AS
work done =W = F.d = F d cosФ (Ф is angle between force F and displacement d) If a body/object is moving on a smooth surface (friction-less surface ) .There is no force acting on that body. F=0 so W=FdcosФ= (0)dcosФ ⇒ W=0
Now if a body is facing some amount of force but under the action of force there is no displacement covered. d=0 so W =FdcosФ= F(0)cosФ ⇒W=0
example: A person is applying a force on rigid wall but wall remains at rest there is no displacement occurs in wall.
The third term upon which work done dependent is angle between force and displacement i.e Ф. If Ф=90° then W= FdcosФ= Fdcos90⇒ W=0 ( as cos 90°=0)
The net force on particle particle q1 is 13.06 N towards the left.
<h3>
Force on q1 due to q2</h3>
F(12) = kq₁q₂/r₂
F(12) = (9 x 10⁹ x 13 x 10⁻⁶ x 7.7 x 10⁻⁶)/(0.25²)
F(12) = -14.41 N (towards left)
<h3>Force
on q1 due to q3</h3>
F(13) = (9 x 10⁹ x 7.7 x 10⁻⁶ x 5.9 x 10⁻⁶)/(0.55²)
F(13) = 1.352 N (towards right)
<h3>Net force on q1</h3>
F(net) = 1.352 N - 14.41 N
F(net) = -13.06 N
Thus, the net force on particle particle q1 is 13.06 N towards the left.
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Answer:
Explanation:
Given
Resistor A has length 
and Resistor B has Length 
and Resistance is given by
Considering 
and A to be constant thus
because 
(a)When they are connected in series
As the current in series is same and power is 
therefore 
as R is greater for second resistor
(b)if they are connected in Parallel
In Parallel connection Voltage is same
resistance of 2 is greater than 1 thus Power delivered by 1 is greater than 2
Answer:
1.90×10²⁰ Electrons
Explanation:
From the question,
Q = It.................... Equation 1
Where Q = charge flowing through the wire, I = current, t = time
Given: I = 4.35 A, t = 7.00 s
Substitute these values into equation 1
Q = 4.35(7.00)
Q = 30.45 C.
But,
1 electron contains 1.6×10⁻¹⁹ C
therefore,
30.45 C = 30.45/1.6×10⁻¹⁹ electrons
= 1.90×10²⁰ Electrons
