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2 years ago

What does personal mission statement?

Physics

1 answer:

KonstantinChe [14]2 years ago

3 0

Answer:

<em>A personal mission statement defines who you are as a person (or as a team member where you work) and identifies your purpose, whether that's in the office or simply in life. </em>

Explanation:

<em>It explains how you aim to pursue that purpose, and why it matters so much to you.</em>

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What will a spring scale read for the weight of a 58.0-kg woman in an elevator that moves in free fall?

solniwko [45]

Answer:

0 N

Explanation:

The elevator is under free fall. So, when a body is under free fall, the acceleration is only due to gravity. So, the acceleration of the elevator or the woman inside it, is acceleration due to gravity in the downward direction.

The spring scale gives the value of the normal force acting on the woman and doesn't give the exact weight of the woman. Under normal conditions, when the spring scale is at rest, then the upward normal force equals the weight and hence weight of a body is equal to the normal force acting on the body.

But, here, the body is not at rest. Weight\tex](mg)[/tex] acts in the downward direction and normal force $(N)$ acts in the upward direction. The woman is moving down with acceleration equal to acceleration due to gravity $(g)$

So, we apply Newton's second law on the woman.

$\textrm{Net force} = \textrm{mass}\times \textrm{acceleration}\\F_{net}=ma$

Net force is equal to the difference of the downward force and upward force.

$F_{net}=mg-N$

Now, $F_{net}=ma$

$mg-N=mg\\N=mg-mg=0$

Therefore, the reading on the spring scale is 0 N.

5 0

3 years ago

What distance will be traveled if you are going at an average speed of 420 km/hr for 45 minutes?

goldenfox [79]

Answer: 12

Explanation: 34

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3 years ago

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

Margarita [4]

Answer:

The final size is approximately equal to the initial size due to a very small relative increase of $1.055\times 10^{- 7}$ in its size

Solution:

As per the question:

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

Distance covered by photon, d = 1 km = 1000 m

Mass of proton, $m_{p} = 1.67\times 10^{- 27} kg$

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

Now,

This is relativistic in nature

The rest mass energy associated with the proton is given by:

$E = m_{p}c^{2}$

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

This energy of proton is $\simeq 250 GeV$

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

Now, the time taken to cover 1 km = 1000 m of the distance:

T = $\frac{1000}{v}$

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

Now, in accordance to the dispersion factor;

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

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