A sinusoidal wave is travelling on a string under tension T = 8.0(N), having a mass per unit length of 1 = 0.0128(kg/m). It’s displacement function is D(x,t) = Acos(kx - t). It’s amplitude is 0.001m and its wavelength is 0.8m. It reaches the end of this string, and continues on to a string with 2 = 0.0512(kg/m) and the same tension as the first string. Give the values of A, k, and , for the original wave, as well as k and  the reflected wave and the transmitted wave.

Answers

Answer 1

Answer:

Answer:

Explanation:

A sinusoidal wave is travelling on a string under tension T = 8.0(N), having a mass per unit length of 1 = 0.0128(kg/m). It’s displacement function is D(x,t) = Acos(kx - t). It’s amplitude is 0.001m and its wavelength is 0.8m. It reaches the end of this string, and continues on to a string with 2 = 0.0512(kg/m) and the same tension as the first string. Give the values of A, k, and , for the original wave, as well as k and  the reflected JJJJJJave and the transmitted wave.

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Two bodies, one hot and the other cold kept in vacuum.what will happen to the tempreture of bodies after some time.

Answers

Hot body will lose heat from it, and that heat will goes out from it through radiation, so it's temperature will decrease after some time.

In same manner, cold body will take the heat, and it's temperature will increase

Hope this helps!

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the

Answers

Complete question:

Answer:

The exit velocity is 629.41 m/s

Explanation:

Given;

initial temperature, T₁ = 1200K

initial pressure, P₁ = 150 kPa

final pressure, P₂ = 80 kPa

specific heat at 300 K, Cp = 1004 J/kgK

k = 1.4

Calculate final temperature;

$T_2 = T_1((P_2)/(P_1))^{(k-1 )/(k)$

k = 1.4

$T_2 = T_1((P_2)/(P_1))^{(k-1 )/(k)}\n\nT_2 = 1200((80)/(150))^{(1.4-1 )/(1.4)}\n\nT_2 = 1002.714K$

Work done is given as;

$W = (1)/(2) *m*(v_i^2 - v_e^2)$

inlet velocity is negligible;

$v_e = \sqrt{(2W)/(m) } = √(2*C_p(T_1-T_2)) \n\nv_e = √(2*1004(1200-1002.714))\n\nv_e = √(396150.288) \n\nv_e = 629.41 \ m/s$

Therefore, the exit velocity is 629.41 m/s

In an electric vehicle, each wheel is powered by its own motor. The vehicle weight is 4,000 lbs. By regenerative braking, its speed is decreased linearly from 60 mph to 30 mph in 10 seconds. Calculate the theoretical maximum energy in kWh that can be recovered during this interval. Ignore all losses.

Answers

Answer:

the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh

Explanation:

Given that;

weight of vehicle = 4000 lbs

we know that 1 kg = 2.20462

m = 4000 / 2.20462 = 1814.37 kg

Initial velocity $V_(i)$ = 60 mph = 26.8224 m/s

Final velocity $V_(f)$ = 30 mph = 13.4112 m/s

now we determine change in kinetic energy

Δk = $(1)/(2)$ m( $V_(i)$ ² - $V_(f)$ ² )

we substitute

Δk = $(1)/(2)$ ×1814.37( (26.8224)² - (13.4112)² )

Δk = $(1)/(2)$ × 1814.37 × 539.5808

Δk = 489500 Joules

we know that; 1 kilowatt hour = 3.6 × 10⁶ Joule

Δk = 489500 / 3.6 × 10⁶

Δk = 0.13597 ≈ 0.136 kWh

Therefore, the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh

The surface pressure of the atmosphere is about 14.7 psi (pounds per square inch). How many pounds per square yard does that amount to

Answers

Answer:

14.7 psi is equal to 19051.2 pounds per square yard.

Explanation:

Dimensionally speaking, a square yard equals 1296 square inches. Therefore, we need to multiply the atmospheric pressure by 1296 to obtain its equivalent in pounds per square yard. That is:

$p = 14.7\,(lbf)/(in^(2))* 1296\,(in^(2))/(yd^(2))$

$p = 19051.2\,(lbf)/(yd^(2))$

14.7 psi is equal to 19051.2 pounds per square yard.

We showed that the length of the pendulum of period 2.000 seconds on the Earth’s surface was 0.99396 meters. What period would this same pendulum have on the surface of Mars? What length would the pendulum be in order to have a period of 2.000 seconds?

Answers

To solve this problem it is necessary to apply the concepts related to the Period based on gravity and length.

Mathematically this concept can be expressed as

$T= 2\pi \sqrt{(l)/(g)}$

Where,

l = Length

g = Gravitational acceleration

First we will find the period that with the characteristics presented can be given on Mars and then we can find the length of the pendulum at the desired time.

The period on Mars with the given length of 0.99396m and the gravity of the moon (approximately $1.62m / s ^ 2)$ will be

$T= 2\pi \sqrt{(l)/(g)}$

$T= 2\pi \sqrt{(0.99396)/(1.62)}$

$T = 4.921seg$

For the second question posed, it would be to find the length so that the period is 2 seconds, that is:

$T= 2\pi \sqrt{(l)/(g)}$

$2= 2\pi \sqrt{(l)/(1.62)}$

$l = 0.16414m$

Therefore, we can observe also that the shorter distance would be the period compared to the first result given.

If a single circular loop of wire carries a current of 61 A and produces a magnetic field at its center with a magnitude of 1.70 10-4 T, determine the radius of the loop.