Answer:
Arrange an annual service. Treat your boiler like your car. ...
Keep your boiler clean. ...
Bleed your radiators. ...
Top up the pressure. ...
Use a Powerflush. ...
Insulate your pipes. ...
Turn the heating on. ...
If all else fails…
Explanation:
It weakened the gravitational force between the oblects.
Planck's constant is 4.39048042 × 10-67 m4 kg2 / s2.
Answer:
The current through the tube is 73.39A.
Explanation:
The relationship between the resistivity
, the electric field
, and the current density
is given by

This equation can be solved for
to get:

Since the current is 

Now, for the tube of mercury
,
, and the area is
; therefore,


Hence, the current through the mercury tube is 73.39A.