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Alecsey [184]
3 years ago
12

At a given instant an object has an angular velocity. It also has an angular acceleration due to torques that are present. There

fore, the angular velocity is changing. Does the angular velocity at this instant increase, decrease, or remain the same (a) if additional torques are applied so as to make the net torque suddenly equal to zero and (b) if all the torques are suddenly removed
Physics
1 answer:
katen-ka-za [31]3 years ago
8 0

a) Constant

b) Constant

Explanation:

a)

We can answer this question by using the equivalent of Newton's second law of motion of rotational motion, which can be written as:

\tau_{net} = I \alpha (1)

where

\tau_{net} is the net torque acting on the object in rotation

I is the moment of inertia of the object

\alpha is the angular acceleration

The angular acceleration is the rate of change of the angular velocity, so it can be written as

\alpha = \frac{\Delta \omega}{\Delta t}

where

\Delta \omega is the change in angular velocity

\Delta t is the time interval

So we can rewrite eq.(1) as

\tau_{net}=I\frac{\Delta \omega}{\Delta t}

In this problem, we are told that at a given instant, the object has an angular acceleration due to the presence of torques, so there is a non-zero change in angular velocity.

Then, additional torques are applied, so that the net torque suddenly equal to zero, so:

\tau_{net}=0

From the previous equation, this implies that

\Delta \omega =0

Which means that the angular velocity at that instant does not change anymore.

b)

In this second case instead, all the torques are suddenly removed.

This also means that the net torque becomes zero as well:

\tau_{net}=0

Therefore, this means that

\Delta \omega =0

So also in this case, there is no change in angular velocity: this means that the angular velocity of the object will remain constant.

So cases (a) and (b) are basically the same situation, as the net torque is zero in both cases, so the object acts in the same way.

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skelet666 [1.2K]

Answer:

ΔT = 17.11 °C

Explanation:

In this case, we have a ship standing on a place with a given mass and it's about to be launched to a lock containing water.

At first, before launch, the ship has a potential energy, and when the ship hits the water after being launched, this potential energy is transformed into kinetic energy.

So, let's calculate first the potential energy of the ship:

E = mgh   (1)

We have the mass, gravity and height, so we need to replace the given data here. Before we do that, let's remember to use the correct units. A ton is 1000 kg, so replacing and converting we have:

E = (55000 ton * 1000 kg/ton) * (9.8 m/s²) * 10 m

E = 5.39x10⁹ J

Now this energy will be the same when the ship hits the water, only that is kinetic energy that will result in the rise of temperature. To get this rise we use the following expression:

E = m * C * ΔT   (2)

We have the energy, the mass of water (assuming density of water as 1 kg/m³) and the specific heat, so, replacing in (2) and solving for ΔT we have:

ΔT = E / m * C    (3)

ΔT = 5.39x10⁹ / 4200 * 75000

<h2>ΔT = 17.11 °C</h2>

Hope this helps

5 0
3 years ago
Where is the frequency of ultrasound in relation to the range of human ability to hear
kykrilka [37]

Answer:

ultra sounds have frequency higher than the upper audible limit of human hearing, for healthy, young adults.

Explanation:

4 0
3 years ago
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An Olympic discus thrower (~100 kg) launches the 2.0 kg discus by spinning rapidly (~4 times per second) with arm outstretched (
vladimir1956 [14]

Answer:

F = 1263.03 N

Explanation:s

given,                      

mass of the disk thrower = 100 Kg

mass of the disk = 2 Kg                

angular speed of the disk  = 4 rev/s

arm outstretched = 1 m                  

centripetal force of the disk in the circular path

F = m ω² r                        

ω = 4 x 2 x π        

ω = 25.13 rad/s

F = m ω² r                      

F = 2 x 25.13² x 1

F = 1263.03 N                                              

hence, centripetal force equal to the F = 1263.03 N

6 0
3 years ago
A train whistle has a sound intensity level of 70. dB, and a library has a sound intensity level of about 40. dB. How many times
kodGreya [7K]

Answer:

The sound intensity of train is 1000 times greater than that of the library.

Explanation:

We have expression for sound intensity level,

            L=10log_{10}\left ( \frac{I}{I_0}\right )

A train whistle has a sound intensity level of 70 dB

We have

           70=10log_{10}\left ( \frac{I_1}{I_0}\right )

A library has a sound intensity level of about 40 dB

We also have

           40=10log_{10}\left ( \frac{I_2}{I_0}\right )

Dividing both equations

           \frac{70}{40}=\frac{10log_{10}\left ( \frac{I_1}{I_0}\right )}{10log_{10}\left ( \frac{I_2}{I_0}\right )}\\\\\frac{7}{4}=\frac{log_{10}\left ( \frac{I_1}{I_0}\right )}{log_{10}\left ( \frac{I_2}{I_0}\right )}\\\\10^7\frac{I_2}{I_0}=10^4\frac{I_1}{I_0}\\\\\frac{I_1}{I_2}=10^3=1000

The sound intensity of train is 1000 times greater than that of the library.

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2 years ago
Find the density of seawater at a depth where the pressure is 500 atm if the density at the surface is 1100 kg/m^3 . Seawater ha
mixer [17]

The density of seawater at a depth where the pressure is 500 atm is 1124kg/m^3

Explanation:

The relationship between bulk modulus and pressure is the following:

B=\rho_0 \frac{\Delta p}{\Delta \rho}

where

B is the bulk modulus

\rho_0 is the density at surface

\Delta p is the variation of pressure

\Delta \rho is the variation of density

In this problem, we have:

B=2.3\cdot 10^9 N/m^2 is the bulk modulus

\rho_0 =1100 kg/m^3

\Delta p = p-p_0 = 500 atm - 1 atm = 499 atm = 5.05\cdot 10^7 Pa is the change in pressure with respect to the surface (the pressure at the surface is 1 atm)

Therefore, we can find the density of the water where the pressure is 500 atm as follows:

\rho = \rho_0 + \Delta \rho = \rho_0+\frac{\rho_0 \Delta p}{B}=\rho_0 (1+\frac{\Delta p}{B})=(1100)(1+\frac{5.05\cdot 10^7}{2.3\cdot 10^9})=1124kg/m^3

Learn more about pressure in a fluid:

brainly.com/question/9805263

#LearnwithBrainly

7 0
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