What is NOT one of the three primary resources that families have to reach financial goals?

A spring is stretched 6 in by a mass that weighs 8 lb. The mass is attached to a dashpot mechanism that has a damping constant o

olya-2409 [2.1K]

Answer:

y= 240/901 cos 2t+ 8/901 sin 2t

Explanation:

To find mass m=weighs/g

m=8/32=0.25

To find the spring constant

Kx=mg (given that c=6 in and mg=8 lb)

K(0.5)=8 (6 in=0.5 ft)

K=16 lb/ft

We know that equation for spring mass system

my''+Cy'+Ky=F

now by putting the values

0.25 y"+0.25 y'+16 y=4 cos 20 t ----(1) (given that C=0.25 lb.s/ft)

Lets assume that at steady state the equation of y will be

y=A cos 2t+ B sin 2t

To find the constant A and B we have to compare this equation with equation 1.

Now find y' and y" (by differentiate with respect to t)

y'= -2A sin 2t+2B cos 2t

y"=-4A cos 2t-4B sin 2t

Now put the values of y" , y' and y in equation 1

0.25 (-4A cos 2t-4B sin 2t)+0.25(-2A sin 2t+2B cos 2t)+16(A cos 2t+ B sin 2t)=4 cos 20 t

So by comparing the coefficient both sides

30 A+ B=8

A-30 B=0

So we get

A=240/901 and B=8/901

So the steady state response

y= 240/901 cos 2t+ 8/901 sin 2t

6 0

2 years ago

A punted football is observed to have velocity components vhorizontal = 15 m/s to the right and vvertical = 1.25 m/s directed do

enot [183]

Answer:

v₀ₓ = 15 m / s, $v_{oy}$ = 5.2 m / s

v = 15.87 m / s , θ = 19.1

Explanation:

This is a projectile launch problem. The horizontal speed that is constant throughout the entire path is worth 15 m / s, instead the vertical speed changes in value due to the acceleration of gravity, let's look for the initial vertical speed

Vy² = $v_{oy}$ ² - 2 g y

$v_{oy}$ ² = $v_{y}$ ² + 2 g y

$v_{oy}$ = √ ( $v_{y}$ ² + 2 gy

Let's calculate

$v_{oy}$ = √ (1.25² + 2 9.8 1.3)

$v_{oy}$ = √ (27.04)

$v_{oy}$ = 5.2 m / s

The initial speed can be calculated by the initial speed

v = √ v₀ₓ² + $v_{oy}$ ²

v = RA (15² + 5.2²)

v = 15.87 m / s

We look for the angle with trigonometry

tan θ = voy / vox

θ = tan⁻¹ I'm going / vox

θ = tan⁻¹ 5.2 / 15

θ = 19.1

The answer is

v₀ₓ = 15 m / s

$v_{oy}$ = 5.2 m / s

5 0

1 year ago

Your latest invention is a car alarm that produces sound at a particularly annoying frequency of 3600 Hz . To do this, the car a

Alex17521 [72]

Answer:

The capacitance and the inductance can choose for a car-alarm circuit are

C = 215.27 μF

L = 9.078 μH

Explanation:

$V =12.0 V$ , $E = 1.55*10^2 J$ , $f = 3600 Hz$

To determine the capacitance can use the equation

$U_c= \frac{1}{2}*C*V^2$

Solve to C'

$C = \frac{U_c*2}{V^2}=\frac{1.55x10^2J*2}{12.0^2V}$

$C=215.27 uF$

To find the inductance can use the frequency of the circuit

$f = \frac{1}{2\pi* \sqrt{C*L} }$

Solve to L'

$L = \frac{1}{4\pi^2*f^2*C}=\frac{1}{4\pi^2*3600^2*215.27 uF}}$

$L = 9.078 uH$

6 0

1 year ago

A bicycle tire rotates 25 times in 10 seconds. What is it’s average angular velocity?

ryzh [129]

<h2>Answer:</h2>

<u>Angular velocity of bicycle tire is 15.78 radians per second.</u>

<h3>Explanation:</h3>

Angular velocity is the change in angular speed of an object with respect to time take for change or it is the rate of change of circular motion.

In the given question the circular displacement is 25 rounds around a central point.

The angular displacement is measured in degrees and 1 round is equal to 360 degrees.

25 Rounds = 25 × 360 = 9000 degrees.

Angular velocity = angular displacement /time = 9000/10 = 900 degrees per second.

In SI,angular velocity is represented in radians per second.

So, 1 radian = 57.29 degrees

Angular velocity = 15.78 radians per second

3 0

2 years ago

jenny's model train is set up on a circular track. There are six telephone poles evenly spaced around the track. It takes the en

d1i1m1o1n [39]

Answer:

T = 60 s

Explanation:

There are 6 poles on the track which are equally spaced

so the angular separation between the poles is given as

$\theta = \frac{2\pi}{6}$

$\theta = \frac{\pi}{3}$

so the angular speed of the train is given as

$\omega = \frac{\theta}{t}$

$\omega = \frac{\pi}{30} rad/s$

now we have time period of the train given as

$T = \frac{2\pi}{\omega}$

$T = \frac{2\pi}{\frac{\pi}{30}}$

$T = 60 s$

5 0

2 years ago