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notsponge [240]
4 years ago
8

An object is thrown directly up (positive direction) with a velocity (vo) of 20.0 m/s and do= 0. How high does it rise (v = 0 cm

/s at top of rise). Remember, acceleration is -9.80 m/s2.d = _________ mA. 1.02B. 2.04C. 20.4D. 40.8
Physics
1 answer:
stepan [7]4 years ago
7 0

The object rises to a height of 20.4 m. So option C is correct.

Explanation:

initial velocity= Vi=20 m/s

final velocity at the top= Vf=0

acceleration= g=-9.8 m/s²

Vf²=Vi²+2gh

0= (20)²+2 (-9.8)h

-200=-9.8h

h=200/9.8

h=20.4 m

Thus the object rises to a height of 20.4 m

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Answer:

velocity = 0.3m/s

speed = 1.21 m/s

Explanation:

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t = 27 + 39 = 66 seconds

The net displacement from the front door to the bench is the distance from the front door to the windmill subtracted by the distance from the windmill to the bench

s = 50 - 30 = 20 m

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The total distance from the front door to the bench is the sum of distance from the front door to the windmill and the distance from the windmill to the bench

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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
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:

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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,

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

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

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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}

Thus the increase in wave packet's width is relatively quite small.

Hence, we can say that:

\Delta t_{o} = \Delta t

where

\Delta t = final width

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