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3 years ago

Can you describe the clothes washing system inside a washing machine and explain how it is an embedded system ?

Physics

2 answers:

Hoochie [10]3 years ago

7 0

Answer:

it functions based on rotatry motion..

Explanation:

because it is stationary....and has the ability to hold things in it...in a kind of random motion

Roman55 [17]3 years ago

4 0

Answer:

A system that is programmed for a particular job function is called an embedded system. A a washing machine system is programmed to wash only clothes, and when it sets to a timer, once its done washing its will end the job.

Explanation:

A system that is pre-programmed for a particular function or job is is called an embedded system.

Let us take washing machine system an example, it is an embedded system programmed for washing only, this is so, because washing machine is only used for washing the clothes and not meant for other things or particular application.

when the machine is pre-programmed, if you set a timer for 10 minutes in the washing machine to wash the clothes , after about 10 minutes has elapsed, the machine will stop washing automatically as a result of the micro-controller inside the machine.

Therefore we say the washing machine is programmed for washing of clothes.

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The order of decreasing photon energy is FM radio, AM radio, yellow light, and microwaves.

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2 years ago

An object accelerates in a direction that is always perpendicular to its motion.

anzhelika [568]

The object will not be able to accelerate perpendicular to direction of motion

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3 years ago

In a game of angry birds you launch a bird with an angle of 53 degrees to horizontal. Unfortunatly, its not a good shot and the

Alisiya [41]

Answer:

The maximum height covered is 3.25 m.

The horizontal distance covered is 9.81 m.

The total time in the air is 1.63 seconds.

Explanation:

The launch speed, $u_0= 10 m/s$ .

Angle of launch with the horizontal, $\theta = 53 ^{\circ}$

So, the vertical component of the initial velocity,

$u_0\sin\theta=10 \sin 53 ^{\circ}\cdots(i)$ .

The horizontal component of the initial velocity,

$u_0\cos\theta=10 \cos 53 ^{\circ}$

Let, t be the time of flight, to the horizontal distance covered

$D=10 \cos (53 ^{\circ})t\cdots(ii)$ .

Not, applying the equation of motion in the vertical direction.

$s= ut +\frac 1 2 at^2$

Where s is the displacement in time t, u is the initial velocity and a is the acceleration.

In this case, $u =10 \sin 53 ^{\circ}$ (from equation (i), s=0 (as the final height is same as the launch height) and $a = -9.81 m/s^2$ (negative sign is due to the downward direction).