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swat32
1 month ago
10

Complete the paragraph to describe the relationship between kinetic energy and braking distance. Use . A car moves at a speed of

50 kilometers/hour. Its kinetic energy is 400 joules. If the same car moves at a speed of 100 kilometers/hour, then its kinetic energy will be joules. The braking distance at the faster speed is the braking distance at the slower speed.
Physics
2 answers:
Softa [3K]1 month ago
7 0

ke prop to v^2

ke1/v1^2=ke2/v2^2

400/50x50=joules/100x100

400x2x2

1600j

Ostrovityanka [3.2K]1 month ago
3 0

Response: A vehicle is traveling at a speed of 50 kilometers/hour, possessing a kinetic energy of 400 joules. If this same vehicle travels at 100 kilometers/hour, its kinetic energy would be 1,600 joules. The braking distance at the higher speed doubles the braking distance at the lower speed.

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A 250 GeV beam of protons is fired over a distance of 1 km. If the initial size of the wave packet is 1 mm, find its final size
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Answer:

The final size is nearly the same as the initial size because the increase in size1.055\times 10^{- 7} is remarkably small

Solution:

According to the problem:

The proton beam energy is E = 250 GeV =250\times 10^{9}\times 1.6\times 10^{- 19} = 4\times 10^{- 8} J

Distance traveled by the photon, d = 1 km = 1000 m

Proton mass, m_{p} = 1.67\times 10^{- 27} kg

Initial size of the wave packet, \Delta t_{o} = 1 mm = 1\times 10^{- 3} m

Now,

This operates under relativistic principles

The rest mass energy for the proton is expressed as:

E = m_{p}c^{2}

E = 1.67\times 10^{- 27}\times (3\times 10^{8})^{2} = 1.503\times 10^{- 10} J

This proton energy is \simeq 250 GeV

Thus, the speed of the proton, v\simeq c

The time to cover 1 km = 1000 m of distance is calculated as:

T = \frac{1000}{v}

T = \frac{1000}{c} = \frac{1000}{3\times 10^{8}} = 3.34\times 10^{- 6} s

According to the dispersion factor;

\frac{\delta t_{o}}{\Delta t_{o}} = \frac{ht_{o}}{2\pi m_{p}\Delta t_{o}^{2}}

\frac{\delta t_{o}}{\Delta t_{o}} = \frac{6.626\times 10^{- 34}\times 3.34\times 10^{- 6}}{2\pi 1.67\times 10^{- 27}\times (10^{- 3})^{2} = 1.055\times 10^{- 7}

Thus, the widening of the wave packet is relatively minor.

Hence, we can conclude that:

\Delta t_{o} = \Delta t

where

\Delta t = final width

3 0
2 months ago
You apply the brakes of your car abruptly and your book starts sliding off the front seat. Three observers sitting in the car ex
Softa [3030]

Answer:

All observers are accurate.

Explanation:

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From their distinct frames of reference, each observer's perspective is valid.

Observer A is in an inertial reference frame.

Observers capable of explaining the book's behavior and its relationship to the car through the interplay of forces and changes in velocity are classified as being in inertial reference frames.

Observer A's observations illustrate this, for she pointed out the relative motion between the book and the car, indicating her position in an inertial reference frame.

Likewise, observers in these inertial reference frames can elucidate object velocity changes based on the forces affecting them from other objects.

This is exemplified by observer B, who notes the car's force impacting the book's velocity.

Observer C occupies a non-inertial reference frame, as Newton's laws of motion do not apply. This scenario arises within non-inertial frames.

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2 months ago
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