The change in the player's internal energy is -491.6 kJ. The number of nutritional calories is -117.44 kCal
For this process to take place, some of the basketball player's perspiration must escape from the skin. This is because sweating relies on a physical phenomenon known as the heat of vaporization.
The heat of vaporization refers to the amount of heat required to convert 1g of a liquid into a vapor without causing the liquid's temperature to increase.
From the given information,
- the work done on the basketball is dW = 2.43 × 10⁵ J
The amount of heat loss is represented by dQ.
where;
∴
Using the first law of thermodynamics:b
dU = dQ - dW
dU = -mL - dW
dU = -(0.110 kg × 2.26 × 10⁶ J/kg - 2.43 × 10⁵ J)
dU = -491.6 × 10³ J
dU = -491.6 kJ
The number of nutritional calories the player has converted to work and heat can be determined by using the relation:

dU = -117.44 kcal
Learn more about first law of thermodynamics here:
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Suvat
we have s, u, v and we want a
the suvat equation with these values in is: v^2 = u^2 - 2as
so a = (-v^2 + u^2)/-2s
plug numbers in
a = (-85^2 + 0^2)/-2*36 = 7225/72 = 100.3... ms^-2
Answer:
6.5e-4 m
Explanation:
We need to solve this question using law of conservation of energy
Energy at the bottom of the incline= energy at the point where the block will stop
Therefore, Energy at the bottom of the incline consists of the potential energy stored in spring and gravitational potential energy=
Energy at the point where the block will stop consists of only gravitational potential energy=
Hence from Energy at the bottom of the incline= energy at the point where the block will stop
⇒
⇒
Also 
where
is the mass of block
is acceleration due to gravity=9.8 m/s
is the difference in height between two positions
⇒
Given m=2100kg
k=22N/cm=2200N/m
x=11cm=0.11 m
∴
⇒
⇒
⇒h=0.0006467m=
Answer:at 21.6 min they were separated by 12 km
Explanation:
We can consider the next diagram
B2------15km/h------->Dock
|
|
B1 at 20km/h
|
|
V
So by the time B1 leaves, being B2 traveling at constant 15km/h and getting to the dock one hour later means it was at 15km from the dock, the other boat, B1 is at a distance at a given time, considering constant speed of 20km/h*t going south, where t is in hours, meanwhile from the dock the B2 is at a distance of (15km-15km/h*t), t=0, when it is 8pm.
Then we have a right triangle and the distance from boat B1 to boat B2, can be measured as the square root of (15-15*t)^2 +(20*t)^2. We are looking for a minimum, then we have to find the derivative with respect to t. This is 5*(25*t-9)/(sqrt(25*t^2-18*t+9)), this derivative is zero at t=9/25=0,36 h = 21.6 min, now to be sure it is a minimum we apply the second derivative criteria that states that if the second derivative at the given critical point is positive it means here we have a minimum, and by calculating the second derivative we find it is 720/(25 t^2 - 18 t + 9)^(3/2) that is positive at t=9/25, then we have our answer. And besides replacing the value of t we get the distance is 12 km.