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4 years ago

Example of an incompressible fluid????

2 answers:

7 0

"Same fluid can behave as compressible and incompressible depending upon flow conditions. ... The most common example of compressible flow concerns is the flow of gases (at low velocities), while the flow of liquids may frequently be treated as incompressible(under low pressure)." -Google

lutik1710 [3]4 years ago

5 0

Gases and liquids that don't have a change in temperature and density

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A Ferris wheel turns at a constant 150.0 revolutions per hour. (a) Express this rate of rotation in units of radians per second.

RUDIKE [14]

Answer:

(a) 0.261 rad/s

(b) 1007.72 m

Explanation:

The angular velocity of the Ferris wheel is 150.0 revolutions per hour.

(a) To calculate the angular velocity of the wheel in units of radians per second, you take into account the following equivalence:

1 hour = 3600 seconds

1 revolution = 2π radians

You use the previous conversion factors:

$150.0\ \frac{rev}{h}*\frac{2\pi \ rad}{1\ rev}*\frac{1\ h}{3600\ s}=0.261\frac{rad}{s}$

In units of radians per seconds the wheel turns at 0.261 rad/s

(b) To find the arc length described by the wheel, you first calculate the angle described by the wheel in the time t, by using the following formula:

$\theta=\omega t$ (1)

ω: angular velocity = 0.261 rad/s

t: time = 4.95 min

You first convert the time to units of seconds

$4.95min*\frac{60s}{1min}=297s$

Next, you replace the values of the parameters in the equation (1):

$\theta=(0.261\frac{rad}{s})(297s)=77.51rad$

Next, you use the following formula for the arc length:

$s=r\theta$ (2)

r: radius of the wheel = 13.0 m

You replace the values of the parameters in the equation (2):

$s=(13.0m)(77.51rad)=1007.72m$

The arc length described by the wheel is 1007.72m

4 0

4 years ago

In a mattress test, you drop a 7.0 kg bowling ball from a height of 1.5 m above a mattress, which as a result compresses 15 cm a

Assoli18 [71]

Answer:

(d) 113.19 J

(e) 113.19 J

(f) 10061 N/m

Explanation:

15 cm = 0.15 m

Let g = 9.8 m/s2

$E_p = mgx = 7*9.8*0.15 = 10.29 J$

(d) By using law of energy conservation with potential energy reference being 0 at the maximum compression point. As the ball falls and come to a stop at the compression point, its potential energy is transferred to elastic energy, which is the work that the mattress does on the ball:

$E_p = E_e$

$E_e = mgh$

where h = 1.5 + 0.15 = 1.65 m is the vertical distance that it falls.

$E_e = 7*9.8*1.65 = 113.19 J$

(e) Before the compression, the potential energy of the mattress is 0. After the compression, the potential energy is 113.19J. So it has increased by 113.19J due to the potential energy transferred from the falling ball.

(f) $E_e = 113.19 = kx^2/2$