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e-lub [12.9K]
3 years ago
11

Which class lever is called force multiplier and why?​

Physics
2 answers:
Degger [83]3 years ago
8 0

Question:

Which class lever is called force multiplier and why?​

Answer:

Lever of second type (or class II)

Explanation:

Lever of second type (or class II) is used as a force multiplier. In second class lever, the effort arm is always longer than the load arm and so its mechanical advantage is always greater than 1. So this lever acts as a force multiplier i.e., by applying a less effort, large load is lifted.

ddd [48]3 years ago
3 0

Answer:5

Explanation:

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A boy is whirling a stone around his head by means of a string. The string makes one complete revolution every second; and the m
Damm [24]

Answer

given,                                                

Tension of string is F                                                  

velocity is increased and the radius is not changed.      

the string makes two complete revolutions every second

consider the centrifugal force acting on the stone          

  = \dfrac{mv^2}{r}                          

now centrifugal force is balanced by tension

T =\dfrac{mv^2}{r}                                

From the above expression we can clearly see that tension is directly proportional to velocity and inversely proportional to radius.

When radius is not changing velocity is increasing means tension will also increase in the string.

8 0
3 years ago
A turtle takes 3.5 minutes to walk 18 m toward the south along a deserted highway. A truck driver stops and picks up the turtle.
Akimi4 [234]

Answer:

3.59 m/s

Explanation:

The average velocity is defined as:

\frac{Total displacement}{Total time}

The turtle first walks 18m south, and then is taken 1,1Km (or 1100m) north. Thus, the total displacement  is 1082m north (1100m north - 18m south).

Now we have to calculate the total time, which will be equal to the sum of the time the turtle walked and the time it was taken by truck.

The walking time is 3.5 minutes. Since 1 minute = 60 seconds, then the walking time is 210 seconds.

To calculate the truck time we use the equation:

Time = \frac{Distance}{Speed}

Where the distance the truck travelled is 1100m and the speed of the truck is 12m/s.

Thus,

Truck time= \frac{1100m}{12m/s}= 91.67s

The total time is the sum of the walking time and the truck time.

Total time = 210s + 91.67s = 301.67s.

As mencioned previously, the average velocity is equal to total displacement/ total time, thus:

Average velocity = \frac{1082m}{301.67s} = 3.59 m/s North

Since the average velocity is a vector, it has a magnitude and a direction. In this case the magnitude is 3.59 m/s and the direction is north since the turtle's final displacement is north of where it started.

8 0
3 years ago
3.00 textbook rests on a frictionless, horizontal tabletop surface. A cord attached to the book passes over a pulley whose diame
sammy [17]

Answer:

a1 = 3.56 m/s²

Explanation:

We are given;

Mass of book on horizontal surface; m1 = 3 kg

Mass of hanging book; m2 = 4 kg

Diameter of pulley; D = 0.15 m

Radius of pulley; r = D/2 = 0.15/2 = 0.075 m

Change in displacement; Δx = Δy = 1 m

Time; t = 0.75

I've drawn a free body diagram to depict this question.

Since we want to find the tension of the cord on 3.00 kg book, it means we are looking for T1 as depicted in the FBD attached. T1 is calculated from taking moments about the x-axis to give;

ΣF_x = T1 = m1 × a1

a1 is acceleration and can be calculated from Newton's 2nd equation of motion.

s = ut + ½at²

our s is now Δx and a1 is a.

Thus;

Δx = ut + ½a1(t²)

u is initial velocity and equal to zero because the 3 kg book was at rest initially.

Thus, plugging in the relevant values;

1 = 0 + ½a1(0.75²)

Multiply through by 2;

2 = 0.75²a1

a1 = 2/0.75²

a1 = 3.56 m/s²

6 0
3 years ago
an aircraft landing on an air craft carrier is brought to a complete stop from an inital velocity of 215km/hr in 2.7 seconds. wh
worty [1.4K]

u= 215 km/hr = 215 * 1000/ 3600 = aprx 60m/s
v=0
t=2.7sec
v= u - at
u= at
60/2.7 = 22.23 m/s^2



Hope it helps
8 0
2 years ago
If javier has a car thats 250,000 kg and a forse of 300 newtons.what would be the accleration.
const2013 [10]

Answer:

The acceleration would be 3.455.

3 0
3 years ago
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