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jenyasd209 [6]
3 years ago
7

What is the major factor controlling how levers work?

Physics
1 answer:
Gekata [30.6K]3 years ago
7 0

Answer:

Displacement between Fulcrum, Job and Force

Explanation:

The employment and use of levers is simple. Levers reduce the force required to do a task by introducing the concept of moment.

There are there components in a simple lever.

1) The Fulcrum

2) The Job (Load)

3) The Force (Effort)

Typically if it weren't for levers then the three points/components would be at the same point. A lever creates a displacement between all these three components create moment. The higher the displacement between the three the lower the force (effort) required to do the job (lift the load).

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The back wall of a home aquarium is a mirror that is a distance of 46.0 cm away from the front wall. The walls of the tank are n
monitta

Answer:

i know the questin but i got to try and find it

Explanation:

5 0
3 years ago
A car travels at a constant speed around a circular track whose radiu is 2.6 km. The goes once arond the track in 360s . What is
AveGali [126]

Answer:

Centripetal acceleration = 0.79 m/s²

Explanation:

<u>Given the following data;</u>

Radius, r = 2.6 km

Time = 360 seconds

<em><u>Conversion:</u></em>

2.6 km to meters = 2.6 * 1000 = 2600 meters

To find the magnitude of centripetal acceleration;

First of all, we would determine the circular speed of the car using the formula;

Circular \; speed (V) = \frac {2 \pi r}{t}

Where;

  • r represents the radius and t is the time.

Substituting into the formula, we have;

Circular \; speed (V) = \frac {2*3.142*2600}{360}

Circular \; speed (V) = \frac {16338.4}{360}

Circular speed, V = 45.38 m/s

Next, we find the centripetal acceleration;

Mathematically, centripetal acceleration is given by the formula;

Centripetal \; acceleration = \frac {V^{2}}{r}

Where;

  • V is the circular speed (velocity) of an object.
  • r is the radius of circular path.

Substituting into the formula, we have;

Centripetal \; acceleration = \frac {45.38^{2}}{2.6}

Centripetal \; acceleration = \frac {2059.34}{2600}

<em>Centripetal acceleration = 0.79 m/s²</em>

3 0
3 years ago
Unele rigle gradate au partea pe care sunt imprimante graditatiile in forma de pana. prezinta aceasta vreo importanta pentru mas
GrogVix [38]
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7 0
3 years ago
In the following diagrams the larger vector has a magnitude of 10, and the smaller
Anton [14]

the sum of vectors with the Pythagorean theorem allows us to find that the maximum magnitude occurs for the case:

d) the two vectors are parallel

Vectors are physical quantities that have modulus and direction, for example: force, velocity, acceleration, etc.

Vector algebra has defined the sum, the product by a scalar and by a vector.  The modulus and the direction of the resulting vector must be encoded.

The sum of two quantities is done using the Pythagorean theorem

                  c² = a² + b²

where c is the resultant called hypotenuse, a and b are the summing vectors called legs; trigonometry is used for the direction.

Let's apply this expression to the present case

a, b) perpendicular vectors

              c² = a² + b²

              c = \sqrt{6^2+10^2}

              c = 11.7

the magnitude is the same in both cases, changing the direction of the vector

c) Antiparallel vectors

             

For this case the vectors are collinear, so the sum reduces to the algebraic addition

             c = a-b

             c = 6 -10

             c = -4

d) parallel vectors

             c = a + b

             c = 4 + 10

             c = 14

We can see that the vectors addition gives their maximum and minimum values ​​when the vectors are collinear.

In conclusion using the vectors addition we find that the correct answer is

d) the two vectors are parallel

learn more about vector addition here:

brainly.com/question/15074838

6 0
3 years ago
Explain why some places can use only certain kinds of energy to generate electricity.
alukav5142 [94]

Answer:

please mark me as a brainleast and follow me please my soster

6 0
3 years ago
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