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3 years ago

What hall voltage is produced by a 0.200-t field applied across a 2.60-cm-diameter aorta when blood velocity is 60.0 cm/s?

1 answer:

amm18123 years ago

8 0

The hall voltage will be calculated using the formula:

E = Blv

where:

>Hall voltage: E = ?
>Magnetic field:
B = 0.200 Tesla or Wb/m^2
>Width of conductor or Diameter of Aorta:
l = 2.60 cm, converting to meter = .0260 m
>Velocity of charge flowing:
v = 60 cm/s, converting to meter = 0.6 m/s

Substituting the given :

E = (0.200 Wb/m^2) * (0.260 m) * (0.6 m/s)
E = (0.200 Wb/m^2) * (0.156 m^2/s)
E = 0.0312 Wb/s

Since 1 volt = 1 Wb/s then,

E = 0.0312 V or 31.2 mV

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max2010maxim [7]

Answer:

Explanation:

point a represents time 0 and position coordinate 30

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3 years ago

A playground merry-go-round has a mass of 50 kg and a diameter of 4.0 m. There are 4 children who want to ride on it. They have

mixer [17]

Answer:

B) I1 = 1680 kg.m^2 I2 = 1120 kg.m^2

C) V = 0.84m/s T = 29.92s

D) ω2 = 0.315 rad/s

Explanation:

The moment of inertia when they are standing on the edge:

$I1 = 1/2*M*R^2 + (m1+m2+m3+m4)*R^2$ where M is the mass of the merry-go-round.

I1 = 1680 kg.m^2

The moment of inertia when they are standing half way to the center:

$I2 = 1/2*M*R^2 + (m1+m2+m3+m4)*(R/2)^2$

I2 = 1120 kg.m^2

The tangencial velocity is given by:

V = ω1*R = 0.84m/s

Period of rotation:

T = 2π / ω1 = 29.92s

Assuming that there is no friction and their parents are not pushing anymore, we can use conservation of the angular momentum to calculate the new angular velocity:

I1*ω1 = I2*ω2 Solving for ω2:

ω2 = I1*ω1 / I2 = 0.315 rad/s

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3 years ago

A torsion-bar spring consists of a prismatic bar, usually of round cross section, that is twisted at one end and held fast at th

ra1l [238]

Answer:

d₁ = 0.29 in

d₂ = 0.505 in

Explanation:

Given:

T = 1500 lbf in

L = 10 in

x = 0.5 L = 5 in

$T_{1} =\frac{T(L-x)}{L} =\frac{1500*(10-5)}{10} =750lbfin$

First case: T = T₁ + T₂

T₂ = T - T₁ = 1500 - 750 = 750 lbf in

If the shafts are in series:

θ = θ₁ + θ₂

θ = ((T₁ * L₁)/GJ) + ((T₂ * L₂)/GJ)

Second case: If d₁ ≠ d₂

θ = ((T₁ * L₁)/GJ₁) + ((T₂ * L₂)/GJ₂) = 0 (eq. 1)

t₁ = t₂

$\frac{16T_{1} }{\pi d_{1}^{3} } =\frac{16T_{2} }{\pi d_{2}^{3} }$ (eq. 2)

T₁ + T₂ = 1500 (eq. 3)

θ₁ first case = θ₁ second case

Replacing:

$\frac{750*5}{G(\frac{\pi }{32})*0.5^{4} } =\frac{T_{1}*3.7 }{G(\frac{\pi }{32})*d_{1} ^{4} }\\T_{1} =16216d_{1} ^{4}$

The same way to θ₂:

$\frac{750*5}{G(\frac{\pi }{32})*0.5^{4} } =\frac{T_{2}*6.3 }{G(\frac{\pi }{32})*d_{2} ^{4} } \\T_{2} =9523.8d_{2} ^{4}$

From equation 2, we have:

d₁ = 0.587 * d₂

From equation 3, we have:

d₂ = 0.505 in

d₁ = 0.29 in

7 0

3 years ago

A dart gun shoots a dart with an angle of 45' above horizontal During the upward part of the trajectory the gravitational accele

ludmilkaskok [199]

Answer:

4. Downward and its value is constant

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