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3 years ago

A weightlifter lifts a 1250N barbell 2m in 3s.How much power was used to lift the barbell

2 answers:

4 0

Work = (force) x (distance)

   = (1,250 N) x (2 m) = 2,500 N-m = 2,500 Joules

Power = (work) / (time to do the work)

   = (2,500 J) / (3 sec) = 2500/3 J/s = 833-1/3 watts

   (1.12 horsepower)

vaieri [72.5K]3 years ago

3 0

Work done is equal to the force applied multiplied by the distance moved in the direction of the force. The weight force acts vertically downwards and the distance moved is vertically upwards so they are in the same line. Hence W = Fx = 1250 * 2 = 2500J. Power is the rate of work done, i.e. work done divided by time. P = W/t = 2500/3 = 833W (3sf)

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Read 2 more answers

A step-up transformer has a primary coil with 100 loops and a secondary coil with 1,500 loops. If the primary coil is supplied w

djyliett [7]

(a) The voltage that is produced in the secondary circuit is 1,800 V.

(b) The current that flows in the secondary circuit is 1 A.

<h3>Voltage in the secondary coil</h3>

Np/Ns = Vp/Vs

where;

Np is number of turns in primary coil
Ns is number of turns in secondary coil
Vp is voltage in primary coil
Vs is voltage in secondary coil

100/1500 = 120/Vs

Vs = (120 x 1500)/100

Vs = 1,800 V

<h3>Current in the secondary coil</h3>

Is/Ip = Vp/Vs

where;

Is is secondary current
Ip is primary current

Is = (IpVp)/Vs

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Is = 1 A

Thus, the voltage that is produced in the secondary circuit is 1,800 V.

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2 years ago

Electrons in a particle beam each have a kinetic energy of 4.0 × 10−17 J. What is the magnitude of the electric field that will

denpristay [2]

Explanation:

Relation between work and change in kinetic energy is as follows.

$W_{net} = \Delta K$

Also, $\Delta K = K_{initial} - K_{final}$

= $(0 - 4.0 \times 10^{-17})$ J

= $-4.0 \times 10^{-17}$ J

Let us assume that electric force on the electron has a magnitude F. The electron moves at a distance of 0.3 m opposite to the direction of the force so that work done is as follows.

w = -Fd

$-4.0 \times 10^{-17} J = -F \times 0.3 m$