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3 years ago

Calculate the rate of energy transfer by a 5 ohms resistor when there is a current of 0.2 A in it.

Physics

1 answer:

vredina [299]3 years ago

5 0

Answer: The rate of energy transfer by a 5 ohms resistor when there is a current of 0.2 A in it.= 1.25 watt.

Explanation:

The rate of energy transfer from one point to another is known as power.
Its SI. unit is <em>Watt.</em>

Formula to compute power:

Power = $I^2 R$ , I = cureent, R = resistance.

Given R = 5 ohm , Current = 0.5 A.

Then, power = $(0.5)^2\times 5=0.25\times5=1.25\ watt$

Hence, the rate of energy transfer by a 5 ohms resistor when there is a current of 0.2 A in it.= 1.25 watt.

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3 years ago

A 2.0 kg, 20-cm-diameter turntable rotates at 100 rpm ons tionless bearings. Two 500 g blocks fall from above, hit the tum ble s

Verdich [7]

There are mistakes in the question.The correct question is here

A 2.0 kg, 20-cm-diameter turntable rotates at 100 rpm on frictionless bearings. Two 500 g blocks fall from above, hit the turntable simultaneously at opposite ends of a diameter, and stick. What is the turntable’s angular velocity, in rpm, just after this event?

Answer:

w=50 rpm

Explanation:

Given data

The mass turntable M=2kg

Diameter of the turntable d=20 cm=0.2 m

Angular velocity ω=100 rpm= 100×(2π/60) =10.47 rad/s

Two blocks Mass m=500 g=0.5 kg

To find

Turntable angular velocity

Solution

We can find the angular velocity of the turntable as follow

Lets consider turntable to be disk shape and the blocks to be small as compared to turntable

$I_{turntable}w=I_{block1}w^{i}+I_{turntable}w^{i}+I_{block2}w^{i}$

where I is moment of inertia

$w^{i}=\frac{I_{turntable}w}{I_{block1}w^{i}+I_{block2}w^{i}+I_{turntable}w^{i}}\\ So\\I_{turntable}=M\frac{r^{2} }{2}\\I_{turntable}=2*(\frac{(0.2/2)}{2} )\\ I_{turntable}=0.01 \\And\\I_{block1}=I_{block2}=mr^{2}\\I_{block1}=I_{block2}=(0.5)*(0.2/2)^{2} \\ I_{block1}=I_{block2}==0.005\\so\\w^{i}=\frac{I_{turntable}w}{I_{block1}w^{i}+I_{block2}w^{i}+I_{turntable}w^{i}}\\w^{i}=\frac{0.01*(10.47)}{0.005+0.005+0.01} \\w^{i}=5.235 rad/s\\w^{i}=5.235*(60/2\pi )\\w^{i}=50 rpm$

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3 years ago

Nimfa-mama [501]

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