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3 years ago

What is the purpose of the commutator in an electric motor?

1 answer:

8 0

<span>Commutator is the moving part of the electric motor and periodically reverses the current direction between the rotor and the external circuit.One end of the coil is attached to each half of the commutator. Commutator are used so when electricity flows through the circuit, the coil will rotate continually in the same direction.</span>

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Does anyone know how to do this?

SIZIF [17.4K]

Explanation:

By Hooke's Law,

F=kx

The only force acting here is weight, and x is the extension of the string (you need to convert this to mm) so

mg=kx

(0.15)(9.81)=k ((420-300)x10^-3)

Then just solve this equation.

8 0

3 years ago

A piece of metal has a mass of 10g and a mass of 2cm

qwelly [4]

Answer:

so whats the questain?!

Explanation:

idgi

3 0

3 years ago

If an object is thrown in an upward direction from the top of a building 160 ft. High at an initial speed of 21.82 mi/h what is

viktelen [127]

To solve this problem we are going to use tow kinematic equations for falling objects.
1. Kinematic equation for final velocity: $V_{f}=V_{i}+gt$
where
$V_{f}$ is the final velocity
$V_{i}$ is the initial velocity
$g$ is the acceleration due to gravity $32 \frac{ft}{s^2}$
$t$ is the time
2. Kinematic equation for distance: $d=V_{i}t+ \frac{1}{2} gt^2$
where
$d$ is the distance
$V_{i}$ is the initial velocity
$V_{f}$ is the final velocity
$g$ is the acceleration due to gravity $32 \frac{ft}{s^2}$
$t$ is the time

First, we are going to convert 21.82 mi/h to ft/s:
$21.82 \frac{mi}{h} =31.21 \frac{ft}{s}$

Next, we are going to use the first equation to find how long it takes for the rock to reach its maximum height.
We know for our problem that the object is thrown in upward direction, so its velocity at its maximum height (before falling again) will be zero; therefore: $V_{f}=0$ . We also know that it initial speed is 31.21 ft/s, so $V_{i}=31.21$ . Lets replace those values in our formula to find $t$ :
$V_{f}=V_{i}+gt$
$0=31.21+(-32)t$
$-32t=-31.21$
$t= \frac{-31.21}{-32}$
$t=0.98$ seconds

Next, we are going to use that time in our second kinematic equation to find the distance the object reach at its maximum height:
$d=V_{i}t+ \frac{1}{2} gt^2$
$d=31.21(0.98)+ \frac{1}{2} (-32)(0.98)^2$
$d=15.22$ ft

Now we can add the height of the building and the maximum height of the object:
$d=160+15.22=175.22$ ft

Next, we are going to use that height (distance) in our second kinematic equation one more time to fin how long it takes for the object to fall from its maximum height to the ground:
$d=V_{i}t+ \frac{1}{2} gt^2$
$175.22=31.21t+ \frac{1}{2} (32)t^2$
$16t^2+31.21t-175.22=0$
$t=2.47$ or $t=-4.43$
Since time cannot be negative, $t=2.47$ is the time it takes the object to fall to the ground.

Finally, we can use that time in our first kinematic equation to find the final speed of the object when it hits the ground:
$V_{f}=V_{i}+gt$
$V_{f}=31.21+(32)(2.47)$
$V_{f}=110.25$ ft/s

We can conclude that the speed of the object when it hits the ground is 110.25 ft/s

5 0

3 years ago

based on the law of conservation of energy. how can we reasonably improve a machines ability to do work? A.move the machine to a

natima [27]

The correct answer is

C. reduce the friction between its moving parts

In fact, by reducing the friction between the moving parts of the machine, it is possible to reduce the energy wasted due to this friction; therefore, more input energy is converted into useful work, and this will improve the efficiency of the machine.

8 0

4 years ago

Read 2 more answers

A 1-kg iron frying pan is placed on a stove. The pan increases from 20°C to 250°C. If the same amount of heat is added to a pan

sergejj [24]

Here mass of the iron pan is given as 1 kg

now let say its specific heat capacity is given as "s"

also its temperature rise is given from 20 degree C to 250 degree C

so heat required to change its temperature will be given as

$Q = ms \Delta T$

$Q = 1*s*(250 - 20)$

$Q = 1*s*230$

now if we give same amount of heat to another pan of greater specific heat

so let say the specific heat of another pan is s'

now the increase in temperature of another pan will be given as

$Q = ms'\Delta T$

$1*s*230 = 1* s' * \Delta T$

now we have

$\Delta T = (\frac{s}{s'})*230$

now as we know that s' is more than s so the ratio of s and s' will be less than 1

And hence here we can say that change in temperature of second pan will be less than 230 degree C which shows that final temperature of second pan will reach to lower temperature

So correct answer is

<u>A) The second pan would reach a lower temperature.</u>

3 0

3 years ago

Read 2 more answers