Desired operation: A + B = C; {A,B,C) are vector quantities.
<span>Issue: {A,B} contain error (measurement or otherwise) </span>
<span>Objective: estimate the error in the vector sum. </span>
<span>Let A = u + du; where u is the nominal value of A and du is the error in A </span>
<span>Let B = v + dv; where v is the nominal value of B and dv is the error in B </span>
<span>Let C = w + dw; where w is the nominal value of C and dw is the error in C [the objective] </span>
<span>C = A + B </span>
<span>w + dw = (u + du) + (v + dv) </span>
<span>w + dw = (u + v) + (du + dv) </span>
<span>w = u+v; dw = du + dv </span>
<span>The error associated with w is the vector sum of the errors associated with the measured quantities (u,v)</span>
Answer:
Real image
Explanation:
The picture is real, but it is reversed and tiny. An picture generated by a pinhole camera has certain features. As compared item, the image created by a pinhole camera is normally pretty small and looks reversed both on the vertically and horizontally axis.
Answer:
The correct answer is - 43%.
Explanation: The increase in CO2 between these two suggested periods is approximately 43%. Even though it is a natural process that the CO2 levels vary in the atmosphere, still this is not the same case nowadays. Nowadays, or rather in the past few decades, apart from the natural increase of CO2 in the atmosphere, it has seen a much more increased levels because of the human activity. The industrial facilities and the vehicles, the cutting of the forests and burning the wood (there's both release of CO2 from the burning of the trees and loss of natural accumulator of the CO2), are just some of the more important human activities that contribute to a significant rise in the CO2 levels.
Answer:
1.84 m
Explanation:
For the small lead ball to be balanced at the tip of the vertical circle just before it is released, the reaction force , N equal the weight of the lead ball W + the centripetal force, F. This normal reaction ,N also equals the tension T in the string.
So, T = mg + mrω² = ma where m = mass of small lead ball, g = acceleration due to gravity = 9.8 m/s², r = length of rope = 1.10 m and ω = angular speed of lead ball = 3 rev/s = 3 × 2π rad/s = 6π rad/s = 18.85 rad/s and a = acceleration of normal force. So,
a = g + rω²
= 9.8 m/s² + 1.10 m × (18.85 rad/s)²
= 9.8 m/s² + 390.85 m/s²
= 400.65 m/s²
Now, using v² = u² + 2a(h₂ - h₁) where u = initial velocity of ball = rω = 1.10 m × 18.85 rad/s = 20.74 m/s, v = final velocity of ball at maximum height = 0 m/s (since the ball is stationary at maximum height), a = acceleration of small lead ball = -400.65 m/s² (negative since it is in the downward direction of the tension), h₁ = initial position of lead ball above the ground = 1.3 m and h₂ = final position of lead ball above the ground = unknown.
v² = u² + 2a(h₂ - h₁)
So, v² - u² = 2a(h₂ - h₁)
h₂ - h₁ = (v² - u²)/2a
h₂ = h₁ + (v² - u²)/2a
substituting the values of the variables into the equation, we have
h₂ = 1.3 m + ((0 m/s)² - (20.74 m/s)²)/2(-400.65 m/s²)
h₂ = 1.3 m + [-430.15 (m/s)²]/-801.3 m/s²
h₂ = 1.3 m + 0.54 m
h₂ = 1.84 m
