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hodyreva [135]
3 years ago
5

How fast must a train accelerates from rest to cover 518 m in the first 7.48 s?

Physics
1 answer:
Rus_ich [418]3 years ago
8 0

Heya!

For this problem, use the formula:

s = Vo * t + (at^2) / 2

Since the initial velocity is zero, the formula simplifies like this:

s = (at^2) / 2

Clear a:

2s = at^2

(2s) / t^2 = a

a = (2s) / t^2

Data:

s = Distance = 518 m

t = Time = 7,48 s

a = Aceleration = ¿?

Replace according formula:

a = (2*518 m) / (7,48 s)^2

Resolving:

a = 1036 m / 55,95 s^2

a = 23,34 m/s^2

The aceleration must be <u>23,34 meters per second squared</u>

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Answer:

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When the mass of the cylinder changes if its external dimensions do not change the angular velocity DOES NOT CHANGE

Explanation:

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The moment of inertia of a cylinder with respect to an axis at the end we can use the parallel axes theorem, it is approximately equal to that of a long bar plus the moment of inertia of the center of mass of the cylinder, this is tabulated

        I = ¼ m r2 + ⅓ m L2

        I = m (¼ r2 + ⅓ L2)

now let's use the concept of density to calculate the mass of the system

        ρ = m / V

        m = ρ V

the volume of a cylinder is

         V = π r² L

          m =  ρ π r² L

let's substitute

        w² = m g (L / 2) / m (¼ r² + ⅓ L²)

        w² = g L / (½ r² + 2/3 L²)

        L >> r

         w = √[g /L (½ r²/L2 + 2/3 ) ]

When the mass of the cylinder changes if its external dimensions do not change the angular velocity DOES NOT CHANGE

4 0
3 years ago
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Nutka1998 [239]
The movement of air flows from high pressure to low pressure
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Three students have been studying relative motion and decide to do an experiment to demonstrate their knowledge. The experiment
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Answer with Explanation:

We are given that

Constant speed of Jane=12.6 m/s

a.When Fred can throw the ball 30  m/s

We have to find the angle relative to the horizontal when he throw the ball in order for Sue to see the ball travel vertically upward.

Let \theta be the angle .

Therefore,

30 cos\theta=12.6

cos\theta=\frac{12.6}{30}=0.42

\theta=cos^{-1}(0.42)=65.165^{\circ}

b.We have to find the height to which ball reach.

v^2-v^2_0=2aS

S=\frac{v^2-v^2_0}{2a}=\frac{0-(30 sin65.165)^2}{2(-9.81)}

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4 0
3 years ago
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A man with a weight of 100 N is located
valentina_108 [34]

Answer:

Given,

  mass of man = 100 N = 10 kg

   height = h = 25m

   since the man does not move anything with his force, work done by him is zero

   work done on the man = gain in potential energy

   P.E=mgh

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  P.E=2.45KJ

Explanation:

so, potential energy gained by man is 2.45 KJ

5 0
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