I hope you get the answer soon dude
Answer:
Explanation:
Start by finding the weight supported by the wheels, as we can then relate that to the total frictional force through the friction coefficient. This is given by:
W=mg
Substitute 1.00×, or 1,000, for m and 9.81 m/ for g. This gives us
W=(1,000)(9.81)
W=9,810 N
Since half the weight is supported by the wheels in question, we divide this by two, and the weight we will use is 4,905 N.
The frictional force is given by:
F=μW
Where μ is the friction coefficient and W is the weight, which we just calculated. The friction coefficient for rubber on dry concrete is given in Table 6.1 as (1.0). Substitute that for μ and 4,905 N for weight and we get the total frictional force as 4,905 N. Now we can find the acceleration by rearranging F=ma to get a=. Substitute our frictional force 4,905 N for F and our total mass 1,000 kg for m to get 4.9 m/.
Potential energy is in short, stored energy
Walking at a speed of 2.1 m/s, in the first 2 s John would have walked
(2.1 m/s) (2 s) = 4.2 m
Take this point in time to be the starting point. Then John's distance from the starting line at time <em>t</em> after the first 2 s is
<em>J(t)</em> = 4.2 m + (2.1 m/s) <em>t</em>
while Ryan's position is
<em>R(t)</em> = 100 m - (1.8 m/s) <em>t</em>
where Ryan's velocity is negative because he is moving in the opposite direction.
(b) Solve for the time when they meet. This happens when <em>J(t)</em> = <em>R(t)</em> :
4.2 m + (2.1 m/s) <em>t</em> = 100 m - (1.8 m/s) <em>t</em>
(2.1 m/s) <em>t</em> + (1.8 m/s) <em>t</em> = 100 m - 4.2 m
(3.9 m/s) <em>t</em> = 95.8 m
<em>t</em> = (95.8 m) / (3.9 m/s) ≈ 24.6 s
(a) Evaluate either <em>J(t)</em> or <em>R(t)</em> at the time from part (b).
<em>J</em> (24.6 s) = 4.2 m + (2.1 m/s) (24.6 s) ≈ 55.8 m
Energy = mass * SHC * temp change
Temp change = energy/SHC * mass
42kJ - 42000J (convert kJ to J as energy is counted as J/kg)
Temp change = 42000/4200*10
= 1C
So it goes up by 1C thus the final temperature would be 26C
Or a more logical way to think of it would be that if it takes 4200J to heat up 1kg of water by 1C (SHC is 4200J/kgC) it would take ten times the amount of energy to heat up 10kg of water by 1C hence it takes 42000J.