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UNO [17]
3 years ago
7

The design of a 60.0 cm industrial turntable requires that it has a kinetic energy of 0.250 j when turning at 45.0 rpm. What mus

t be the moment of inertia of the turntable about the rotation axis
Physics
1 answer:
Aneli [31]3 years ago
5 0

Answer:

The moment of inertia of the turntable about the rotation axis is 0.0225 kg.m²

Explanation:

Given;

radius of the turnable, r = 60 cm = 0.6 m

rotational kinetic energy, E = 0.25 J

angular speed of the turnable, ω = 45 rpm

The rotational kinetic energy is given as;

E_{rot} = \frac{1}{2} I \omega ^2

where;

I is the moment of inertia about the axis of rotation

ω is the angular speed in rad/s

\omega = 45 \frac{rev}{\min} \times \frac{2 \pi \ rad}{1 \ rev} \times \frac{1 \ \min}{60 \ s} \\\\\omega = 4.712 \ rad/s

E = \frac{1}{2} I \omega ^2\\\\I = \frac{2E}{\omega ^2} \\\\I = \frac{2 \ \times \ 0.25}{(4.712)^2} \\\\I = 0.0225 \ kg.m^2

Therefore, the moment of inertia of the turntable about the rotation axis is 0.0225 kg.m²

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What is the strength of the electric field 0.1 mmmm below the center of the bottom surface of the plate
Aleksandr [31]

Complete Question

A thin, horizontal, 12-cm-diameter copper plate is charged to 4.4 nC . Assume that the electrons are uniformly distributed on the surface. What is the strength of the electric field 0.1 mm above the center of the top surface of the plate?

Answer:

The  values is  E =248.2 \  N/C

Explanation:

From the question we are told that

   The  diameter is  d =  12 \  cm  =  0.12 \ m

    The charge  is  Q =  4.4 nC  =  4.4 *10^{-9} \  C

    The  distance from the center  is  k =  0.1 \ mm   =  1*10^{-4} \  m

Generally the radius is mathematically represented as

        r =  \frac{d}{2}

=>     r =  \frac{0.12}{2}

=>       r =  0.06 \  m

Generally electric field is mathematically represented as  

       E =  \frac{Q}{ 2\epsilon_o } [1 - \frac{k}{\sqrt{r^2 +  k^2 } } ]

substituting values  

      E =  \frac{4.4 *10^{-9}}{ 2* (8.85*10^{-12}) } [1 - \frac{(1.00 *10^{-4})}{\sqrt{(0.06)^2 +  (1.0*10^{-4})^2 } } ]

     E =248.2 \  N/C

4 0
4 years ago
magine two carts, one with twice the mass of the other, that are going to have a head-on collision. In order for the two carts t
scoray [572]

Answer:

Twice as fast

Explanation:

Solution:-

- The mass of less massive cart = m

- The mass of Massive cart = 2m

- The velocity of less massive cart = u

- The velocity of massive cart = v

- We will consider the system of two carts to be isolated and there is no external applied force on the system. This conditions validates the conservation of linear momentum to be applied on the isolated system.

- Each cart with its respective velocity are directed at each other. And meet up with head on collision and comes to rest immediately after the collision.

- The conservation of linear momentum states that the momentum of the system before ( P_i ) and after the collision ( P_f ) remains the same.

                             P_i = P_f

- Since the carts comes to a stop after collision then the linear momentum after the collision ( P_f = 0 ). Therefore, we have:

                             P_i = P_f = 0

- The linear momentum of a particle ( cart ) is the product of its mass and velocity as follows:

                             m*u - 2*m*v = 0

Where,

                 ( u ) and ( v ) are opposing velocity vectors in 1-dimension.

- Evaluate the velcoity ( u ) of the less massive cart in terms of the speed ( v ) of more massive cart as follows:

                          m*u = 2*m*v

                              u = 2*v

Answer: The velocity of less massive cart must be twice the speed of more massive cart for the system conditions to hold true i.e ( they both come to a stop after collision ).

8 0
3 years ago
The drill used by most dentists today is powered by a small air-turbine that can operate at angular speeds of 350000 {\rm rpm} .
damaskus [11]

Answer:

5833.33

Explanation:

\alpha = Angular acceleration

\theta = Number of revolutions

\omega_i = Initial angular speed = 0

t = Time taken = 2 s

Final angular speed

\omega_f=\dfrac{350000}{60}=5833.33\ rps

From the equation of rotational motion we have

\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{5833.33-0}{2}\\\Rightarrow \alpha=2916.665\ rev/s^2

\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\Rightarrow \theta=0\times t+\dfrac{1}{2}\times 2916.665\times 2^2\\\Rightarrow \theta=5833.33\ rev

The number of revolutions is 5833.33

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3 years ago
Four spheres with positive and negative charges hang from strings.
Arada [10]
Be because it’s the answer
5 0
3 years ago
Match each method of erosion with the correct definition
Katena32 [7]
In this item, we are asked to determine the correct definition for the terms. It is unfortunate that they meanings to which we could match them are not given. With that, the answers written here are the well-researched ones.

Solution --> This type of erosion happens when the components found in the rocks (i.e. limestone) are dissolved through the acid components of acid rain.

Abrasion --> This is a process by which the rocks that are eroded to the riverbanks wear down other components of the bed.

Saltation --> This process involves the movement of the particles through a series of bounces that are short and are along the surface of the ground. This results to dislodging the other particles. 
7 0
4 years ago
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