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Juli2301 [7.4K]
3 years ago
14

Suppose you have a 115 kg wooden crate resting on a wood floor, with coefficient of static friction 0.500 between these wood sur

faces. (Enter the magnitudes.) (a) What maximum force (in N) can you exert horizontally on the crate without moving it
Physics
1 answer:
alexdok [17]3 years ago
4 0

Answer:

maximum force (in N) can you exert horizontally on the crate without moving it = 564.075 N

Explanation:

Given data:

mass m= 115 kg

coefficient of friction μ =0.5

according to newton law of motion the net force acted on body is given by

F_{net}=ma=F_{max}- F_{f}

here the body is in rest ,

F_{net}=0\\\Rightarrow F_{max}=F_{f}\\\text{frriction force}  F_{f} = \mu\times N\\

N= it is the normal reaction force

N= mg = 115×9.81

therefore,

F_{f} = 0.5\times115\times9.81 = 564.075 N = F_{max}

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An aluminum cup contains 225 g of water and a 40-g copper stirrer, all at 27°C. A 470-g sample of silver at an initial temperat
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Answer:

Mal = 0.232 kg = 232 g

Explanation:

mass of water (Mw) = 225 g = 0.225 kg

mass of copper stirrer (Mcu) = 40 g = 0.04 kg

initial temperature of water (Tw) = 27 degrees

mass of silver (Mag) = 470 g = 0.47 kg

initial temperature of silver (Ts) = 85 degrees

final temperature of the mixture (T) = 32 degrees

find the mass of the aluminum cup (Mal)

applying the conservation of energy

((Mal.cAl) + (Mw.cW) + (Mcu.cCu))(ΔTw) = (Mag.cAg)(ΔTag)

we require the specific heat capacities of water (cW) , aluminium (cAl) , copper (cCu) and silver (cAg) which are as follows

water (cW) =4186 J/kg.K

aluminium (cAl) = 900 J/kg.K

copper (cCu) = 387 J/kg.K

silver (cAg) = 234 J/kg.K

now we can put in our values to get the mass of the Aluminium cup (Mag)

((Mal.900) + (0.225 x 4186) + (0.04 x 387))(32-27) = (0.47 x 234)(85-32)

(900 . Mal + 957.33) x 5 = 5828.94

900 .Mal + 957.33 = 1165.79

900.Mal = 1165.79 - 957.33 = 208.5

Mal = 0.232 kg = 232 g

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Both options 5 and 6

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Let us consider option 6,

In option 6 body is moving down with initial velocity lower than that of final velocity which gets accelerated. Therefore the acceleration is positive in this case.

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