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3 years ago

What can be done to increase speed of rotation of a DC motor

Physics

1 answer:

nexus9112 [7]3 years ago

7 0

Answer: There are multiple options:

By varying the supply voltage.

By varying the flux, and by varying the current through the field winding.

By varying the armature voltage, and by varying the armature resistance.

Helpful Link:

https://www.elprocus.com/what-are-the-best-ways-to-control-the-speed-of-dc-motor/

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Two resistors with values of 6.0 ohms and 12 ohms are connected in parallel. This combination is connected in series with a 2.0

ivanzaharov [21]

Answer:

Explanation:

According to ohm's law;

E = IRt where;

E is the source voltage = 24volts

I is the total current flowing in the circuit = ?

Rt is the total effective resistance in the circuit.

To find Rt, we will resolve the resistors in parallel first.

Since 6ohms and 12ohms resistors are in parallel, their effective resistance will give;

1/R = 1/6+1/12

1/R= 2+1/12

1/R = 3/12

3R = 12

R = 4ohms.

This resistor will now be in series with the 2.0ohms resistor to finally have;

Rt = 4+2

Rt = 6ohms

From the ohms law formula;

I = E/Rt

I = 24/6

I = 4Amperes

The total current in the circuit is 4A

This same currents will flow in the 2ohms resistor since same current flows in a series connected resistors.

3 0

3 years ago

Read 2 more answers

I need help ASAP please :)

rusak2 [61]

Density offers a convenient means of obtaining the mass of a body from its volume or vice versa; the mass is equal to the volume multiplied by the density (M = Vd), while the volume is equal to the mass divided by the density (V = M/d).

M = V d

M = 1.4 * 2 = 2.8 kg

7 0

2 years ago

A particle is moving with velocity v(t) = t2 _ 9t + 18 with distance, s measured in meters, left or right of zero, and t measure

Alex787 [66]

V = t^2 - 9t + 18

position, s
s = t^3 /3 - 4.5t^2 +18t + C

   t = 0, s = 1 => 1=C => s = t^3/3 -4.5t^2 + 18t + 1

Average velocity: distance / time

   distance: t = 8 => s = 8^3 / 3 - 4.5 (8)^2 + 18(8) + 1 = 27.67 m
   Average velocity = 27.67 / 8 = 3.46 m/s

t = 5 s

   v = t^2 - 9t + 18 = 5^2 - 9(5) + 18 = -2 m/s
   speed = |-2| m/s = 2 m/s

Moving right
   V > 0 => t^2 - 9t + 18 > 0
   (t - 6)(t - 3) > 0

   => t > 6 and t > 3 => t > 6 s => Interval (6,8)

    => t < 6 and t <3 => t <3 s => interval (0,3)



Going faster and slowing dowm

acceleration, a = v' = 2t - 9
   a > 0 => 2t - 9 > 0 => 2t > 9 => t > 4.5 s
   Then, going faster in the interval (4.5 , 8) and slowing down in (0, 4.5)

4 0

3 years ago

A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from 0 to 5.40 m/s

tekilochka [14]

Answer:

a) 378Ns

b) 477.27N

Explanation:

Impulse is the defined as the product of the applied force and time taken. This is expressed according to the formula

I = Ft = m(v-u)

m is the mass = 70kg

v is the final velocity = 5.4m/s

u is the initial velocity = 0m/s

Get the impulse

I = m(v-u)

I = 70(5.4-0)

I = 70(5.4)

I = 378Ns

b) Average total force is expressed as

F = ma (Newton's second law)

F = m(v-u)/t

F = 378/0.792

F = 477.27N

Hence the average total force experienced by a 70.0-kg passenger in the car during the time the car accelerates is 477.27N

3 0

3 years ago

You set your stationary bike on a high 80-N friction-like resistive force and cycle for 30 min at a speed of 8.0 m/s . Your body

stellarik [79]

A) The change in internal chemical energy is $1.15\cdot 10^7 J$

B) The time needed is 1 minute

Explanation:

First of all, we start by calculating the power output of you and the bike, given by:

$P=Fv$

where

F = 80 N is the force that must be applied in order to overcome friction and travel at constant speed

v = 8.0 m/s is the velocity

Substituting,

$P=(80)(8.0)=640 W$

The energy output is related to the power by the equation

$P=\frac{E}{t}$

where:

P = 640 W is the power output

E is the energy output

$t = 30 min \cdot 60 = 1800 s$ is the time elapsed

Solving for E,

$E=Pt=(640)(1800)=1.15\cdot 10^6 J$

Since the body is 10% efficient at converting chemical energy into mechanical work (which is the output energy), this means that the change in internal chemical energy is given by

$\Delta E = \frac{E}{0.10}=\frac{1.15\cdot 10^6}{0.10}=1.15\cdot 10^7 J$