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Katen [24]
3 years ago
12

What is your REASONING for this CLAIM? EXPLAIN WHY your claim is true.

Physics
1 answer:
jok3333 [9.3K]3 years ago
7 0

I DONT know FiGURE it out YOURSELF

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A constant torque of 3 Nm is applied to an unloaded motor at rest at time t = 0. The motor reaches a speed of 1,393 rpm in 4 s.
irakobra [83]

Answer:

The moment of inertia of the motor is 0.0823 Newton-meter-square seconds.

Explanation:

From Newton's Laws of Motion and Principle of Motion of D'Alembert, the net torque of a system (\tau), measured in Newton-meters, is:

\tau = I\cdot \alpha (1)

Where:

I - Moment of inertia, measured in Newton-meter-square seconds.

\alpha - Angular acceleration, measured in radians per square second.

If motor have an uniform acceleration, then we can calculate acceleration by this formula:

\alpha = \frac{\omega - \omega_{o}}{t} (2)

Where:

\omega_{o} - Initial angular speed, measured in radians per second.

\omega - Final angular speed, measured in radians per second.

t - Time, measured in seconds.

If we know that \tau = 3\,N\cdot m, \omega_{o} = 0\,\frac{rad}{s }, \omega = 145.875\,\frac{rad}{s} and t = 4\,s, then the moment of inertia of the motor is:

\alpha = \frac{145.875\,\frac{rad}{s}-0\,\frac{rad}{s}}{4\,s}

\alpha = 36.469\,\frac{rad}{s^{2}}

I = \frac{\tau}{\alpha}

I = \frac{3\,N\cdot m}{36.469\,\frac{rad}{s^{2}} }

I = 0.0823\,N\cdot m\cdot s^{2}

The moment of inertia of the motor is 0.0823 Newton-meter-square seconds.

5 0
3 years ago
A wave with a greater amplitude will transfer . . . . \
BabaBlast [244]

less mass is more mass but less energy in more mass. less mass has more energy

6 0
3 years ago
HELP ASAP WITH number 6 plz
Alika [10]
The answer would be c
6 0
3 years ago
The nucleus of an atom is composed of unchanged particles as well as of the following
Ivan
I suppose that you wanted write "uncharged". The particles without electrical charge present in the nucleus are called neutrons.
7 0
3 years ago
Water is pumped steadily out of a flooded basement at a speed of 5.4 m/s through a uniform hose of radius 0.83 cm. The hose pass
Gala2k [10]

To solve this problem it is necessary to apply the concepts related to the flow as a function of the volume in a certain time, as well as the potential and kinetic energy that act on the pump and the fluid.

The work done would be defined as

\Delta W = \Delta PE + \Delta KE

Where,

PE = Potential Energy

KE = Kinetic Energy

\Delta W = (\Delta m)gh+\frac{1}{2}(\Delta m)v^2

Where,

m = Mass

g = Gravitational energy

h = Height

v = Velocity

Considering power as the change of energy as a function of time we will then have to

P = \frac{\Delta W}{\Delta t}

P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)

The rate of mass flow is,

\frac{\Delta m}{\Delta t} = \rho_w Av

Where,

\rho_w = Density of water

A = Area of the hose \rightarrow A=\pi r^2

The given radius is 0.83cm or 0.83 * 10^{-2}m, so the Area would be

A = \pi (0.83*10^{-2})^2

A = 0.0002164m^2

We have then that,

\frac{\Delta m}{\Delta t} = \rho_w Av

\frac{\Delta m}{\Delta t} = (1000)(0.0002164)(5.4)

\frac{\Delta m}{\Delta t} = 1.16856kg/s

Final the power of the pump would be,

P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)

P = (1.16856)((9.8)(3.5)+\frac{1}{2}5.4^2)

P = 57.1192W

Therefore the power of the pump is 57.11W

6 0
3 years ago
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