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balandron [24]
3 years ago
8

A circuit element maintains a constant resistance. If the current through the circuit element is doubled, what is the effect on

the power dissipated by the circuit element
Physics
1 answer:
Morgarella [4.7K]3 years ago
3 0

Answer:

<h3>This shows that the power dissipated by the circuit element is four times its original power if the current is doubled.</h3>

Explanation:

The formula for calculating the power expended in a circuit is P =  I²R... 1

i is the current (in amperes)

R is the resistance (in ohms)

If  a circuit element maintains a constant resistance and the current through the circuit element is doubled, then new current I₂ = 2I

New power dissipated P₂ = (I₂)²R

P₂ = (2I)²R

P₂ = 4I²R ... 2

Dividing equation 2 by 1 will give;

P₂/P = 4I²R/I²R

P₂/P = 4

P₂ = 4P

This shows that the power dissipated by the circuit element is four times its original power if the current is doubled.

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When a carpenter shuts off his circular saw, the 10.0 inch diameter blade slows from 4250 rpm to 0.00 in 4.00 s. (a) What is the
MaRussiya [10]

Answer:

(a) \alpha=-111.26rad/s

(b) s=4450.6in

(c) 8.66in

Explanation:

First change the units of the velocity, using these equivalents 1rev=2\pi rad and 1 min =60s

4250rpm(\frac{2\pi rad}{1rev})(\frac{1 min}{60 s} )=445.06rad/s

The angular acceleration \alpha the time rate of change of the angular speed \omega according to:

\alpha=\frac{\Delta \omega}{\Delta t}

\Delta  \omega=\omega_i-\omega_f

Where \omega_i is the original velocity, in the case the velocity before starting the deceleration, and \omega_f is the final velocity, equal to zero because it has stopped.

\alpha=\frac{\Delta \omega}{\Delta t} =\frac{\omega_i-\omega_f}{4}\frac{0-445.06}{4} =\frac{-445.06}{4} =-111.26rad/s

b) To find the distance traveled in radians use the formula:

\theta = \omega_i t + \frac{1}{2} \alpha t^2

\theta = 445.06 (4) + \frac{1}{2}(-111.26) (4)^2=1780.24-890.12=890.12rad

To change this result to inches, solve the angular displacement \theta for the distance traveled s (r is the radius).

\theta=\frac{s}{r} \\s=\theta r

s=890.12(5)=4450.6in

c) The displacement is the difference between the original position and the final. But in every complete rotation of the rim, the point returns to its original position. so is needed to know how many rotations did the point in the 890.16 rad of distant traveled:

\frac{890.12}{2\pi}=141.6667

The real difference is in the 0.6667 (or 2/3) of the rotation. To find the distance  between these positions imagine a triangle formed with the center of the blade (point C), the initial position (point A) and the final position (point B). The angle \gamma=\frac{2\pi}{3}=\frac{360^o}{3}=120 is between the two sides known. Using the theorem of the cosine we can find the missing side of the the triangle(which  is also the net displacement):

c^2=a^2+b^2-2abcos(\gamma)

c^2=5^2+5^2-2(5)(5)cos(\frac{2\pi}{3} )\\c^2=25+25+25\\c^2=75\\c=5\sqrt{3}=8.66in

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4 years ago
When velocity is positive and acceleration is negative, what happens to the object’s motion?
xenn [34]

Answer:

Option (D) : The object slows down.

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3 years ago
Statistically, consumers tell more people about negative experiences in business encounters than about positive experiences. A)
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3 years ago
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5 0
3 years ago
What is the energy in joules of a photon with a frequency of 3.16e 12 s-1?
erica [24]
We have: Energy(E) = Planck's constant(h) × Frequency(∨)
Here, Planck's constant(h) = 6.626 × 10⁻³⁴ J/s
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Substitute the values into the expression:
E = (6.626 × 10⁻³⁴)(3.16 × 10¹²) J
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In short, Your Final answer would be 2.093 × 10⁻²¹ J

Hope this helps!
5 0
3 years ago
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