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rusak2 [61]
4 years ago
13

A 75-m-long train begins uniform acceleration from rest. The front of the train has a speed of 18 m/swhen it passes a railway wo

rker who is standing 125 m from where the front of the train started. What will be the speed of the last car as it passes the worker?
Physics
1 answer:
earnstyle [38]4 years ago
3 0

Answer:22.76 m/s

Explanation:

Given

Train length(L)=75 m

Front of train after travelling 125 m is 18 m/s

Time taken by the front of train to cover 125 m

v^2-u^2=2as

18^2-0=2\times a\times 125

a=1.296 m/s^2

Speed of the last part of train when it passes the worker i.e. front of train has to travel has to travel  a distance of 125+75=200 m

v^2-u^2=2as

v^2=2\times 1.296\times 200

v=\sqrt{518.4}=22.76 m/s

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Elodia [21]

As stated in the statement, we will apply energy conservation to solve this problem.

From this concept we know that the kinetic energy gained is equivalent to the potential energy lost and vice versa. Mathematically said equilibrium can be expressed as

\Delta KE = \Delta PE

\frac{1}{2}mv_f^2-\frac{1}{2} mv_0^2 = mgh_2-mgh_1

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v_{f,i} = initial and final velocity

g = Gravity

h = height

As the mass is tHe same and the final height is zero we have that the expression is now:

\frac{1}{2}v_f^2-\frac{1}{2} v_0^2 = gh_2

\frac{1}{2} (v_f^2-v_0^2) = gh_2

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v_f = \sqrt{2(9.8)(23.5)+13.6^2}

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7 0
3 years ago
3. Neha travels 4 m towards the north, 3m towards the east, and then 7 m towards south, find the displacement and total distance
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Answer:

Explanation:

Explanation: total displacement =3√2m. and total distance covered=14m. I hope this is right and helps u.

7 0
3 years ago
A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?
KATRIN_1 [288]

Complete Question

Planet D has a semi-major axis = 60 AU and an orbital period of 18.164 days. A piece of rocky debris in space has a semi major axis of 45.0 AU.  What is its orbital period?

Answer:

The value  is  T_R  = 11.8 \  days  

Explanation:

From the question we are told that

   The semi - major axis of the rocky debris  a_R = 45.0\  AU

   The semi - major axis of  Planet D is  a_D  = 60 \  AU

    The orbital  period of planet D is  T_D = 18.164 \  days

Generally from Kepler third law

          T \  \ \alpha \ \ a^{\frac{3}{2} }

Here T is the  orbital period  while a is the semi major axis

So  

        \frac{T_D}{T_R}  =  \frac{a^{\frac{3}{2} }}{a_R^{\frac{3}{2} }}

=>     T_R  = T_D *  [\frac{a_R}{a_D} ]^{\frac{3}{2} }  

=>     T_R  = 18.164  *  [\frac{ 45}{60} ]^{\frac{3}{2} }

=>      T_R  = 11.8 \  days  

   

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A bicycle wheel has a diameter of 63.0 cm and a mass of 1.75 kg. Assume that the wheel is a hoop with all of the mass concentrat
Masteriza [31]

Answer:

F2 = 834 N

Explanation:

We are given the following for the bicycle;

Diameter; d1 = 63 cm = 0.63 m

Mass; m = 1.75 kg

Resistive force; F1 = 121 N

For the sprocket, we are given;

Diameter; d2 = 8.96 cm = 0.0896 m

Radius; r2 = 0.0896/2 = 0.0448 m

Radial acceleration; α = 4.4 rad/s²

Now moment of inertia of the wheel which is assumed to be a hoop is given by; I = m(r1)²

Where r1 = (d1)/2 = 0.63/2

r1 = 0.315 m

Thus, I = 1.75 × 0.315²

I = 0.1736 Kg.m²

The torque is given by the relation;

I•α = F1•r1 - F2•r2

Where F2 is the force that must be applied by the chain to give the wheel an acceleration of 4.40 rad/s².

Thus;

0.1736 × 4.4 = (121 × 0.315) - (0.0448F2)

>> 0.76384 = 38.115 - (0.0448F2)

>> 0.0448F2 = 38.115 - 0.76384

>> F2 = (38.115 - 0.76384)/0.0448

>> F2 = 833.73 N

Approximately; F2 = 834 N

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