A 7.1 m garden hose with mass 6.9 kg is coiled up on the ground and a distance 28.5 m from your garden. You pull on one end of the hose with force of magnitude 13.6 N in a single direction. You continue to pull until your end of the hose is at the edge of the garden. It is very reasonable to note that the hose is very thin compared to its length, and as a first approximation, assume that the hose straightens out to its total length 7.1 m.

Part (a) What is the speed of the hose in m/s as you reach the garden, neglecting friction?
Part (b) What is the change in energy of the hose in joules? là 25% Part (c) What is the change in internal energy of the hose in joules? - 25% Part (d) The hose is rather stiff and energy must go into forcing the material to flex. If the material of the hose has a specific heat of 2.5 kJ/kg °C, what is the change in temperature of the hose in °C?

(a) v = 10.6m/s

(b) E = 387.6J

(c) U = 387.6J

(d) ΔT = 0.022°C

Explanation:

(a) In order to calculate the speed of the hose we will use the work-energy theorem. In simple terms it is a theorem that states that the work done is equal to the change in energy

Workdone = Force × distance

W = Fd

The above equation states that the total workdone is equal to the applied force times the distance over which the force was applied. In this problem a force of 13.6N is applied on the hose, moving one end of it over a distance of 28.5m from where the hose is located to the edge of the garden. Since the point of of application of the force moves the end over that distance, we only take this into consideration and not the length of the whole hose. So

workdone = 13.6 × 28.5 = 387.6J

This workdone is converted into the kinetic energy given to hose in moving it over this distance.

So

W = 1/2×mv²

v² = 2×W/m

v = √(2×W/m) = √(2×387.6/6.9) = 10.6m/s

(b) ΔE = 387.6 (work-energy theorem)

(c) From the first law of thermodynamics

Q = ΔU + W

Q = Heat supplied or taken from the hose

ΔU = change in internal energy of the hose

W = workdone by or on the hose = –387.6 (work is done on the hose. That's the reason for the negative sign)

So substituting their value,

0 = ΔU + (–387.6)

ΔU = 387.6J

(d) ΔU = mCv×ΔT

Given Cv = 2.5kJ/kg°C = 2500J/kg°C

387.6 = 6.9×2500×ΔT

387.6 = 17250ΔT

ΔT = 387.6/17250

ΔT = 0.022°CΔ

answer: question1

1,000 tons

question2

3: 8

question 3

store d

step-by-step explanation: