When distance<span> is increased the amount of </span>force<span> needed will depend on the </span>mass<span> of the object. </span>
Explanation:
a straight line under the letter
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:
Answer:12.5 W
Explanation:
work = force * distance
175 J = 25N * d
d = 7 meters ◄
power = work / time
P = 175J / 14s = 12.5 W ◄