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>
Hi the answer is cumulonimbus clouds. Hope this helps.
Explanation:
v=140 m/s
t=50s
u=0
change of velocity = v-u
140-0=140 m/s
for acceleration:
from Newton's first law of motion:
v=u+at
140=0+a50
50a=140
a=140/50
a=2.8 m/s²
Answer:
The magnitude of the force of friction equals the magnitude of my push
Explanation:
Since the crate moves at a constant speed, there is no net acceleration and thus, my push is balanced by the frictional force on the crate. So, the magnitude of the force of friction equals the magnitude of my push.
Let F = push and f = frictional force and f' = net force
F - f = f' since the crate moves at constant speed, acceleration is zero and thus f' = ma = m (0) = 0
So, F - f = 0
Thus, F = f
So, the magnitude of the force of friction equals the magnitude of my push.
The number of charge drifts are 3.35 X 10⁻⁷C
<u>Explanation:</u>
Given:
Potential difference, V = 3 nV = 3 X 10⁻⁹m
Length of wire, L = 2 cm = 0.02 m
Radius of the wire, r = 2 mm = 2 X 10⁻³m
Cross section, 3 ms
charge drifts, q = ?
We know,
the charge drifts through the copper wire is given by
q = iΔt
where Δt = 3 X 10⁻³s
and i = 
where R is the resistance
R = 
ρ is the resistivity of the copper wire = 1.69 X 10⁻⁸Ωm
So, i = 
q = 
Substituting the values,
q = 3.14 X (0.02)² X 3 X 10⁻⁹ X 3 X 10⁻³ / 1.69 X 10⁻⁸ X 0.02
q = 3.35 X 10⁻⁷C
Therefore, the number of charge drifts are 3.35 X 10⁻⁷C
