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Anon25 [30]
3 years ago
6

A city bus travels six blocks east and eight blocks north east wok is 100 m long is a bus travels as distance and 15 minutes wha

t is the average speed on the bus
Physics
1 answer:
mariarad [96]3 years ago
7 0

Answer:

the average speed on the bus =  1.55 m/s

Explanation:

From the information we are being provided with ;

a city block travels 6 blocks due east

and 8 blocks due north

the length of the individual blocks = 100m

As such their respective distance will be

distance (d) traveled due east direction by bus = 6 × 100 = 600 m

distance (d) traveled towards the north direction by bus = 8 × 100 = 800 m

The total distance covered  = (600+ 800)m

= 1400 m

However, it requires 15 minutes to travel this distance by bus.

Thus,  

the average speed = distance/ time

the average speed on the bus = 1400 m / (15 × 60) s

the average speed on the bus =  1400 m / 900 s

the average speed on the bus =  1.55 m/s

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A spiral spring has a length of 14 cm when a force of 4 N is hung on it. A force of 6 N extends
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2 years ago
A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The s
aleksklad [387]

Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is U_{s} = 0

And, initial potential gravitational potential energy of the rod is U_{g} = \frac{mgL}{2}.

It is given that,

       mass of the bar = 0.795 kg

            g = 9.8 m/s^{2}

           L = length of the rod = 0.2 m

Initial total energy T = \frac{mgL}{2}

Now, when the rod is in horizontal position then final total energy will be as follows.

            T = \frac{1}{2}kx^{2} + I \omega^{2}

where,    I = moment of inertia of the rod about the end = \frac{mL^{2}}{3}

Also,    \omega = \frac{\nu}{L}

where,    \nu = speed of the tip of the rod

              x = spring extension

The initial unstrained length is x_{o} = 0.1 m

Therefore, final length will be calculated as follows.

              x' = \sqrt{(0.2)^{2} + (0.1)^{2}} m

Then,  x = x' - x_{o}

          x = \sqrt{(0.2)^{2} + (0.1)^{2}} m - 0.1 m

             = 0.1236 m

       k = 25 N/m

So, according to the law of conservation of energy

       \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}

      \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}

Putting the given values into the above formula as follows.

   \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}

  \frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}

          v = 2.079 m/s

Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

7 0
3 years ago
Two nuclei join to form a larger nucleus during which process
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