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kap26 [50]
3 years ago
14

A 4kg box accelerated from rest by a force across the floor at a rate of 2m/s^2 for 7 seconds. What is the net work done on the

box?

Physics
2 answers:
AlladinOne [14]3 years ago
7 0

Answer and work is shown in the image attached.

please mark me brainliest :)

Illusion [34]3 years ago
6 0

The solution in the first answer is excellent.

Here's another way to do it:

Force = (mass) x (acceleration)

Force = (4 kg) x (2 m/s²)

Force = 8 Newtons

Distance = (1/2) (a) (T²)

D = (1/2) (2 m/s² ) (7 sec)²

D = 49 meters

Work = (force) x (distance)

Work = (8 N) x (49 m)

<em>Work = 392 Joules</em>   yay!

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7 0
3 years ago
You plug a string of 100 lights in series into a 120 V power outlet, and each light has a resistance of 3.00 _. If each light ha
UNO [17]
V = IR

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4 years ago
Beams used in Heavy Timber construction are sometimes firecut. This is done to: a) allow air circulation at the beam’s end in an
ladessa [460]

Answer:

option C

Explanation:

The correct answer is option C

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8 0
3 years ago
What would be the result of an alpha particle coming into a magnetic field?
Aneli [31]
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8 0
3 years ago
Read 2 more answers
Light of wavelength 480 nm illuminates a pair of slits separated by 0.27 mm. If a screen is place 1.7 m from the slits, determin
Dahasolnce [82]

Answer:

  Δy = 6.05 mm

Explanation:

The double slit phenomenon is described by the expression

      d sin θ = m λ                constructive interference

      d sin θ = (m + ½) λ       destructive interference

      m = 0,±1, ±2, ...

As they tell us that they measure the dark stripes, we are in a case of destructive interference, let's use trigonometry to find the sins tea

      tan θ = y / x

      y = x tan θ

In the interference experiments the measured angle is very small so we can approximate the tangent

      tan θ = sin θ / cos θ

     cos θ = 1

     tan θ = sin θ

     y = x sin θ

We substitute in the destructive interference equation

     d (y / x) = (m + ½) λ

    y = (m + ½) λ x / d

The first dark strip occurs for m = 0 and the third dark strip for m = 2. Let's find the distance for these and subtract it

 m = 0

      y₀ = (0+ ½) 480 10⁻⁹ 1.7 / 0.27 10⁻³

      y₀ = 1.511 10⁻³ m

 m = 2

     y₂ = (2 + ½) 480 10⁻⁹ 1.7 / 0.27 10⁻³

     y₂ = 7.556 10⁻³ m

The separation between these strips is Δy

     Δy = y₂-y₀

     Δy = (7.556 - 1.511) 10⁻³

     Δy = 6.045 10⁻³ m

     Δy = 6.05 mm

5 0
3 years ago
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