Answer:
1977.696 J
Explanation:
Given;
Weight of the box = 28.0 kg
Force applied by the boy = 230 N
angle between the horizontal and the force = 35°
Therefore,
the horizontal component of the force = 230 × cosθ
= 230 × cos 35°
= 188.405 N
Coefficient of kinetic friction, μ = 0.24
Force by friction, f = μN
here,
N = Normal force = Mass × acceleration due to gravity
or
N = 28 × 9.81 = 274.68 N
therefore,
f = 0.24 × 274.68
or
f = 65.9232 N
Now,
work done by the boy, W₁ = 188.405 N × Displacement
= 188.405 N × 30
= 5652.15 J
and,
the
work done by the friction, W₂ = - 65.9232 N × Displacement
= - 65.9232 N × 30 m
= - 1977.696 J
[ since the friction force acts opposite to the direction of motion, therefore the workdone will be negative]
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>
The answer would be in the chart or graph A is 1 B is 2
Answer:
Protons, Electrons, and neutrons
