Answer:
15.7m/s
Explanation:
To solve this problem, we use the right motion equation.
Here, we have been given the height through which the ball drops;
Height of drop = 14.5m - 1.9m = 12.6m
The right motion equation is;
V² = U² + 2gh
V is the final velocity
U is the initial velocity = 0
g is the acceleration due to gravity = 9.8m/s²
h is the height
Now insert the parameters and solve;
V² = 0² + 2 x 9.8 x 12.6
V² = 246.96
V = √246.96 = 15.7m/s
S = d/t, s = 120, d = 2000, t = ?
Input the values,
120 = 2000/t
Make t the subject of the formula by cross multiplying, Therefore,
120t = 2000
Divide both sides by 120
t = 16.7hrs to 1 decimal place.
If you're asked to convert it, you can.
Hi
I think the answer is:
GRAVITATIONAL POTENTIAL ENERGY TRANSFORMS INTO KINETIC ENERGY.
HOPE IT HELPS.
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:
