At point x = 0, the particle accelerates. Since there will be change of velocity at that point. The the force of the particle will change from negative sign to positive sign according to the given figure, we can therefore conclude that the particle will have a turning point at point x = 0.
Given that a 2.0 kg particle moving along the z-axis experiences the force shown in a given figure.
Force is the product of mass and acceleration. While acceleration is the rate of change of velocity. Both the force and acceleration are vector quantities. They have both magnitude and direction.
If the particle's velocity is 3.0 m/s at x = 0 m, that mean that the particle experience change of velocity at point x = 0. Since the the force of the particle will change from negative sign to positive sign according to the given figure, we can therefore conclude that the particle will have a turning point at point x = 0.
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Based on the answer provided, it seems the writer wanted you to assume that the energy loss per plank is constant. This is not the same as the bullet losing <span><span>1/nth</span><span>1/nth</span></span><span> of its velocity per plank (however, the fact that the question does not mention this assumption arguably makes the question ambiguous).
</span><span>With this assumption, the energy loss becomes
</span><span>
ΔE = <span>1/2 </span>m<span>v2 </span>− <span>1/2 </span>m <span><span>(<span>v−<span>v/n</span></span>) </span><span>2
</span></span></span>
and the number of planks <span>NN</span><span> becomes
</span>
N = <span><span><span>1/2</span>m<span>v2 /</span></span><span>ΔE </span></span>= <span><span>n2/ </span><span>2n−1
</span></span>
Otherwise, if you assume that the bullet loses <span><span>1/<span>nth</span></span><span>1/<span>nth</span></span></span><span> of its velocity per plank, then the answer is </span><span><span>N=∞</span></span><span><span>
</span>
</span>
Answer:
clc
clear all
close all
format long
A=load('xyg1.mat');
x=A(:,1);
y=A(:,2);
[z,N,R2]=polyfitsystem(x,y,0.95)
function [z,N,R2]=polyfitsystem(x,y,R2)
for N=1:20
z=polyfit(x,y,N);
SSR=sum((y-polyval(z,x)).^2);
SST=sum((y-mean(y)).^2);
s=1-SSR/SST;
if(s>=R2)
R2=s;
break;
end
end
xx=linspace(min(x),max(x));
plot(x,y,'o',xx,polyval(z,x));
xlabel('x');
ylabel('y(x)');
title('Plot of y vs x');
end
Explanation:
For one you’ll smell better, won’t be thinking when can i get my next cigarette, stop lung damage, skin will start to clear up
Answer:
An egg raised above the ground has potential energy due to the force of gravity. When dropped, the egg's potential energy is converted into the kinetic energy of motion. :)
Explanation:
