<span><u>Answer
</u>
The mass of 220 lb football has less than 288 lb football. So, it will be easier to move it since it will require less force. The heavy football will have a bigger momentum. Since 288 lb has more weight than 220 lb, it will have bigger inertia making it difficult for the players to stop it.
This makes it easier to tackle 220 lb football than 288 lb football.
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Given:
Ca = 3Cb (1)
where
Ca = heat capacity of object A
Cb = heat capacity f object B
Also,
Ta = 2Tb (2)
where
Ta = initial temperature of object A
Tb = initial temperature of object B.
Let
Tf = final equilibrium temperature of both objects,
Ma = mass of object A,
Mb = mass of object B.
Assuming that all heat exchange occurs exclusively between the two objects, then energy balance requires that
Ma*Ca*(Ta - Tf) = Mb*Cb*(Tf - Tb) (3)
Substitute (1) and (2) into (3).
Ma*(3Cb)*(2Tb - Tf) = Mb*Cb*(Tf - Tb)
3(Ma/Mb)*(2Tb - Tf) = Tf - Tb
Define k = Ma/Mb, the ratio f the masses.
Then
3k(2Tb - Tf) = Tf - Tb
Tf(1+3k) = Tb(1+6k)
Tf = [(1+6k)/(1+3k)]*Tb
Answer:

where
Answer:
the length of the now stationary spacecraft = 89.65m
Explanation:
In contraction equation, Length contraction L is the shortening of the measured length of an object moving relative to the observer’s frame.
Thus, it has a formula;
L = L_o(√(1 - (v²/c²))
Where in this question;
L = 71m and v = 0.610 c
Thus;
71 = L_o (√(1 - ((0.61c)²/c²))
c² will cancel out to give;
71 = L_o (√(1 - 0.61²)
71 = L_o (√(1 - 0.61²)
71 = 0.792L_o
L_o = 71/0.792
L_o = 89.65m
Answer:
The correct answer would be B. 18 to 26%.
The muscle efficiency is calculated by dividing mechanical work output by total metabolic cost.
It is estimated that human muscles have an efficiency of about 18% to 26%.
The efficiency is low because most of the energy is lost when food energy is converted into ATP (adenosine triphophate).
In addition, there is second energy loss when energy in the form of ATP is converted into the mechanical energy such as rowing, cycling et cetera.
Answer:
assume nitrogen is an ideal gas with cv=5R/2
assume argon is an ideal gas with cv=3R/2
n1=4moles
n2=2.5 moles
t1=75°C <em>in kelvin</em> t1=75+273
t1=348K
T2=130°C <em>in kelvin</em> t2=130+273
t2=403K
u=пCVΔT
U(N₂)+U(Argon)=0
<em>putting values:</em>
=>4x(5R/2)x(Tfinal-348)=2.5x(3R/2)x(T final-403)
<em>by simplifying:</em>
Tfinal=363K