Answer:
Electromagnetic induction
Explanation:
The process of generating electric current with a magnetic field. It occurs whenever a magnetic field and an electric conductor move relative to one another so the conductor crosses lines of force in the magnetic field.
Here mass of the iron pan is given as 1 kg
now let say its specific heat capacity is given as "s"
also its temperature rise is given from 20 degree C to 250 degree C
so heat required to change its temperature will be given as



now if we give same amount of heat to another pan of greater specific heat
so let say the specific heat of another pan is s'
now the increase in temperature of another pan will be given as


now we have

now as we know that s' is more than s so the ratio of s and s' will be less than 1
And hence here we can say that change in temperature of second pan will be less than 230 degree C which shows that final temperature of second pan will reach to lower temperature
So correct answer is
<u>A) The second pan would reach a lower temperature.</u>
Answer:
W ≅ 292.97 J
Explanation:
1)What is the work done by tension before the block goes up the incline? (On the horizontal surface.)
Workdone by the tension before the block goes up the incline on the horizontal surface can be calculated using the expression;
W = (Fcosθ)d
Given that:
Tension of the force = 62 N
angle of incline θ = 34°
distance d =5.7 m.
Then;
W = 62 × cos(34) × 5.7
W = 353.4 cos(34)
W = 353.4 × 0.8290
W = 292.9686 J
W ≅ 292.97 J
Hence, the work done by tension before the block goes up the incline = 292.97 J
Answer:
15.065ft
Explanation:
To solve this problem it is necessary to consider the aerodynamic concepts related to the Drag Force.
By definition the drag force is expressed as:

Where
is the density of the flow
V = Velocity
= Drag coefficient
A = Area
For a Car is defined the drag coefficient as 0.3, while the density of air in normal conditions is 1.21kg/m^3
For second Newton's Law the Force is also defined as,

Equating both equations we have:



Integrating


Here,






Replacing:




B. kinetic energy increases and potential energy decreases