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sveta [45]
2 years ago
10

A bus slows down uniformly from 75.0 km/h to 0 km/h in 21 s. How far does it travel before stopping?

Physics
2 answers:
RoseWind [281]2 years ago
7 0

1 hour = 3,600 seconds
1 km = 1,000 meters

75 km/hour = (75,000/3,600) m/s =  20-5/6 m/s

The average speed of the bus while it slows down is

                                   (1/2)(20-5/6 + 0) = 10-5/12 m/s .

Traveling at an average of 10-5/12 m/s for 21 seconds,
the bus covers

                         (10-5/12) x (21) =  218.75 meters .

                  
soldi70 [24.7K]2 years ago
4 0

As per the question,the bus slows down uniformly from 75 km/h to 0 km/h.

Hence  the initial velocity of the bus[ u]=75 km/hr.

             the final velocity of the bus [v]=0 km/h

             the taken by the bus to stop [t]=21 s

we know that 1 km= 1000 m and 1 hour =3600 second.

Hence 75 km/h =75*\frac{1000 m}{3600s}

                          =20.83 m/s

we are asked to calculate the stopping distance.

From  equation of kinematics we know that-

                                    v= u +at   where a is the acceleration of the particle.

we have v= 0 km/h  = 0 m/s and u= 75 km/h =20.83 m/s

              t = 21 s

  Putting these values in above equation  we get-

                                 0 = 20.83 +a×21

                                ⇒ a×21 = -20.83 m/s

                                ⇒a=  -[20.83]÷21

                                     = -0.9919 m/s^2    [ Here negative sign indicates that particle is decelerating ]

Again from the equation of kinematics we  know that-

                                 s = ut +\frac{1}{2} at^2

Here s is the distance traveled. Putting the above quantities we get-

                                      s = 20.83 *21 -\frac{1}{2} 0.9919 *[21^2]

                                            s = 218.7 metre.  [ans]

Hence the bus will stop after a distance of 218.7 m.


                                 




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Explanation:

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There are two blocks, one of 400 kg on a horizontal surface and other of 100 kg on top of it tied to a vertical wall by a string. If we try to push the first block, it will not move freely, because two friction forces appear: one exerted by the surface and the other exerted by the contact between both blocks. Let's call them Fr1 and Fr2 respectively. The block 2 is attached to the wall by a string, so it won't simply move with the block 1.  

Please find the free body diagrams in the figure provided below.

The equilibrium condition for the mass 1 is

\displaystyle F_a-F_{r1}-F_{r2}=m.a=0

The mass m1 is being pushed by the force Fa so that slipping with the mass m2 barely occurs, thus the system is not moving, and a=0. Solving for Fa

\displaystyle F_a=F_{r1}+F_{r2}.....[1]

The mass 2 is tried to be pushed to the right by the friction force Fr2 between them, but the string keeps it fixed in position with the tension T. The equation in the horizontal axis is

\displaystyle F_{r2}-T=0

The friction forces are computed by

\displaystyle F_{r2}=\mu \ N_2=\mu\ m_2\ g

\displaystyle F_{r1}=\mu \ N_1=\mu(m_1+m_2)g

Recall N1 is the reaction of the surface on mass m1 which holds a total mass of m1+m2.

Replacing in [1]

\displaystyle F_{a}=\mu \ m_2\ g\ +\mu(m_1+m_2)g

Simplifying

\displaystyle F_{a}=\mu \ g(m_1+2\ m_2)

Plugging in the values

\displaystyle F_{a}=0.25(9.8)[400+2(100)]

\boxed{F_a=1470\ N}

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Answer:

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In the opposite case, when the fish looked at the face of the man, the angle of greater reason why it seems to be further away

Explanation:

This exercise can be analyzed with the law of refraction that establishes that a ray of light when passing from one medium to another with a different index makes it deviate from its path,

      n₁ sin θ₁ = n₂ sin θ₂

where n₁ and n₂ are the refractive indices of the incident and refracted means and the angles are also for these two means.

In this case, the index of seawater replacement is 1.33, the index of refraction of air is 1, which is why the angle of replacement is less than the incident angle, so the fish seems to be closer

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In the opposite case, when the fish looked at the face of the man, the angle of greater reason why it seems to be further away

4 0
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Vinil7 [7]

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The speed vf of the object when a time t has passed is given by:

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The stadium is h=32 m high. A pair of glasses is dropped from the top and reaches the ground at a time:

\displaystyle t_1=\sqrt{\frac{2\cdot 32}{9.8}}=2.56\ sec

The pen is dropped 2 seconds after the glasses. When the glasses hit the ground, the pen has been falling for:

t_2=2.56 - 2 = 0.56\ sec

Therefore, it has traveled down a distance:

\displaystyle y=\frac{9.8\cdot 0.56^2}{2} = 1.54\ m

Thus, the height of the pen is:

h_p = 32 - 1.54\Rightarrow h_p=30.46\ m

8 0
2 years ago
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