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JulsSmile [24]
3 years ago
8

An alternating current is supplied to an electronic component with a warning that the voltage across it should never exceed 12 V

. What is the highest rms voltage that can be supplied to this component while staying below the voltage limit in the warning?
Physics
1 answer:
EastWind [94]3 years ago
8 0

Answer:

<em>The highest rms voltage will be 8.485 V</em>

<em></em>

Explanation:

For alternating electric current, rms (root means square) is equal to the value of the direct current that would produce the same average power dissipation in a resistive load

If the peak or maximum voltage should not exceed 12 V, then from the relationship

V_{rms} = \frac{V_{p} }{\sqrt{2} }

where V_{rms} is the rms voltage

V_{p} is the peak or maximum voltage

substituting values into the equation, we'll have

V_{rms} = \frac{12}{\sqrt{2} } = <em>8.485 V</em>

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Which of the following is NOT correct? . . a A changing electric field can produce a changing magnetic field . . b A steady magn
postnew [5]
The correct answer of the given question above would be option B. The statement that is not correct is that, a steady magnetic field produces a steady current. The rest of the statements are all correct. <span>An unchanging/static magnetic field (relative to a wire/circuit) induces zero current.</span>
4 0
3 years ago
Read 2 more answers
A bird watcher meanders through the woods, walking 1.93 km due east, 1.03 km due south, and 3.84 km in a direction 52.8 ° north
Sedaia [141]

Answer:

Magnitude of displacement = 2.07 km

Magnitude of average velocity = 1.17 kmph

Explanation:

Let east represent positive x axis and north represent positive y axis.

A bird watcher meanders through the woods, walking 1.93 km due east, 1.03 km due south, and 3.84 km in a direction 52.8 ° north of west.

1.93 km due wast

           s ₁ = 1.93 i km

1.03 km due south

           s₂ = -1.03 j km

3.84 km in a direction 52.8 ° north of west

           s₃ = -3.84 cos 52.8 i + 3.84 sin 52.8 j = -2.32 i + 3.06 j km

Total displacement

          s = s ₁+  s₂+ s₃ = 1.93 i - 1.03 j -2.32 i + 3.06 j = -0.39 i + 2.03 j

  Magnitude of displacement, =\sqrt{(-0.39)^2+2.03^2}=2.07km

Time taken = 1.771 hour

Magnitude of average velocity, =\frac{2.07}{1.771}=1.17km/hr

7 0
3 years ago
Can you make the work output of a machine greater than the work input?
drek231 [11]
Nearly equal the output work is greater than the input work because of friction.All machines use some amount of input work to overcome friction.The only way to increase the work output is to increase the work you put into the machine.You cannot get more work out of a machine than you put into it.
4 0
3 years ago
Compute the kinetic energy of a proton (mass 1.67×10−27kg ) using both the nonrelativistic and relativistic expressions for spee
JulsSmile [24]

Answer:

The non-relativistic kinetic energy of a proton is 6.76\times10^{-12}\ J

The relativistic kinetic energy of a proton is 7.25\times10^{-12}\ m/s

Explanation:

Given that,

Mass of proton m=1.67\times10^{-27}\ kg

Speed v= 9.00\times10^{7}\ m/s

We need to calculate the kinetic energy for non relativistic

Using formula of kinetic energy

K.E=\dfrac{1}{2}mv^2

Put the value into the formula

K.E=\dfrac{1}{2}\times1.67\times10^{-27}\times(9.00\times10^{7})^2

K.E=6.76\times10^{-12}\ J

We need to calculate the kinetic energy for relativistic

Using formula of kinetic energy

K.E=mc^2(\sqrt{(\dfrac{1}{1-\dfrac{v^2}{c^2}})}-1)

K.E=1.67\times10^{-27}\times(3\times10^{8})^{2}\cdot\left(\sqrt{\frac{1}{1-\frac{\left(9.00\times10^{7}\right)^{2}}{(3\times10^{8})^{2}}}}-1\right)

K.E=7.25\times10^{-12}\ m/s

Hence, The non-relativistic kinetic energy of a proton is 6.76\times10^{-12}\ J

The relativistic kinetic energy of a proton is 7.25\times10^{-12}\ m/s

7 0
3 years ago
Q2. You push a crate up a ramp with a force of 10 N. Despite your pushing, the crate slides down the ramp 4 m. How much work did
Ksivusya [100]

Answer:

40 J

Explanation:

From the question given above, the following data were obtained:

Force (F) = 10 N

Distance (s) = 4 m

Workdone (Wd) =?

Work done is simply defined as the product of force and distance moved in the direction of the force. Mathematically, we can express the Workdone as:

Workdone = force × distance

Wd = F × s

With the above formula, we can obtain the workdone as follow:

Force (F) = 10 N

Distance (s) = 4 m

Workdone (Wd) =?

Wd = F × s

Wd = 10 × 4

Wd = 40 J

Thus, 40 J of work was done.

5 0
3 years ago
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