Ask question

Ask question

All categories

3 years ago

12

Suppose that a car weighing 4000 pounds is supported by four shock absorbers Each shock absorber has a spring constant of 6500 l

bs/foot, so the effective spring constant for the system of 4 shock absorbers is 26000 lbs/foot. Assume no damping and determine the period of oscillation of the vertical motion of the car.
Required:
a. Assume no damping and determine the period of oscillation of the vertical motion of the car.
b. After 10 seconds the car body is 1/3 foot above its equilibrium position and at the high point in its cycle. What were the initial conditions?

1 answer:

scoundrel [369]3 years ago

7 0

Answer:

0.43622 seconds

0.9158 foot

-4.43 ffot/sec

Explanation:

we first find the period of oscillation

= 2π√w/gk

= 2π√4000/32x2600

= 2π√0.00481

= 2π0.0694

= 0.43622

b. we find the angular velocity

2π/T

= 2π/0.43622

= 14.41 rad/sec

we find displacement

rom the calculation in the attachment

Ф = -144.1

initial condition

1*cos(-144.1 rad)

= 0.9158 foot

initial velocity of the car

= (-1)(14.41)sin(ω(0)-144.1)

= -4.44 foot/sec

You might be interested in

Two wheels have the same mass and radius of 4.4 kg and 0.48 m, respectively. One has (a) the shape of a hoop and the other (b) t

abruzzese [7]

Answer:

1) $\sum T_{externalhoop}=0.4418Nm$

2) $\sum T_{externaldisc}=0.2209Nm$

Explanation:

Using the second equation of angular motion we have

$\theta =\omega _{o}t+\frac{1}{2}\alpha t^{2}$

Since the wheels start from rest we ahve $omega _{o}=0$

Applying the given values in the equation we have

$14=\frac{1}{2}\alpha \times 8^{2}\\\\\therefore \alpha =\frac{28}{64}=0.4375rad/s^{2}$

Now by newton's second law of motion in angular motion we have

$\sum T_{external}=I\alpha$

1) For Hoop We have

$I_{hoop}=Mr^{2}\\\\\therefore I_{hoop}=4.4\times(0.48)^{2}=1.01376kgm^{2}$

Thus

$\sum T_{external}=1.01376\times 0.4375$

$\sum T_{external}=0.4418Nm$

2)For disc We have

$I_{disc}=\frac{Mr^{2}}{2}\\\\\therefore I_{hoop}=2.2\times(0.48)^{2}=0.506kgm^{2}$

Thus

$\sum T_{external}=0.506\times 0.4375$

$\sum T_{external}=0.2209Nm$

5 0

4 years ago

A newly discovered particle, the SPARTYON, has a mass 945 times that of an electron. If a SPARTYON at rest absorbs an anti-SPART

Gennadij [26K]

Answer:

$\nu =1166\times 10^{20}Hz$

Explanation:

We have given the rest mass of SPARTYON = 945 times of mass of electron

We know that mass of electron $=9.11\times 10^{-31}kg$

So mass of SPARTYON $=945\times 9.11\times 10^{-31}=8608.95\times 10^{-31}kg$

Speed of light $c=3\times 10^8m/sec$

According to Einstein equation energy is given by

$E=mc^2=8608.95\times 10^{-31}\times (3\times 10^{8})^2=77480.55\times 10^{-15}j$

Now according to planks's rule

Energy is given by

$E=h\nu$ , here h is plank's constant $h=6.6\times 10^{-34}$

So $77480.55\times 10^{-15}=6.6\times 10^{-34}\nu$

$\nu =1166\times 10^{20}Hz$

3 0

3 years ago

1. A 24 kg rock is resting on the floor. How much potential energy does it have? Explain.

In-s [12.5K]

Relative to the floor its resting on, its height is zero, so it potential energy is zero.

8 0

3 years ago

In what way is Speed related to Kinetic Energy?

Papessa [141]

Answer:

Explanation:

In order to find kinetic energy, you need information about the velocity. Speed is the magnitude of the velocity.

Kinetic energy is 1/2mv^2

m being mass

v being velocity

8 0

3 years ago

. A 65 kg climber on top of Mt. Everest (8800 m high). How much Pontential Energy does he have?

Kruka [31]

Answer:

He has 5,609,404 J of potential energy

Explanation:

P.E. = M*G*H

65 * 9.8^2 * 8800

7 0

3 years ago