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pishuonlain [190]
2 years ago
14

Using the formula for work, Match the Force × Distance problems with their correct "Work Performed" answer. The first example ha

s been completed for you.
Force × Distance = Work Performed

10 newtons 6 meters =60 Nm

1. 62 Nm
40 pounds 100 feet

2. 4000ft.-lbs.
2.5 pounds 24 inches

3. 5 ft.-lbs
62 newtons 1 meter

4. 200ft.-lbs.
1 ton 20 feet

5. 40,000 ft.-lbs.
400 pounds 0.5 feet
Physics
1 answer:
Alenkinab [10]2 years ago
3 0
1. 62NM = 62 newtons 1 meter
2. 4000ft.-lbs = 40 pounds 100 feet
3. 5ft.-lbs = 2.5 pounds 24 inches
4. 200ft.-lbs = 400 pounds 0.5 feet
5. 40,000 ft.-lbs = 1 ton 20 feet
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Answer:

Answers can be seen below

Explanation:

First we must explain the essential when we clear equations, and that is that if the term we need to clear is accompanied by other terms that are being added up, then those terms go to the other side of the equation to subtract if those terms are subtracting, then they go to the other side to add, if those terms are found multiplying then they go to the other side of the equation to divide and if those other terms are found dividing then they go to the other side of the equation to multiply.

(Primero debemos explicar lo esencial cuando despejamos ecuaciones, y es que si el término que necesitamos despejar va acompañado de otros términos que se están sumando, entonces esos términos van al otro lado de la ecuación para restar si esos términos están restando, luego van al otro lado para sumar, si esos términos se encuentran multiplicando luego van al otro lado de la ecuación a dividir, y si esos términos se encuentran dividiendo, pasan al otro lado de la ecuación a multiplicar.)

1 )  

t=\frac{v}{a} ; d=s*(t-t_{0} )

2)

k=\frac{2*U}{x^{2} }; T_{2}=\frac{P_{2}*V_{2}*T_{1}  }{P_{1}*V_{1}  }  \\

3)

L=\frac{F}{\pi*r*P}; d=\frac{w}{F*cos(o)}

4)

t^{2}=\frac{2*x}{g}  ; V_{2}=\frac{A_{1}*V_{1} }{A_{2} }  \\

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h=\frac{V}{\pi *r^{2} } ; r=\frac{t}{F*sin(o)}

6)

h=\frac{m}{(1/2)*\pi *r^{2} }  ; h_{2}=\frac{F_{2}*(1/2)*b_{1} *h_{1} }{F_{1}*(1/2)*b_{2}*h_{2}   }

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b=\frac{mg-ma}{v}; m=\frac{F+kx}{g*cos(o)}

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N=\frac{mg*sin(o)-F}{u} ; x^{2}=\frac{W+\frac{1}{2}k*x_{1}^{2}   }{\frac{1}{2}*k }

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x=x_{o} +\frac{v^{2-v_{o}^{2}  } }{2a}  ;  m=\frac{P*A-F_{1}-F_{2} }{g}

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x_{o} = x-\frac{F}{k} ;  u=\frac{cos(o)-\frac{a}{g} }{sin(o)}

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t=\frac{d}{v} +t_{o} ; t_{o} = t-(\frac{v-v_{o} }{a} )

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y_{1}=y-\frac{u}{mg}  ; x^{2} = \frac{2W}{k}+x_{o} ^{2}

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