Why is phenyl weakly activating?
In the following diagram, which carbocation is more stable, A or B?



The answer is A because the positive charge on the positively charged carbon can be delocalized into the attached benzene ring (I've drawn one of the possible resonance structures to illustrate this).

The same thinking can be applied to your question. Look at resonance structures C and D (we can draw many more) for the attack of an electrophile on a benzene ring.  If there is a second benzene attached and it is attached ortho or para to where the electrophile attacks, then the positive charge can be delocalized into the second benzene ring.  

This additional delocalization will stabilize the system. Hence, the second benzene is an ortho-para activating substituent.  However, since this stabilization involves electrons involved in pi bonds, the stabilization (activating effect) will be weaker than when a substituent with a "more readily available" lone pair of electrons (amino, hydroxyl, etc.) is involved.