Answer:
The correct answer is - Denying the pleasure of an immediate reward so that you can receive a greater reward later.
Explanation:
Delayed gratification is a gratification method or the process that helps in improving self-control and helps in goals that take time to achieve. It is referred to as the ability to stop or prevent oneself from something more rewarding or pleasurable immediately, in order to get something more pleasurable or rewarding later.
For instance, an individual watch tv before a big exam or studying for the exam so can watch tv after getting good grades.
The correct answer is - option D.
Answer:
the thermal energy generated in the loop = 
Explanation:
Given that;
The length of the copper wire L = 0.614 m
Radius of the loop r = 
r = 
r = 0.0977 m
However , the area of the loop is :



Change in the magnetic field is 
Then the induced emf e = 
e = 
e = 2.74 × 10⁻³ V
resistivity of the copper wire
Ω m
diameter of the wire = 1.08 mm
radius of the wire = 0.54 mm = 0.54 × 10⁻³ m
Thus, the resistance of the wire R = 
R = 
R = 1.13× 10⁻² Ω
Finally, the thermal energy generated in the loop (i.e the power) = 
= 
= 
The diameter of the football player's piston is 0.55 m
Given that the mass of the cheerleader(m) is 55 kg, mass of football player to be hold (M) is 130 kg, height of the players (h) is 1.30 m, radius of the piston corresponding to the diameter (r) is 0.09 m, Diameter of football player's piston (R), P1 is Pressure on the cheerleader's side, P2 is Pressure on the football player's side
Using Pascal's law,
This law states that if there is a change at a point of a body immersed in a fluid then that change will spread thoroughly to each and every point of the body.
The formula of hydraulic system is,
P1= P2
F1/A1 = F2/A2
mg/πr^2 = 4Mg/πR^2
m/r^2=4M/R^2
R^2=4M×r^2/m
By plugging the values, we get.,
R^2=(4×130×0.09^2)/55
R^2=4.21/55=0.076
R=√0.076 = 0.275 m
Hence, diameter of football player's piston is 0.55 m
Learn more about Hydraulic lift here
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Work,
in thermodynamics, is the amount of energy that is transferred from one system
to another system without transfer of entropy. It is equal to the external
pressure multiplied by the change in volume of the system. It is expressed as
follows:<span>
W = PdV
Integrating and assuming that P is not affected
by changes in V or it is constant, then we will have:
W = P (V2 - V1)
Substituting the given values:
P = 1.0 atm = 101325 Pa
(V2 - V1) = 0.50 L =
W = 101325 N/m^3 ( 0.50) (1/1000) m^3
W = 50.66 N-m or 50.66 J
<span>
So, in the expansion process about 50.66 J of work is being done.
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