In general, the electric force in an electric field is exerted outward from a positive atom, and inward for the negative atom. Therefore based on the figure you specified, we can say that the <span>electric force exerted by paperclip 1 on paperclip 2 is repulsive.</span>
Answer:
<h2>The water will move faster</h2>
Explanation:
This happens as part of the consequences of the continuity of an incompressible fluid, of which water is an example.
so as the water emerges from the wider section the flow velocity increases and vice versa.
Hence the flow velocity in indirectly proportional to the area
The continuity equation is given below for our analysis
-------------------1
where
A1= the area of the wider section of the tank
V1= the flow velocity from the wider section
A2= the area of the hole at the bottom
V2= velocity of flow at the bottom.
Answer:
<em>T</em><em>he value of work being done on the object is 958J.</em>
Explanation:
<em>Work</em><em> done</em><em> </em><em>is </em><em>equal</em><em> to</em><em> </em><em>force</em><em> </em><em>multiply</em><em> by</em><em> </em><em>distance,</em><em> </em><em>but </em><em>when</em><em> </em><em>the </em><em>angle</em><em> </em><em>is </em><em>between</em><em> </em><em>the</em><em> </em><em>force</em><em> </em><em>and</em><em> </em><em>distance</em><em> </em><em>work</em><em> </em><em>done=</em><em>Force</em><em> (</em><em>cos</em><em> </em><em>theta)</em><em> </em><em>×</em><em> </em><em>distance</em>
<em>Work</em><em> done</em><em> </em><em>=</em><em> </em><em>Force(</em><em>cos </em><em>theta</em><em>)</em><em> </em><em>×</em><em> </em><em>distance</em>
<em>Work</em><em> done</em><em> </em><em>=</em><em> </em><em>2</em><em>5</em><em>N</em><em>(</em><em>cos </em><em>4</em><em>0</em><em>.</em><em>0</em><em>°</em><em>)</em><em> </em><em>×</em><em> </em><em>5</em><em>0</em><em>m</em>
<em>Work</em><em> done</em><em> </em><em>=</em><em> </em><em>2</em><em>5</em><em>N</em><em>(</em><em>0</em><em>.</em><em>7</em><em>6</em><em>6</em><em>0</em><em>)</em><em> </em><em>×</em><em> </em><em>5</em><em>0</em><em>m</em>
<em>Work</em><em> done</em><em> </em><em>=</em><em> </em><em>19.15N </em><em>×</em><em> </em><em>5</em><em>0</em><em>m</em>
<em>Work</em><em> done</em><em> </em><em>=</em><em> </em><em>957.5J </em><em>=</em><em> </em><em>9</em><em>5</em><em>8</em><em>J</em>
<em>T</em><em>herefore</em><em> the</em><em> </em><em>value</em><em> of</em><em> </em><em>work</em><em> </em><em>being</em><em> </em><em>done </em><em>on </em><em>the</em><em> </em><em>object</em><em> </em><em>is </em><em>9</em><em>5</em><em>8</em><em>J</em><em>.</em>
Answer:
No, the net force on the skydiver is zero
Explanation:
According to Newton's Second Law, the net force on an object is equal to the product between the mass of the object and its acceleration:

where
F is the net force
m is the mass of the object
a is the acceleration
In this problem, the acceleration of the skydiver is zero:
a = 0
This implies that also the net force on the skydiver is zero, according to the previous equation:
F = 0
So, the net force on the skydiver is zero. This occurs because the air resistance, which points upward, exactly balances the force of gravity on the skydiver, acting downwards.
Answer:
The power consumed by the air filter is 9.936 watts
Explanation:
It is given that, the secondary coil of a step-up transformer provides the voltage that operates an electrostatic air filter.
Turn ratio of the transformer, 
Voltage of primary coil, 
Current in the secondary coil, 
The power consumed by the air filter is :
...........(1)
For a transformer, 
So, 


So, the power consumed by the air filter is 9.936 watts. Hence, this is the required solution.