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Nat2105 [25]
3 years ago
9

An inclined plane makes an angle of 30.0º with the horizontal. 

Physics
1 answer:
Wittaler [7]3 years ago
4 0

The force required to push the box upward is 145.3N and the force to pus the box downward is -109.3N

Data given;

  • mass = 15kg
  • angle = 30 degree
  • acceleration = 1.2 m/s^2
  • acceleration due to gravity = 9.8 m/s^2

<h3>Force against gravity</h3>

To move the plane upward, the box will move against gravity.

F = F - mgcos X\\&#10;ma = F - mg cosX

Let's solve for F

ma = F - mgcosX\\&#10;15*1.2 = F - 15*9.8 * cos 30\\&#10;18 = F - 127.30\\&#10;F = 18 + 127.30\\&#10;F = 145.3N

<h3>Force towards gravity</h3>

When the force pushes the box down the inclined plane, it moves towards gravity.

F = F + mgcosX\\&#10;ma = F + mgcosX\\&#10;15*1.2 = F + 15*9.8*cos30\\&#10;F = -109.3N

The force required to push the box upward is 145.3N and the force required to push the box downward is -109.3N

Learn more on force across an inclined plane here;

brainly.com/question/11888124

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A force of 100. newtons is used to move an object a distance of 15 meters with a power of 25 watts. Find the time it takes to do
Flauer [41]
<h3>It takes 60 seconds to do the work</h3>

<em><u>Solution:</u></em>

Given that,

Force = 100 newtons

Distance = 15 meters

Power = 25 watts

To find: time it takes to do the work

<em><u>Find the work done:</u></em>

work = force \times displacement\\\\work = 100\ newtons \times 15\ meters\\\\work = 1500\ joule

<em><u>Find the time taken</u></em>

power = \frac{work}{time}\\\\25\ watts = \frac{1500\ joule}{time}\\\\time = \frac{1500\ joule}{25\ watt}\\\\time = 60\ second

Thus it takes 60 seconds to do the work

3 0
3 years ago
To apply Problem-Solving Strategy 12.2 Sound intensity. You are trying to overhear a most interesting conversation, but from you
Ivenika [448]

Answer:

r₂ = 0.316 m

Explanation:

The sound level is expressed in decibels, therefore let's find the intensity for the new location

            β = 10 log \frac{I}{I_o}

let's write this expression for our case

           β₁ = 10 log \frac{I_1}{I_o}

           β₂ = 10 log \frac{I_2}{I_o}

           

          β₂ -β₁ = 10 ( log \frac{I_2}{I_o} - log \frac{I_1}{I_o})

          β₂ - β₁ = 10 log \frac{I_2}{I_1}

          log \frac{I_2}{I_1} = \frac{60 - 20}{10} = 3

           \frac{I_2}{I_1} = 10³

           I₂ = 10³ I₁

having the relationship between the intensities, we can use the definition of intensity which is the power per unit area

           I = P / A

           P = I A

the area is of a sphere

          A = 4π r²

           

the power of the sound does not change, so we can write it for the two points

          P =  I₁ A₁ =  I₂ A₂

          I₁ r₁² = I₂ r₂²

we substitute the ratio of intensities

          I₁ r₁² = (10³ I₁ ) r₂²

         r₁² = 10³ r₂²

         

         r₂ = r₁ / √10³

         

we calculate

          r₂ = \frac{10.0}{\sqrt{10^3} }

          r₂ = 0.316 m

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