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Alchen [17]
4 years ago
5

A 2kg box is pushed along a flat frictionless surface with an applied force of 53.91newton j+19.62 j were j is horizontal and j

is vertical if the box moves 10m horizontal in 90 seconds what is the work and power that is required to complete the movement
Physics
1 answer:
tamaranim1 [39]4 years ago
5 0

Since the box doesn't move vertically at all, no work is done by the vertical component of the force.

If 53.91 Newtons is the horizontal component of the force (very unclear in the question), then the work done is

<u>Work = (force) x (distance)</u>

Work = (53.91 N) x (10 m)

<em>Work = 539.1 Joules</em>

<u>Power = (work done) / (time to do the work)</u>

Power = (539.1 joules) / (90 sec)

Power = (539.1/90) (joule/sec)

<em>Power = 5.99 watts</em>

=====================================

Note:  If the mass of the box is only 2 kg, and you push it along the surface with a constant force of 53⁺ Newtons, and the surface really is frictionless, then that box is gonna cover a WHOOOOOLE LOT more than 10 meters in 90 seconds.  I get 109,168 meters !

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Answer:

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Explanation:

Coulombs law states that "the force of attraction between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them."

Mathematically, F = kq1q2/r²

Where q1 and q2 are the charges

r is the distance between the charges.

According to the law, the force between two charged objects is related to the inverse of the square of the distance separating them.

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3 years ago
f a single circular loop of wire carries a current of 45 A and produces a magnetic field at its center with a magnitude of 1.50
Lelu [443]

Answer:

Radius of the loop is 0.18 m or 18 cm

Explanation:

Given :

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Consider R be the radius of the circular wire.

The magnetic field at the center of the current carrying circular wire is determine by the relation:

B=\frac{N\mu_{0} I}{2R}

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Substitute the suitable values in the above equation.

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3 years ago
During heavy rain, a section of a mountainside measuring 2.5 km horizontally, 0.80 km up along the slope, and 2.0 m deep slips i
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Answer:

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Explanation:

Given that,

length = 2.5 km

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Width of valley = 0.40 km

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Put the value into the formula

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