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3 years ago

Construction, principal and working of electric motor

1 answer:

5 0

A power supply A basic motor has a DC power source, in general. It supplies power for the motor or field coils armature. Field Magnet can be a permanent magnet or electromagnet The magnetic field tends to generate a torque on the spinning armature coil by virtue of Fleming's lefthand rule.

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An object of mass 9.00 kg attached to an ideal massless spring is pulled with a steady horizontal force across a frictionless le

frosja888 [35]

Answer:

Acceleration of the object will be $1.05m/sec^2$

Explanation:

We have given mass of the object m = 9 kg

spring constant k = 95 N/m

Spring is stretched by 10 cm

So x = 10 cm = 0.1 m

We know that force is given by $F=kx=95\times 0.1=9.5N$

From newton's law we also know that force is given by

F = ma , here m is mass and a is acceleration

So $9.5=9\times a$

$a=1.05m/sec^2$

5 0

4 years ago

A 260-kg glider is being pulled by a 1,940-kg jet along a horizontal runway with an acceleration of a= 2.20 m/s^2 to the right.

lisov135 [29]

Answer:

a) The magnitude of the thrust provided by the jet's engines is 4840 newtons.

b) The magnitude of the tension in the cable connecting the jet and glider is 572 newtons.

Explanation:

a) By Newton's laws we construct the following equations of equilibrium. Please notice that both the glider and the jet experiments has the same acceleration:

Jet

$\Sigma F = F - T = m_{J}\cdot a$ (1)

Glider

$\Sigma F = T = m_{G}\cdot a$ (2)

Where:

$F$ - Thrust of jet engines, measured in newtons.

$T$ - Tension in the cable connecting the jet and glider, measured in newtons.

$m_{G}$ , $m_{J}$ - Masses of the glider and the jet, measured in kilograms.

$a$ - Acceleration of the glider-jet system, measured in meters per square second.

If we know that $m_{G} = 260\,kg$ , $m_{J} = 1,940\,kg$ and $a = 2.20\,\frac{m}{s^{2}}$ , then the solution of this system of equations:

By (2):

$T = (260\,kg)\cdot \left(2.20\,\frac{m}{s^{2}} \right)$

$T = 572\,N$

By (1):

$F = T+m_{J}\cdot a$

$F = 572\,N+(1,940\,kg)\cdot \left(2.20\,\frac{m}{s^{2}} \right)$

$F = 4840\,N$

The magnitude of the thrust provided by the jet's engines is 4840 newtons.

b) The magnitude of the tension in the cable connecting the jet and glider is 572 newtons.

7 0

3 years ago

When a 2.50 kg object is hung vertically on a certain light spring described by hookes law the spring stretches 2.76 cm.

Readme [11.4K]

F=K*X,
F=M*a

M*a=K*X

2.5*9.81=K*0.0276

24.525=K*0.0276

24.525/0.0276=K

K= 888.6 N/m ---- force constant

assuming 2.5 refers to the new extension, just divide F/ 0.025
to get

981N/m

8 0

4 years ago

What does the star release, and in what form do we see it ?

Dimas [21]

Stars are made of very hot gas. This gas is mostly hydrogen and helium, which are the two lightest elements. Stars shine by burning hydrogen into helium in their cores, and later in their lives create heavier elements.

3 0

4 years ago

what is the property of an object due to its mass by which it resists any change in its position unless overcome by force ?

UkoKoshka [18]

Answer:

Mass as a Measure of the Amount of Inertia

All objects resist changes in their state of motion. All objects have this tendency - they have inertia.

Explanation:

hope this helps

5 0

3 years ago

Construction, principal and working of electric motor​

Construction, principal and working of electric motor