Novice question about low-pass filter
If you only use the capacitor you would have a load, like a resistor, only this time frequency dependent. The thing is that this load doesn't change the signal.
Enter the resistor. Now you have a voltage divider. For low frequencies the capacitor's "resistance" will be high compared to the resistor, so most of the signal will arrive at the output. For high frequencies the capacitor's "resistance" is low compared to the resistor, and the divider will attenuate the input signal: only a small portion of it will appear at the output.
I put the capacitor's "resistance" in quotes because it's not just a real resistance, it's more complex, in every meaning of the word..