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Dmitry [639]
3 years ago
13

the train is travelling at 15 m/s. It accelerates at 3 m/s² for 20 seconds. How far does the train travel in this time?

Physics
2 answers:
lord [1]3 years ago
7 0
So I believe that you have to use s=ut+1/2at2 so s=15 into 20+1/2 into 3 into 400 this is equal to 900m btw idk if it’s true perhaps you can check on snapsolve
Anvisha [2.4K]3 years ago
5 0
18meters per 20 seconds so
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Answer:

C. balancing the amount of energy that is taken in with the amount of energy that is released by the body

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3 years ago
A uniform crate with a mass of 22 kg must be moved up along the 15° incline without tipping. The force P is horizontal. Determin
Paul [167]

Answer:

F_x=208.25\ N

Explanation:

Given that,

Mass of a crate is 22 kg

It moved up along the 15 degrees incline without tipping.

We need to find the corresponding magnitude of force P. The force P is acting in horizontal direction.

It means that the horizontal component of force is given by :

F_x=F\cos\theta\\\\F_x=mg\cos\theta\\\\F_x=22\times 9.8\times \cos(15)\\\\F_x=208.25\ N

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3 years ago
Rank the following objects by their accelerations down an incline (assume each object rolls without slipping) from least to grea
Alexxx [7]

Answer:

acceleration are

     hollow cylinder < hollow sphere < solid cylinder < solid sphere

Explanation:

To answer this question, let's analyze the problem. Let's use conservation of energy

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          Em₀ = U = m g h

Final point. To get off the ramp

          Em_f = K = ½ mv² + ½ I w²

notice that we include the kinetic energy of translation and rotation

         

energy is conserved

        Em₀ = Em_f

        mgh = ½ m v² +1/2 I w²

angular and linear velocity are related

         v = w r

         w = v / r

we substitute

          mg h = ½ v² (m + I / r²)

          v² = 2 gh   \frac{m}{m+ \frac{I}{r^2} }

          v² = 2gh    \frac{1}{1 + \frac{I}{m r^2} }

this is the velocity at the bottom of the plane ,, indicate that it stops from rest, so we can use the kinematics relationship to find the acceleration in the axis ax (parallel to the plane)

         v² = v₀² + 2 a L

where L is the length of the plane

         v² = 2 a L

         a = v² / 2L

we substitute

         a = g \ \frac{h}{L} \  \frac{1}{1+ \frac{I}{m r^2 } }

let's use trigonometry

         sin θ = h / L

         

we substitute

         a = g sin θ   \ \frac{h}{L} \  \frac{1}{1+ \frac{I}{m r^2 } }

the moment of inertia of each object is tabulated, let's find the acceleration of each object

a) Hollow cylinder

      I = m r²

we look for the acerleracion

      a₁ = g sin θ    \frac{1}{1 + \frac{mr^2 }{m r^2 } }1/1 + mr² / mr² =

      a₁ = g sin θ    ½

b) solid cylinder

       I = ½ m r²

       a₂ = g sin θ  \frac{1}{1 + \frac{1}{2}  \frac{mr^2}{mr^2} } = g sin θ   \frac{1}{1+ \frac{1}{2} }

       a₂ = g sin θ   ⅔

c) hollow sphere

     I = 2/3 m r²

     a₃ = g sin θ   \frac{1}{1 + \frac{2}{3} }

     a₃ = g sin θ \frac{3}{5}

d) solid sphere

     I = 2/5 m r²

     a₄ = g sin θ  \frac{1 }{1 + \frac{2}{5} }

     a₄ = g sin θ  \frac{5}{7}

We already have all the accelerations, to facilitate the comparison let's place the fractions with the same denominator (the greatest common denominator is 210)

a) a₁ = g sin θ ½ = g sin θ      \frac{105}{210}

b) a₂ = g sinθ ⅔ = g sin θ     \frac{140}{210}

c) a₃ = g sin θ \frac{3}{5}= g sin θ       \frac{126}{210}

d) a₄ = g sin θ \frac{5}{7} = g sin θ      \frac{150}{210}

the order of acceleration from lower to higher is

   

     a₁ <a₃ <a₂ <a₄

acceleration are

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2 years ago
a skinny girl devours food that is equivalent to 600 kcal, as she is anorexic she repents and wants to lose twice as many calori
blondinia [14]

Answer:

Explanation:

Work done in lifting the weight once = mgh

= 20 x 9.8 x (1.9+1.7)

= 705.6 J  

= 705.6 / 4.2 calorie

= 168 cals

Total energy to be spent = 600 x 10³ cals

No of times weight is required to be lifted

= 600 x 10³ / 168

= 3.57  x 10³ times

Total time to be taken = 2 x 3.57 x 10³

= 7.14 x 10³ s

=119 minutes .

4 0
3 years ago
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