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OLEGan [10]
3 years ago
12

A 1.5 wire carries a 2 a current when a potential difference of 87 v is applied. What is the resistance of the wire

Physics
1 answer:
Dmitry_Shevchenko [17]3 years ago
3 0

Ohm's law states that V = IR

                                   87 = 2 x R

                                    R = 87/2 ohms

Hope this helps :)

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Carbon is allowed to diffuse through a steel plate 15 mm thick. The concentrations of carbon at the two faces are 0.65 and 0.30
beks73 [17]

Answer:

T=575.16K

Explanation:

To solve the problem we proceed to use the 1 law of diffusion of flow,

Here,

J=-D\frac{\Delta C}{\Delta x}

\Delta C is the rate in concentration

\Delta xis the rate in thickness

D is the diffusion coefficient, where,

D= D_0 exp(\frac{Q_d}{RT})

Replacing D in the first law,

J=-(D_0 exp(\frac{-Q_D}{RT}))\frac{\Delta }{\Delta x}

clearing T,

T=\frac{Q_d}{R*ln(\frac{J*\Delta x}{D_0*\Delta C})}

Replacing our values

T=-\frac{80000}{8.31*ln(\frac{(6.2*10^{-7})(-15*10^{-3})}{(1.43*10^{-9})(0.65-0.30)})}

T=-\frac{80000}{-138.09}

T=575.16K

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3 years ago
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3 years ago
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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3 years ago
1. Una pelota rueda hacia la derecha siguiendo una trayectoria en línea recta de modo que recorre una distancia de 10m en 5 s ,
devlian [24]

Answer:

https://youtu.be/ymHHdoCGJOU

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3 years ago
What is the weight of a 225kg space probe on the moon and the acceleration of gravity on the moon is 1.62
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<span>364N should be your answer.. hope this helps

</span>
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3 years ago
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