## How do you calculate the power needed to move an object?

Learning the Formula. Multiply mass times acceleration. The force (F) required to move an object of mass (m) with an acceleration (a) is given by the formula F = m x a. So, force = mass multiplied by acceleration.

## What is the rate at which work is done in lifting a 35?

Explanation: The rate at which work is done is also the power dissipated in lifting the object at that constant speed. Therefore, the the rate at which work is done in lifting the object vertically at a constant speed of 5 m/s is 1,700 W.

## Does the power expended depend on how fast it is lifted?

No. Work is independent of time. Power is dependent on time, so lifting it faster would require more power.

## What are the methods of improving power factor?

The simplest way to improve power factor is to add PF correction capacitors to the electrical system. PF correction capacitors act as reactive current generators. They help offset the non-working power used by inductive loads, thereby improving the power factor.

## What is 3phase power factor?

p.f. = average power factor or the three separate phases. 1.732 = a constant necessary with 3 phase. In a three phase circuit, the use of the constant 1.732 results from the fact that not all three phases are producing the same amount of power at the same time.

## Is power factor good or bad?

Going one step further, Power Factor (PF) is the ratio of working power to apparent power, or the formula PF = kW / kVA. A high PF benefits both the customer and utility, while a low PF indicates poor utilization of electrical power.

## What are the effects of Power Factor?

A lower power factor causes a higher current flow for a given load. As the line current increases, the voltage drop in the conductor increases, which may result in a lower voltage at the equipment. With an improved power factor, the voltage drop in the conductor is reduced, improving the voltage at the equipment.

## Does improving power factor save money?

The financial savings & benefits of power factor correction can by substantial and generally are achievable from the following areas: Reduction in power consumption kw/h losses. Removal of penalties on electricity bill in the form of reactive power charges. Reduction in capacity charges / authorised supply capacity.

## What are the advantages of improving power factor?

The reduction in current flow resulting from improved power factor may allow the circuit to carry new loads, saving the cost of upgrading the distribution network when extra capacity is required for additional machinery or equipment, saving your company thousands of dollars in unnecessary upgrade costs.

## How can we reduce the power factor?

Some strategies for correcting your power factor are: Minimize operation of idling or lightly loaded motors. Avoid operation of equipment above its rated voltage. Replace standard motors as they burn out with energy-efficient motors.

## How do you calculate power factor on a calculator?

The following formula can be used to solve for power factor:

1. PF = cos θ = PS.
2. PF = P(W)(V(V) × I(A))
3. PF = P(W)(√3 × V(V) × I(A))
4. capacitance(µF) = 1,000,000 × Q(VAR)(2 × π × 60(Hz) × V(V)2)