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Alex787 [66]
3 years ago
11

A bucket filled with water has a mass of 70kg and is hanging from a rope that is wound around a 0.054 m radius stationary cylind

er. If the cylinder does not rotate and the bucket hangs straight down, what is the magnitude of the torque the bucket produces around the center of the cylinder?
Physics
1 answer:
guajiro [1.7K]3 years ago
3 0

Answer:

T = 37.08 [N*m]

Explanation:

We must remember that torque is defined as the product of a force by a distance. This distance is measured from the point of application of force to the center of rotation of the rotating body.

The force is equal to the product of mass by gravitational acceleration.

F=m*g\\F=70*9.81\\F=686.7[N]

Now the torque can be calculated:

T=F*r\\T=686.7*0.054\\T=37.08[N*m]

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luda_lava [24]

Answer:

the third stage was 480 km long

Explanation:

Stage 1:

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Stage 2:

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Stage 3:

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Let the Distance at the stage 3 be x

Average speed of the train route = 100 km/h

So

\frac{ \text{speed at stage 1} + \text{speed at stage 2} + \text{speed at stage 3}}{3} = 0

\frac{ \text{speed at stage 1} + \text{speed at stage 2} + \text{speed at stage 3}}{3} = 100

Lets find the speed at stage 1

Speed =  \frac{Distance }{Time}

Speed =  \frac{80}{1}

Speed 1= 80 km/hr

The speed at stage 2

Speed =  \frac{Distance }{Time}

Speed =  \frac{200}{2}

Speed 2  = 100 km/hr

The speed at stage 3

Speed =  \frac{Distance }{Time}

Speed =  \frac{x}{4}

Speed 3  = \frac{x}{4}

we kow that average is ,

\frac{ \text{speed 1} + \text{speed 2} + \text{speed 3}}{3} = 100

\frac{ 80 + 100+ \frac{x}{4} }{3} = 100

\frac{ 180 + \frac{x}{4} }{3} = 100

\frac{ \frac{720 +x}{4} }{3} = 100

\frac{720 +x}{4} \times \frac{1}{3} = 100

\frac{720 +x}{12} = 100

720 +x = 100 \times 12

720 +x = 1200

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x = 480

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3 years ago
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#28

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Explain newton 2nd law ???<br><br>​
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Explanation:

Newton's second law simply says that the net force on an object is equal to the object's mass times its acceleration.

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If the forces are equal (balanced), then the net force is 0 N, so Newton's second law tells us that the rope's acceleration is 0 m/s².

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