Answer:
The time required by the Athlete to work off 1.00 lb of body fat = 0.296 minute
Explanation:
1 lb of body fat = 4.1 k cal
1 k cal = 4.184 Kilo joule
1 lb of body fat = 4.1 × 4.184 = 17.1544 Kilo joule
Athlete expends 3480 Kilo joule in one hour
⇒ Time required to expand 3480 Kilo joule = 60 minute
⇒ Time required to expand 1 Kilo joule = 
⇒ Time required to expand 17.1544 Kilo joule = × 17.1544 = 0.296 min
Therefore the time required by the Athlete to work off 1.00 lb of body fat = 0.296 minute
Maybe, but she hasn't proved it yet. As another example, somebody may bring a jar of gas to your laboratory, and you test the gas in the jar and you find nitrogen atoms, oxygen atoms, argon atoms, and krypton atoms, and there isn't a single compound in it. The gas in the jar is a MIXTURE of gases . . . a mixture that we call "Air".
Answer:
9.8 m/s
Explanation:
The work done by the force pushing the cart is equal to the kinetic energy gained by the cart:
where
W is the work done
is the final kinetic energy of the cart
is the initial kinetic energy of the cart, which is zero because the cart starts from rest, so we can write:
But the work is equal to the product between the pushing force F and the displacement, so
So, the final kinetic energy of the cart is 480 J. The formula for the kinetic energy is
(1)
where m is the mass of the cart and v its final speed.
We can find the mass because we know the weight of the cart, 98.0 N:
Therefore, we can now re-arrange eq.(1) to find the final speed of the cart:
Answer:
V= 14.2m/s
Explanation:
We know that acceleration= dv/dt
So 16m/s²=dv/ dt = v dv/ds
So this wil be
Integral of 16m/s² ds [at 2,2]= integral of v dv at[ 0, v]
So 16[s (3/2)/3/2] at ( s,3) = v²/2
At s= 6m
So v² = 64/3( 6^1.5-3^1.5)
= 14.2m/s
