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morpeh [17]
3 years ago
9

You are walking back toward the back of a bus that is moving forward with a constant velocity. Describe your motion relative to

the bus and relative to a point on the ground.
Physics
1 answer:
HACTEHA [7]3 years ago
7 0

Answer: Relative to the bus, you are stationary. Relative to a point on the ground, you are walking forward with velocity  equal to the velocity of the bus.

As a person walks toward the back of a bus, they are walking towards a stationary object according to their own perspective. However, from an outside perspective, the bus is moving forward with a constant velocity and so relative to that point in motion, you are actually walking backwards with respect to it. This difference in motion means you would not be able have the same experience as the person on the bus when getting off because when you get off you would stop while they continue going forward. The last sentence could be considered a footnote because it only applies if this was an actual situation rather than just an analogy question in a physics test.

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Black_prince [1.1K]
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Which energy source is one that is used to boil water to make steam in power stations
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Answer:

heat energy is used in boiling water and to make steam at power stations

Explanation:

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A celebrated Mark Twain story has motivated contestants in the Calaveras County Jumping Frog Jubilee, where frog jumps as long a
IrinaK [193]

The frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.

To find the answer, we need to know about the time of flight and range of projectile motion.

<h3>What's the expression of range of a projectile motion?</h3>
  • Range = U²× sin(2θ)/g
  • U= initial velocity, θ= angle of projectile and g= acceleration due to gravity
  • U=√{Range×g/sin(2θ)}
  • Here, range= 2.20m, = 36.5°
  • U= √{2.20×9.8/sin(73)}

U= √{2.20×9.8/sin(73)} = 22.5m/s

<h3>What's the expression of time of flight in projectile motion?</h3>
  • Time of flight= (2×U×sinθ)/g
  • So, T= (2×22.5×sin36.5°)/9.8

= 2.73 s

Thus, we can conclude that the frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.

Learn more about the range and time period of projectile motion here:

brainly.com/question/24136952

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2 years ago
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In one experiment, the students allow the block to oscillate after stretching the spring a distance A. If the potential energy s
atroni [7]

Answer:

    K = m g (A - A2)

Explanation:

In a block spring system the total energy is the sum of the potential energy plus the kinetic energy, for maximum elongation all the energy is potential

         Em = U₀ = m g A

For when the system is at an ele

Elongation A2 less than A, energy has two parts

        Em = K + U₂

       K = Em –U₂

We substitute

     K = m g A - m gA2

    K = m g (A - A2)

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3 years ago
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