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user100 [1]
3 years ago
15

On a Saturday afternoon, you decide to pay a neighborhood kid to mow your lawn. The kid usesa manual push lawn mower with a mass

of m=12 kg. The handle of the lawnmower is tiltedθ=53 degrees with respect to the vertical and the coefficient of kinetic friction between thelawnmower and the ground is μ​k​=0.16. Assume that the kid is pushing parallel to the handle.A.Find an expression for much power the kid supplies to the lawnmower (P​kid​) if thelawnmower is moving at a constant speed of v​o​=1.5 m/s.B.Find the numerical value for P​kid
Physics
1 answer:
Mars2501 [29]3 years ago
6 0

Answer with Explanation:

We are given that

A.Mass,m=12 kg

\theta=53^{\circ}

\mu_k=0.16

Speed,v=1.5m/s

Net force in x direction must be zero

F_{net}=0

Fsin\theta-f=0

Fsin\theta=f

Net force in y direction

N-mg-Fcos\theta=0

N=mg+Fcos\theta

f=\mu_kN=\mu_k(mg+Fcos\theta)

Fsin\theta=\mu_k(mg+Fcos\theta)

Fsin\theta=\mu_kmg+\mu_kFcos\theta

Fsin\theta-\mu_kFcos\theta=\mu_kmg

F(sin\theta-\mu_kcos\theta)=\mu_kmg

F=\frac{\mu_kmg}{sin\theta-\mu_kcos\theta}

Power,P=Fv

P=\frac{\mu_kmg}{sin\theta-\mu_kcos\theta}v

Where g=9.8m/s^2

B.Substitute the values

P=\frac{0.16\times 12\times 9.8}{sin53-0.16cos53}\times 1.5

P=40.17W

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Answer:

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Explanation:

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PE=77*9.81*42.6sin42.3^{o}=21656.7095 Nm

PE=21.6567095 KNm

We also know that Kinetic energy is given by0.5mv^{2} where v is the velocity and substituting v for 20.3 we obtain

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Answer:

The temperature of air will increase by \Delta T=41044.967\ K

Explanation:

Given:

  • no. of person in a theater, n=501
  • volume of air in the theater, V=8\times 10^3\ m^3
  • rate of heat given off by each person, P=110\ J.s^{-1}
  • duration of movie, t=2\ hr=7200\ s
  • initial pressure in the theater, p_i=1.01\times 10^5\ Pa
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<u>The total quantity of heat released by the total people in the theater during the movie:</u>

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Q=(\frac{p_i.V}{R.T}) \times c\times (T_f-T_i)

396792000=(\frac{1.01\times 10^5\times 8\times 10^3}{287\times 293}) \times 1.0061\times (T_f-293)

T_f=41337.967\ K

Change in temperature of air:

\Delta T=41044.967\ K

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