Explanation:
The magnitude of the electric field between the plates is given by
E = -ΔV/d
minus sign indicates Potential decreases in the direction of electric field
where
ΔV is the potential difference between the plates
D is the distance between the plates.
The work done when carrying an electrical charge on an equipotential surface between one position to the other is zero W= q(V-V)=0 The electric field lines of force are always perpendicular to an equipotential surface. That conductor in an equipotential surface as direction E is at right angles to an eauipotential surface The intensity of the electric field along an equipotential surface is always zero. Equipotential surfaces never collide with each other as this would mean that at that point, there are two alternative values that are not true.
Answer:
The work done on the wagon is 37 joules.
Explanation:
Given that,
The force applied by Charlie to the right, F = 37.2 N
The force applied by Sara to the left, F' = 22.4 N
We need to find the work done on the wagon after it has moved 2.50 meters to the right. The net force acting on the wagon is :
Work done on the wagon is given by the product of net force and displacement. It is given by :
W = 37 Joules
So, the work done on the wagon is 37 joules. Hence, this is the required solution.
The time taken for the athlete to finish the race is 20 s (Option A)
<h3>What is power? </h3>
Power is simply defined as the rate at which work is done. It can be expressed mathematically as
Power (P) = work (W) / time (t)
But
Work = weight × distance
Therefore,
Power = (weight × distance ) / time
<h3>How to determine the time </h3>
- Mass (m) = 55 Kg
- Acceleration due to gravity (g) = 9.8 m/s²
- Weight = mg = 55 × 9.8 = 539 N
- Power (P) = 5.4 KW = 5.4 × 1000 = 5400 W
- Distance (d) = 200 m
- Time (t) =?
Power = (weight × distance ) / time
5400 = (539 × 200) / t
5400 = 107800 / t
Cross multiply
5400 × t = 107800
Divide both side by 5400
t = 107800 / 5400
t = 20 s
Learn more about power:
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Answer:
(A) 140 j/sec (b) 1.26 K
Explanation:
We have given the heat heat flowing into the refrigerator = 40 J/sec
Work done = 40 W
(a) So the heat discharged from the refrigerator 
(b) Total heat absorbed =140 j/sec 
Let the temperature be 
Heat absorbed per hour =504000 ![[tex]=400\times 10^3\times \Delta T](https://tex.z-dn.net/?f=%5Btex%5D%3D400%5Ctimes%2010%5E3%5Ctimes%20%5CDelta%20T)
So 
Infrared waves are used in heat lamps and other heat sensing devices. Infrared waves or commonly known as Infrared radiations (IR) is the type of electromagnetic radiation we encounter most in our everyday life. Heat lamps are electrical devices which emit infrared radiation.
