It is the foundation for issues with weight and balance. According to this rule, a lever is balanced when the weight multiplied by the arm on one side of the fulcrum, which serves as the pivot point for the device, equals the weight multiplied by the arm on the opposing side.
The lever is balanced, in other words, when the sum of the moments about the fulcrum is zero.
The situation in which the positive moments (those attempting to turn the lever clockwise) equal the negative moments is known as this (those that try to rotate it counterclockwise).
Moving the weights closer to or away from the fulcrum, as well as raising or lowering the weights, can alter the balance point, or CG, of the lever.

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3 0

1 year ago

A large water tank is 3.70 m high and filled to the brim, the top of the tank open to the air. A small pipe with a faucet is att

ipn [44]

h =(3.7 - .58)m = 3.12m

Now put PE into KE and we have to use the formula:

√2gh (g = gravity and h = height) therefor:

√2 x 9.8 x 3.12

= 7.82m/s

I hope this helps!

7 0

2 years ago

Two blocks of masses 3.0 kg and 5.0 kg are connected by a spring and rest on a frictionless surface. They are given velocities t

miskamm [114]

Answer:

-0.7 m/sec

Explanation:

Mass of first block = m1 =3.0 kg

Mass of second block = m2= 5.0 kg

Velocity of first block = V1= 1.2 m/s

Velocity of second block = V2 = ?

Momentum of Center of mass MVcom is sum of both blocks momentum and is given by

MVcom= m1v1+m2v2

Where

M= mass of center of mass

Vcom= Velocity of center of mass=0 m/s (because center of mass is at rest , so Vcom = 0 m.sec)

Putting values, we get;

0= 3×1.2+5v2

==> v2= 3.6/5= - 0.7 m/s

-ve sign indicates that block 2 is moving in opposite direction of block 1

3 0

3 years ago

Electrical energy is used to turn the blades of a fan. The amount of energy transformed is seen here: 750 j electrical energy is

MAVERICK [17]

Here is the energy that is left after the quantity of energy is transformed: 750 j of electrical energy is changed into 400 j of kinetic or mechanical energy, which is then turned into 0.32 j of efficient energy.

To run the fan, electrical energy is utilized.

Here, under the specified circumstances, 750 J of electrical energy is utilized to operate the fan, which is transformed into 400 J of kinetic energy. As a result, 350 J of energy is wasted due to various frictional and resistive losses.

Therefore, we may conclude that only 400 J of the 750 J available energy is used to power the fan, with the remaining energy being wasted as a result of friction.

Additionally, we can state that this fan's effectiveness will be

n = Useful ÷ Total

n = 400 ÷ 750

n = 8 ÷ 25

n = 0.32

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4 0

10 months ago