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victus00 [196]
3 years ago
13

What is the efficiency of the machine in Figure 7.2 if the weight of the box is 200 N and required 98 N of effort Force?

Physics
1 answer:
zepelin [54]3 years ago
3 0
this question is tricky but the answer is C
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A 20-kg cart travels at 20 m/s in a circle with a radius of 20 m. What is the centripetal force?
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Centripetal force = (mass) x (speed)² / (radius)

                           = (20 kg) x (20 m/s)² / (20 m)

                           = (20 x 20 / 20)  (kg-meter/sec²)

                           = 20 newtons


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When a body of mass 0.25 kg is attached to a vertical massless spring, it is extended 5.0 cm from its unstretched length of 4.0
lora16 [44]

Answer:

d=0.165m

Explanation:

Given

m=0.25kg,x_{1}=5cm*\frac{1m}{100cm}=0.05m,x_{2}=4cm*\frac{1m}{100cm}=0.04m,v=2\frac{rev}{s}

The tension of the spring is

F_{k}=K*x_{1}=m*g

K=\frac{m*g}{x_{1}}

K=\frac{0.25kg*9.8m/s^2}{0.05m}=49N/m

The force in the spring is equal to centripetal force so

F_{c}=\frac{m*v^2}{r}

v=w*r=2\pi*r

But Fc is also

Fc=KxΔr

F_{c}=K*(r-x_{2})

Replacing

m*4\pi^2*r=K*(r-x_{2})

0.25kg*4\pi^2*r=49*(r-0.04m)

r=0.205m

total distance is

d=0.205-0.04=0.165m

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