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3 years ago

How would you determine how much error there is between a vector addition and the real results

1 answer:

6 0

Desired operation: A + B = C; {A,B,C) are vector quantities.

Issue: {A,B} contain error (measurement or otherwise) 

Objective: estimate the error in the vector sum. 

Let A = u + du; where u is the nominal value of A and du is the error in A 
Let B = v + dv; where v is the nominal value of B and dv is the error in B 
Let C = w + dw; where w is the nominal value of C and dw is the error in C [the objective] 

C = A + B 

w + dw = (u + du) + (v + dv) 

w + dw = (u + v) + (du + dv) 

w = u+v; dw = du + dv 

The error associated with w is the vector sum of the errors associated with the measured quantities (u,v)

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Here are the observations

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3 years ago

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Ainat [17]

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3 years ago

A spider spins a web with silk threads of density 1300 kg/m3 and diameter 3.0 μm . a typical tension in the radial threads of su

Tema [17]

Answer:

Explanation:

The velocity of a wave in a string is equal to:

v = √(T / (m/L))

where T is the tension and m/L is the mass per length.

To find the mass per length, we need to find the cross-sectional area of the thread.

A = πr² = π/4 d²

A = π (3.0×10⁻⁶ m)²

A = 2.83×10⁻¹¹ m²

So the mass per length is:

m/L = ρA

m/L = (1300 kg/m³) (2.83×10⁻¹¹ m²)

m/L = 3.68×10⁻⁸ kg/m

So the wave velocity is:

v = √(T / (m/L))

v = √(7.0×10⁻³ N / (3.68×10⁻⁸ kg/m))

v ≈ 440 m/s

The speed of sound in air at sea level is around 340 m/s. So the spider will feel the vibration in the thread before it hears the sound.

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3 years ago

A phone cord is 4.89 m long. The cord has a mass of 0.212 kg. A transverse wave pulse is produced by plucking one end of the tau

slega [8]

Answer:

Tension in Cord=174 N

Explanation:

Given Data

L (Phone Cord Length)=4.89 m

m (Cord Mass)=0.212 Kg

T (Time for four trips)=0.617 s

Tension=?

Solution

V=λ×f

$V=\frac{8*4.89}{0.617}\\ V=63.4m/s$

$Sigma=\frac{mass}{length}\\ Sigma=\frac{0.212}{4.89}\\ Sigma=0.0433 \frac{kg}{m}$

$Wave Speed=\sqrt{\frac{Tension}{Sigma} }\\ \\V=\sqrt{\frac{T}{Sigma} }\\ V^{2}=\frac{T}{Sigma}\\ T=V^{2}*Sigma\\ T=(63.4)^{2}*(0.0433)\\ T=174 N$

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3 years ago

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Gemiola [76]

I can't decide between A and B, but B seems more likely to me. Even though the molecules don't look like they're moving, the area of contact is slightly more compressed.

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4 years ago