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A centrifuge rotor (hollow disk) is rotating at 10,000 rpm is shut off and brought to rest by a constant frictional torque of 1.

Y_Kistochka [10]

Answer:

The no of revolutions rotor turn before coming to rest is 1,601.1943 and time taken is equal to 19.21 seconds

Explanation:

here we know that torque = I×α

α= angular acceleration

I = moment of inertia of hollow disc = m× $k^{2}$

given that m=4.37kg

k=0.0710m

torque=1.2Nm

$w_{o}=10000\times \frac{\pi}{30} rad /s$

$\alpha =\frac{torque}{I}$

from the above equation we can calculate the angular acceleration of the hollow disc .

since $w^{2}-w_o^{2} =2\alpha$ $\theta$

from this above equation $\theta=\frac{w^{2}-w_{o}^{2}}{2\alpha }$

no of revolutions = $\frac{\theta}{2\pi}$ = 1,601.1943.

Now to calculate time we know that time = $\frac{2\theta}{w+w_{o}}$

so upon calculating we will be getting t=19.21 seconds

5 0

3 years ago

A 13 kg hanging sculpture is suspended by a 95-cm-long, 5.0 g steel wire. When the wind blows hard, the wire hums at its fundame

Artyom0805 [142]

Answer:

$f=81.96 \ Hz$

Explanation:

Givens

$L=95cm$

$m_{sculpture} =13kg$

$m_{wire}=5g$

The frequency is defined by

$f=\frac{v}{\lambda}$

Where $v$ is the speed of the wave in the string and $\lambda$ is its wave length.

The wave length is defined as $\lambda = 2L = 2(0.95m)=1.9m$

Now, to find the speed, we need the tension of the wire and its linear mass density

$v=\sqrt{\frac{T}{\mu} }$

Where $\mu=\frac{0.005kg}{0.95m}= 5.26 \times 10^{-3}$ and the tension is defined as $T=m_{sculpture} g=13kg(9.81 m/s^{2} )=127.53N$

Replacing this value, the speed is

$v=\sqrt{\frac{127.53N}{5.26 \times 10^{-3} } }=155.71 m/s$

Then, we replace the speed and the wave length in the first equation

$f=\frac{v}{\lambda}\\f=\frac{155.71 m/s}{1.9m}\\ f=81.96Hz$

Therefore, the frequency is $f=81.96 \ Hz$

3 0

3 years ago

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What information do you need to describe an object's location

stira [4]

longitude and latitude<span />

4 0

3 years ago

A car, traveling at , encounters a dip in the road. The radius of curvature at the bottom of the dip is . Each of the car’s four

labwork [276]

Answer:

spring deflection is x = (v2 / R + g) m / 4

Explanation:

We will solve this problem with Newton's second law. Let's analyze the situation the car goes down a road and finds a dip (hollow) that we will assume that it has a circular shape in the lower part has the car weight, elastic force and a centripetal acceleration

Let's write the equations on the Y axis of this description

Fe - W = m $a_{c}$

Where Fe is elastic force, W the weight and $a_{c}$ the centripetal acceleration. The elastic force equation is

Fe = - k x

4 (k x) - mg = m v² / R

The four is because there are four springs, R is theradio of dip

We can calculate the deflection (x) of the springs

x = (m v2 / R + mg) / 4

x = (v2 / R + g) m / 4

5 0

3 years ago

How much heat is needed to raise the temperature of 50.0 g of water by 25.0°C

love history [14]

Answer:

Explanation:

In order to be able to solve this problem, you will need to know the value of water's specific heat, which is listed as

c

=

4.18

J

g

∘

C

Now, let's assume that you don't know the equation that allows you to plug in your values and find how much heat would be needed to heat that much water by that many degrees Celsius.

Take a look at the specific heat of water. As you know, a substance's specific heat tells you how much heat is needed in order to increase the temperature of

1 g

of that substance by

1

∘

C

.

In water's case, you need to provide

4.18 J

of heat per gram of water to increase its temperature by

1

∘

C

.

What if you wanted to increase the temperature of

1 g

of water by

2

∘

C

? You'd need to provide it with

increase by 1

∘

C



4.18 J

+

increase by 1

∘

C



4.18 J

=

increase by 2

∘

C



2

×

4.18 J

To increase the temperature of

1 g

of water by

n

∘

C

, you'd need to supply it with

increase by 1

∘

C



4.18 J

+

increase by 1

∘

C



4.18 J

+

...

=

increase by n

∘

C



n

×

4.18 J

Now let's say that you wanted to cause a

1

∘

C

increase in a

2-g

sample of water. You'd need to provide it with

for 1 g of water



4.18 J

+

for 1 g of water



4.18 J

=

for 2 g of water



2

×

4.18 J

To cause a

1

∘

C

increase in the temperature of

m

grams of water, you'd need to supply it with

for 1 g of water



4.18 J

+

for 1 g of water



4.18 J

+

,,,

=

for m g of water



m

×

4.18 J

This means that in order to increase the temperature of

m

grams of water by

n

∘

C

, you need to provide it with

heat

=

m

×

n

×

specific heat

This will account for increasing the temperature of the first gram of the sample by

n

∘

C

, of the the second gram by

n

∘

C

, of the third gram by

n

∘

C

, and so on until you reach

m

grams of water.

And there you have it. The equation that describes all this will thus be

q

=

m

⋅

c

⋅

Δ

T

, where

q

- heat absorbed

m

- the mass of the sample

c

- the specific heat of the substance

Δ

T

- the change in temperature, defined as final temperature minus initial temperature

In your case, you will have

q

=

100.0

g

⋅

4.18

J

g

∘

C

⋅

(

50.0

−

25.0

)

∘

C

q

=

10,450 J

Rounded to three sig figs and expressed in kilojoules, t

Explanation:

3 0

3 years ago

Read 2 more answers

Picture is down below Thank you!​

Picture is down below Thank you!