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Vanyuwa [196]
3 years ago
14

A boy is pulling his two sisters on a sled.

Physics
1 answer:
OlgaM077 [116]3 years ago
4 0

Answer:

Explanation:

The oly way we can figure this out is if the boy is pulling the sled at a constant velocity. If not, we need a value for acceleration, and you don't have that here. If the boy is pulling the sled at a constant velocity, then the value for acceleration is 0, making this a really simple problem. I'm going with that, since there is no way to answer you otherwise. If velocity is not constant, please either repost the question or put it in the notes section under the question as it stands. If acceleration is 0, then

F - f = ma becomes

F - f = m(0) which is

F - f = 0 and

F = f which says that the applied force is the same as the frictional force. We need then to find the frictional force, which has an equation of

f = μF_n where normal force is the same as the weight of the 2 girls. We will find that, then:

Each girl's mass is different so the normal force/weight equation is

w = (30.0)(9.8) + (40.0)(9.8) to get

w = 290 + 390 and

w = 680. Plug that into the frictional force equation:

f = (.120)(680) so

f = 82N

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since Q = m * c * ( T eq - Ti ) , where m = mass, c = specific heat, T eq = equilibrium temperature and Ti = initial temperature

and denoting w as water and co as copper :

m w * c w * (T eq - Tiw) = - m co * c co * (T eq - Ti co) =  m co * c co * (T co - Ti eq)

m w = m co * c co * (T co - Ti eq) / [ c w * (T eq - Tiw) ]

We take the specific heat of water as c= 1 cal/g °C = 4.186 J/g °C . Also the specific heat of copper can be found in tables → at 25°C c co = 0.385 J/g°C

if we assume that both specific heats do not change during the process (or the change is insignificant)

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