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dezoksy [38]
1 year ago
7

A copper wire 225m long must experience a voltage drop of less than 2.0v when a current of 3.5 a passes through it. compute the

minimum diameter of the wire in millimeters (mm) if the electrical conductivity of copper is 60.0 x106 (ohm-m)-1 . answer should be in millimetres and carried out 2 decimal places (x.xx mm)
Physics
1 answer:
MA_775_DIABLO [31]1 year ago
5 0

Answer:

V = I * R

R = 2 / 3.5 = .571 ohms     maximum resistance of wire

R = ρ L / A  where R is proportional to L and inversely proportional to A

A = ρ L / R     minimum area of wire

ρ = 1 / μ  =     1.67E-8 ohm-m      resistivity inverse of conductivity

A = 1.67E-8 ohm-m * 225 m / .571 ohm = 6.68E-6 m^2

A = 6.68 mm^2       since 1 mm^2 = 10-6 m^2   or 1 mm = 10-3 m

A = Π r^2 = 6.68 mm^2

r = (6.68 / 3.14)^1/2 mm = 2.13 mm      radius of wire

d = 2 * r = 4.26 mm

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B. helium, xenon, neon

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A simple pendulum consisting of a bob of mass m attached to a string of length L swings with a period T. If the pendulum is take
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To solve this problem we will use the definition of the period in a simple pendulum, which warns that it is dependent on its length and gravity as follows:

T =2\pi \sqrt{\frac{L}{g}}

Here,

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Since the body is in constant free fall, that is, a point where gravity tends to be zero:

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2 years ago
A person has legs of length L = 1.10 m. (a) If the maximum angle between the legs during walking is ϕ = 50o , what is the person
vampirchik [111]

Answer:

a). Maximum Length L=0.929m

b). T=0.83 Hz or 1.2s

c). Longer, the effortless waling T=2.1 Hz or t=0.475s

d). t=1.2s V=0.774 \frac{m}{s}

t=0.475s V=1.95 \frac{m}{s}

Explanation:

Length legs=L=1.1m

angle=50

the step that give the person forms a triangle whose two sides are known and the angle that forms between them, then using trigonometry as the image

Divide the original triangle in two and form a right triangle so the angle is 25 and the L is hypotenuse and the opposite is the step length

a).

sin(\alpha) =\frac{op}{h}

op=h*sin(\frac{\alpha }{2})\\ op=1.1m*sin(\frac{50}{2})\\op=0.464m

Length of the step

L=0.464m*2

L=0.928m

b).

period=T

T=\frac{1}{time}=\frac{1}{t}\\ T=\frac{1}{1.2s}\\T=0.83 s^{-1}\\ T=0.83Hz

c).

T1=2\pi *\sqrt{\frac{L}{g}} \\T1=2\pi *\sqrt{\frac{1.1m}{9.8\frac{m}{s^{2}}}}\\ T1=2.1 Hz

The period is the inverse of the time of the motion so, the T1 is faster that the T because

t=\frac{1}{T1}=t= \frac{1}{2.1}=0.47s

d).

The speed is the relation between the distance with time so:

Vt=\frac{0.928m}{1.2s} \\Vt=0.773 \frac{m}{s} \\Vt=\frac{0.928m}{0.475s} \\Vt=1.953 \frac{m}{s}

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