Answer:
Explanation:
the directions may change
Or they will repel and become opposite sides
Answer:
Explanation:
From the question;
We will make assumptions of certain values since they are not given but the process to achieve the end result will be the same thing.
We are to calculate the following task, i.e. to determine the electric field at the distances:
a) at 4.75 cm
b) at 20.5 cm
c) at 125.0 cm
Given that:
the charge (q) = 33.3 nC/m
= 33.3 × 10⁻⁹ c/m
radius of rod = 5.75 cm
a) from the given information, we will realize that the distance lies inside the rod. Provided that there is no charge distribution inside the rod.
Then, the electric field will be zero.
b) The electric field formula 

E = 1461.95 N/C
c) The electric field E is calculated as:

E = 239.76 N/C
Answer:
To find the mass using density and volume we just multiply them against each other which causes ml to cancel and just leaves us with grams which represents how much the item weights.



Therefore, our final answer is that our pencil weight 3.5 grams
<u><em>Hope this helps! Let me know if you have any questions</em></u>
Answer:
A.
Explanation:
momentum depends on weight and speed
Answer:
the filling stops when the pressure of the pump equals the pressure of the interior air plus the pressure of the walls.
Explanation:
This exercise asks to describe the inflation situation of a spherical fultball.
Initially the balloon is deflated, therefore the internal pressure is equal to the pressure of the air outside, atmospheric pressure, when it begins to inflate the balloon with a pump this creates a pressure in the inlet valve and as it is greater than the pressure inside, the air enters it, this is repeated in each filling cycle, manual pump.
When the ball is full we have two forces, the one created by the external walls and the one aired by the pressure of the pump, these forces are directed towards the inside, but the air molecules exert a pressure towards the outside, which translates into a force. When these two forces are equal, the pump is no longer able to continue introducing air into the balloon.
Consequently the filling stops when the pressure of the pump equals the pressure of the interior air plus the pressure of the walls.