Answer:
Intensity=
Explanation:
Given:

The sound level
in dB with intensity 
and reference intensity
is given by:

Plugging in values.

Dividing both sides by 10.


The above can be written as

Multiplying both sides by 



∴ 
Intensity =
I believe Action potential is the brief wave of positive charge that sweeps down the axon. Axon is part of the neuron that conducts impulses from the dendrites towards the cell body along the neuron. The action potential is brief since the sodium channels can only stay open for a very brief amount of time. As it travels along the neuron there is a change in polarity across the membrane of the axon .
Answer:
d) It will be cut to a fourth of the original force.
Explanation:
The magnitude of the electrostatic force between the charged objects is

where
k is the Coulomb's constant
q1 and q2 are the charges of the two objects
r is the separation between the two objects
In this problem, the initial distance is doubled, so
r' = 2r
Therefore, the new electrostatic force will be

So, the force will be cut to 1/4 of the original value.
Here's the part you need to know:
(Weight of anything) =
(the thing's mass)
times
(acceleration of gravity in the place where the thing is) .
Weight = (mass ) x (gravity) .
That's always true everywhere.
You should memorize it.
For the astronaut on Saturn . . .
Weight = (mass ) x (gravity) .
Weight = (68 kg) x (10.44 m/s²)
= 709.92 newtons .
__________________________________
On Earth, gravity is only 9.8 m/s².
So as long as the astronaut is on Earth, his weight is only
(68 kg) x (9.8 m/s²)
= 666.4 newtons .
Notice that his mass is his mass ... it doesn't change
no matter where he goes.
But his weight changes in different places, because
it depends on the gravity in each place.