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3 years ago

What type of motor operates at a constant steady-state speed regardless of the load?

1 answer:

7 0

The type of motor that allows for constant speed regardless of load are called Geared Speed Control motors. This type of motor has a tachometer feedback device attached at the rear of the motor that gives constant feedback to the speed controller giving the advantage of constant speed regardless of load. The tachometer allows for varied frequency delivery to the motor to maintain pre-set output speed.

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A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 35 ft/s. Its height

Svetllana [295]

Answer:

Part A:

(a): -121.26 ft/s.

(b): -121.13 ft/s.

(c): -121.052 ft/s.

(d): -121.026 ft/s.

Part B:

-121.00 ft/s.

Explanation:

Given that the height of the balloon after t seconds is

$\rm y(t) = 35 t-26t^2.$

The average velocity of an object is defined as the total distance traveled by the object divided by the time taken in covering that distance.

$\rm v_{av} = \dfrac{y_2-y_1}{t_2-t_1}$

where,

$\rm y_2,\ y_1$ are the positions of the object at time $\rm t_1$ and $\rm t_2$ respectively.

For the average velocity for the time period beginning when t=3 and lasting .01 sec.

For this case,

$\rm t_1 = 3\ sec.\\t_2 = 3+0.01\ sec = 3.01\ sec.\\\\Therefore, \\\\y_1 = 35\cdot 3-26\cdot 3^2=-129\ ft.\\y_2 = 35\cdot 3.01-26\cdot 3.01^2=-130.2126\ ft.\\\\\Rightarrow v_{av}=\dfrac{-130.2126-(-129)}{3.01-3}=-121.26\ ft/s.$

For the average velocity for the time period beginning when t=3 and lasting .005 sec.

For this case,

$\rm t_1 = 3\ sec.\\t_2 = 3+0.005\ sec = 3.005\ sec.\\\\Therefore, \\\\y_1 = 35\cdot 3-26\cdot 3^2=-129\ ft.\\y_2 = 35\cdot 3.005-26\cdot 3.005^2=-129.60565\ ft.\\\\\Rightarrow v_{av}=\dfrac{-129.60565-(-129)}{3.005-3}=-121.13\ ft/s.$

For the average velocity for the time period beginning when t=3 and lasting .002 sec.

For this case,

$\rm t_1 = 3\ sec.\\t_2 = 3+0.002\ sec = 3.002\ sec.\\\\Therefore, \\\\y_1 = 35\cdot 3-26\cdot 3^2=-129\ ft.\\y_2 = 35\cdot 3.002-26\cdot 3.002^2=-129.2421\ ft.\\\\\Rightarrow v_{av}=\dfrac{-129.2421-(-129)}{3.002-3}=-121.052\ ft/s.$

For the average velocity for the time period beginning when t=3 and lasting .001 sec.

For this case,

$\rm t_1 = 3\ sec.\\t_2 = 3+0.001\ sec = 3.001\ sec.\\\\Therefore, \\\\y_1 = 35\cdot 3-26\cdot 3^2=-129\ ft.\\y_2 = 35\cdot 3.001-26\cdot 3.001^2=-129.121\ ft.\\\\\Rightarrow v_{av}=\dfrac{-129.121-(-129)}{3.001-3}=-121.026\ ft/s.$

The instantaneous velocity of the balloon at the given time is defined as the rate of change of its position at that time.

$\rm v(t) = \dfrac{dy}{dt}\\=\dfrac{d}{dt}\left ( 35 t-26t^2\right )\\\\=35-26\times 2t.\\\\At\ t=3,\\\\v(t)=35-26\times 2\times 3=-121.00\ ft/s.$

<u>Note:</u><em> The negative sign with all the velocities indicates that the direction of these velocities are downwards.</em>

8 0

3 years ago

Two narrow, parallel slits separated by 0.850 mm are illuminated by 600-nm light, and the viewing screen is 2.80 m away from the

CaHeK987 [17]

Answer:

a) 7.947 radians

b) $\mathbf{\frac{I}{I_{max}}=0.4535}$

Explanation:

y = Distance from central bright fringe = 2.5 mm

λ = Wavelength = 600 nm

L = Distance between screen and source = 2.8 m

d = Slit distance = 0.85 mm

$tan\theta =\frac{y}{L}\\\Rightarrow tan\theta =\frac{2.5}{2800}=0.000892\\\Rightarrow \theta=tan^{-1}0.000892=0.05115^{\circ}$

$\Delta r= dsin\theta\Rightarrow \Delta r=dsin0.05115=7.589\times10^{-7}$

a) Phase difference

$\phi=\frac{2\pi}{\lambda}\Delta r\\\Rightarrow \phi=\frac{2\pi}{600\times 10^{-9}}7.589\times10^{-7}=7.947\ rad$

∴ Phase difference between the two interfering waves on a screen at a point 2.50 mm from the central bright fringe is 7.947 radians

b) $\frac{I}{I_{max}}=cos^2\frac{\phi}{2}\\\Rightarrow \frac{I}{I_{max}}=cos^2\frac{7.947}{2}\\\Rightarrow \mathbf{\frac{I}{I_{max}}=0.4535}$

∴ Ratio of the intensity at this point to the intensity at the center of a bright fringe $\mathbf{\frac{I}{I_{max}}=0.4535}$

8 0

3 years ago

In order to do work, the force vector must be

aliina [53]

in the same direction as the displacement vector and the motion

5 0

3 years ago

What is an example of a non contact force

olchik [2.2K]

Answer: Gravitational force

Explanation:

A non contact force can be described as a force applied to an object by another body that is not in direct contact with it.

For example, an object thrown upwards will return back due to the force of gravity acting on it. So, it means Gravitational force is acting on the body without necessarily being in contact with that body.

8 0

3 years ago

If I work out rotational energy to be 102.2J which equals kg.M/s^2, and I hadn't factored time into it, would that be Joules per

Marina CMI [18]

Answer:

0.057 joules is needed to create the total rotational energy each second.

Explanation:

The energy rate is the ratio of total energy to time, which coincides with the definition of power at constant rate:

$\dot W = \frac{\Delta E}{\Delta t}$