FFT of signal with starting value zeros
I expand and modify your code:

N = 100;  %number of sample
fs = 100; % Sample frequency (Hz)
t = 0:1/fs:(N-1)/fs;                    
x = sin(2*pi*10*t);  % 10 Hz component
%x(:)=1;    
f = 0:fs/N:(N-1)*fs/N;    

subplot(2,2,1);
plot(t,x);
title('full signal'); grid on;
subplot(2,2,3);
plot(f(1:N/2),abs(fft(x)(1:N/2)));
title('spectrum'); grid on;

x1=x;     %x1 - zero pad signal
x1(1:ceil(N/3))=0;
x1(ceil(2*N/3):N)=0;

subplot(2,2,2);
plot(t,x1,'b',t,x1,'*');
title('zero pad signal'); grid on;
subplot(2,2,4);
plot(f(1:ceil(N/2)),abs(fft(x1)(1:ceil(N/2))));
title('spectrum'); grid on;




You can see signals before and after zero padding are quite different. And the spectrums are quite different too. There is strong part of signal x1 with frequency 10 Hz, and you can see strong peak on spectrum.

You can uncomment line with x(:)=1; First full signal will be DC (constant), second (zero padding) - rectangle. Quite different signals and quite different spectrum.

UPDATE
About phase spectrum. Update to code:

F=fft(x);
angle(F(11))  % phase = -1.5708
F1=fft(x1);
angle(F1(11)) % phase =  -1.6018


Phase for x makes sense only for bin with non zero amplitude. Non zero amplitude numbers are 11 and 91. Phase for any other index are results of calculations errors of FFT function.
As x1 has  big part of x the phase spectrum for this bins has approximately the same value (as you can see).  

Hint. You can present x1 as product of $sin$ function and rectangle function. So spectrum of x1 is a convolution of spectrum $sin$ and rectangle.