W = F * d
W = 10N * 2.5 m
W = 25 N m
So the answer you want is the third one down.
        
             
        
        
        
The release of free energy drives the spontaneous reaction.
Spontaneity can be <span>determined
using the change in </span>Gibbs free energy
(the thermodynamic potencial): 
delta G=delta H – T*delta
S
where delta H is the enthalpy and delta S is the entropy.
The direction (the sign) of delta G depends of the changes
of enthalpy and entropy. If delta G is negative then the process is
spontaneous.
In our case, both delta H and delta S are negative values, the
process as said is spontaneous which means that it may proceed in the forward
direction.
 
        
             
        
        
        
Answer:
Therefore, light travelling at 3.0x10^8 meters per second takes 500 seconds (8 minutes, 20 seconds) to reach the Earth, which is 1.5x10^11 meters away from the sun
Explanation:
 
        
             
        
        
        
Answer:
the ship's energy is greater than this and the crew member does not meet the requirement
Explanation:
In this exercise to calculate kinetic energy or final ship speed in the supply hangar let's use the relationship
                 W =∫ F dx = ΔK
                  
Let's replace
              
           ∫ (α x³ + β) dx = ΔK
          α x⁴ / 4 + β x = ΔK
            
Let's look for the maximum distance for which the variation of the energy percent is 10¹⁰ J
          x (α x³ + β) =  - K₀
 - K₀
            = K₀ + x (α x³ + β)
  = K₀ + x (α x³ + β)
 Assuming that the low limit is x = 0, measured from the cargo hangar
      
Let's calculate
          = 2.7 10¹¹ + 7.5 10⁴ (6.1 10⁻⁹ (7.5 10⁴) 3 -4.1 10⁶)
  = 2.7 10¹¹ + 7.5 10⁴ (6.1 10⁻⁹ (7.5 10⁴) 3 -4.1 10⁶)
         Kf = 2.7 10¹¹ + 7.5 10⁴ (2.57 10⁶ - 4.1 10⁶)
         Kf = 2.7 10¹¹ - 1.1475 10¹¹
         Kf = 1.55 10¹¹ J
In the problem it indicates that the maximum energy must be 10¹⁰ J, so the ship's energy is greater than this and the crew member does not meet the requirement
We evaluate the kinetic energy if the System is well calibrated
                 W = x F₀ =  –K₀
 –K₀
                  = K₀ + x F₀
 = K₀ + x F₀
We calculate
                = 2.7 10¹¹ -7.5 10⁴ 3.5 10⁶
 = 2.7 10¹¹ -7.5 10⁴ 3.5 10⁶
                 = (2.7 -2.625) 10¹¹
 = (2.7 -2.625) 10¹¹
                = 7.5 10⁹ J
 = 7.5 10⁹ J