Can someone pls give me the answer to this?

What is the answer to life the universe and everything (worth 95 points!)

taurus [48]

I would say nothing, time is endless and the cycle of life is endless, not only on earth but almost anywhere, people try to find answers like what’s at the bottom of the ocean and stuff like that but there’s no point in finding out because it has no benefit, life is made for different reasons so there’s not one answer to it

5 0

3 years ago

Read 2 more answers

La viscosidad de un liquido es igual a 0.04 N s/m^2 . Este valor en Dinas s/cm^2 se encuentra en el literal

nadezda [96]

Answer:

0.4 Dinas*s/cm^2

Explanation:

Tenemos una viscosidad:

V = 0.04 N*s/m^2

Y queremos reescribir esto en Dinas*s/cm^2

Primero transformemos la unidad del denominador, es decir, tenemos que pasar de 1/m^2 a 1/cm^2

Para ello, usamos que:

1m = 100cm

entonces:

(1m/100cm) = 1

Si elevamos ambos lados al cuadrado, obtenemos:

(1m/100cm)^2 = 1

Ahora podemos multiplicar el valor de la viscosidad por esto (que es igual a 1)

V = 0.04 N*s/m^2*((1m/100cm)^2 = 0.00004 N*s/cm^2

Ahora debemos convertir de Newtons a Dinas

Sabemos que:

1 N = 100,000 dinas

1 = (100,000 dinas/1N)

Entonces, de vuelta podemos multiplicar nuestra viscosidad por (100,000 dinas/1N), que es igual a 1 (asi que no cambia el valor, solo sirve para cambiar las unidades)

0.00004 N*s/cm^2 = (100,000 dinas/1N)*(0.00004 N*s/cm^2)

= (100,000 dinas)*(0.00004 s/cm^2)

= 0.4 Dinas*s/cm^2

5 0

3 years ago

Thermodynamics 3-Hacmi 2 m3 olan kapalı bir kap 200 kPa basınçta 4 kg doymuş sıvı ve doymuş buhar karışımı su bulundurmaktadır.

Rufina [12.5K]

I want say the best answer is C a sister-in-law Tran

7 0

3 years ago

1. 6.1 PSPICEMULTISIM The current in a 50μH inductor is known to be iL=18te−10tAfor t≥0. 1. Find the voltage across the inductor

Andre45 [30]

Answer:

a. Voltage across the inductor for t > 0 is 0.9e^-10t(1-10t)

b. Power = -59.3μW

c. Inductor is delivering power.

d. Energy = 5934.3nJ

e. Time = 100ms; Energy = 1095941.025nJ

Explanation:

Given

Current; iL=18te^(−10t)A or t≥0.

L.= inductor = 50μH

a. The voltage, V across the inductor for t>0 is calculated as follows;

V = L(di/dt)

Where L = 50μH

di/dt = 18(e^-10t + (-10)te^-10t)

di/dt = 18e^-10t(1 - 10t)

Substitute 50μH for L and 18e^-10t(1 - 10t) for di/dt in V = L(di/dt)

V = 50μH * 18e^-10t(1 - 10t)

V = 50 * 10^-6(18e^-10t(1 - 10t))

V = 0.9e^-10t(1-10t)

Hence, the voltage across the inductor for t > 0 is 0.9e^-10t(1-10t)

b. Find the power (in microwatts) at the terminals of the inductor when t=200 ms.

Given that t = 200ms = 200 * 10^-3s = 0.2s

Power, p is calculated using the following formula;

p = Li(di/dt)

p = 50 * 10^-6(18te^-10t)18e^-10t(1-10t)

p = 50 * 10^-6 * (18 * 0.2 * e^-(10*0.2)) * (18 * e^(-10 * 0.2) * (1-10*0.2)

p = -5.93E5W

p = -59.3μW

c. Is the inductor absorbing or delivering power at 200 ms?

Because of the negative sign, the inductor is delivering power.

d. Find the energy (in microjoules) stored in the inductor at 200 ms.

Energy is calculated as ½Li²

= ½ * 50 * 10^-6 * (18te^-10t)²

= ½ * 50 * 10^-6 * (18 * 0.2 * e ^ (-10 * 0.2))²

= 0.0000059342669999498J

= 5934.3nJ

e. Find the maximum energy (in microjoules) stored in the inductor and the time (in milliseconds) when it occurs.

Calculating the derivation in (a)

di/dt = 0

18e^-10t(1-10t) = 0

1 - 10t = 0

-10t = -1

t = 1/10

t = 100ms

To calculate the energy, first we need to calculate the current

I(t=100) = 18 * 0.1 * e^(-10(0.1)

I = 0.662182994108596

I = 6621.82mA

The energy is calculated as follows;

w = ½ * 50 * 10^-6 * (6.621)²

w = 0.001095941025

w = 1095941.025nJ

8 0

3 years ago

A smooth concrete pipe with an 18-inch diameter has water flows. The flow rate is 10 ft3/s. Determine the pressure drop in a 100

Verdich [7]

Answer:

i)pressure drop at 100-ft horizontal section = 0.266 psi

ii)pressure drop at +2ft and -2ft change in elevation = 1.13 psi and -0.601 psi

Explanation:

Flow rate = 10 ft^3/s

diameter of pipe ( D ) = 18 inches

Area = π * D^2

elevation ( L ) = 100 ft

i) Calculate pressure drop at 100-ft horizontal section

applying Bernoulli's equation

Δp = pg ( Z₂ - Z₁ ) + f $\frac{L}{D} \frac{pV^2}{2}$ ---------- ( 1 )

where : p = 1.94 slug/ft^3 , g = 32.2 ft/s^2, Z₂ = Z₁ , f = 0.0185, L=100ft , D = 18inches, V = 5.66 ft/s ( i.e. flowrate / A )

Input given values into equation 1

Δp ( pressure drop ) = 0.266 psi

ii) Calculate pressure drop at ± 2ft change in elevation

Δp = pg ( Z₂ - Z₁ ) + f $\frac{L}{D} \frac{pV^2}{2}$ ---------- ( 2 )

where : Z₂ - Z₁ = ± 2 ft , p = 1.94 slug/ft^3 , g = 32.2 ft/s^2, Z₂ = Z₁ , f = 0.0185, L=100ft , D = 18inches, V = 5.66 ft/s

input given values into equation above

Δp = ( 1.13 psi , -0.601 psi )

3 0

3 years ago