} -0.765]] =0$

$706.93 \ v_1^2 - 8657.49 =0$

$706.93 \ v_1^2 = 8657.49$

$v_1^2 = \frac{8657.49}{706.93 }$

$v_1 ^2 = 12.25$

$v_1 = \sqrt{ 12.25$

$v_1 = 3.5 \ m/s$

6 0

3 years ago

A sound technician is testing the sound acoustics in a theatre for an upcoming music concert. He sets up speakers in different l

velikii [3]

Consertive alex dggghh

4 0

3 years ago

Which statement does not describe the X Games?

nikdorinn [45]

the answer Is C. ,, thank you :)

5 0

3 years ago