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d1i1m1o1n [39]
3 years ago
8

While running these tests, crall notices a similarity in the velocity measured at the ground independent of whether the tennis b

all is thrown upward or downward. she decides to see if her hypothesis that the final speed is independent of the sign of the initial velocity is correct. crall and whipple decide to run the experiment several times using different initial speeds of the elevator each time comparing the final velocity of the tennis ball when the elevator is traveling upward vs. downward at the same speed?
Physics
2 answers:
lara [203]3 years ago
6 0

At the ground the ball will always have velocity along the direction of gravity. If upward motion is taken positive it will always have negative velocity at the ground because, if the ball was given an initial upward velocity then gravity will decelerate it and bring it down with a negative final velocity. If the ball is given an initial downward velocity then the ball will be further accelerated by gravity in the downward direction only, again maintaining negative direction. The magnitude however in both cases will be different. the final velocity at the ground will have higher magnitude in case of elevator moving downwards.

lora16 [44]3 years ago
6 0

Answer:

The velocities will be different

Explanation:

Thinking process:

The downward motion is more or less the same. This is because the gravitational pull on the object. In a vacuum, the terminal speed of an object is the same throughout.

Throwing the ball downward simply converts the potential energy to kinetic energy. However, throwing the ball up needs more energy to overcome the gravitational pull on the ball. Hence the velocity of the ball will be different.

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Consider a collection of charges in a given region and suppose all other charges are distant and have a negligible effect. Furth
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Answer:

E. Some charges in the region are positive, and some are negative.

Explanation:

Electric potential is given as;

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3 0
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A 220 kg crate hangs from the end of a rope of length L = 14.0 m. You push horizontally on the crate with a varying force F to m
kifflom [539]

<u>Answer</u>:

(a) magnitude of F = 797 N

(b)the total work done  W = 0

(c)work done by the gravitational force =  -1.55 kJ

(d)the work done by the pull  = 0

(e) work your force F does on the crate = 1.55 kJ

<u>Explanation</u>:

<u>Given</u>:

Mass of the crate, m =  220 kg

Length of the rope, L = 14.0m

Distance, d =  4.00m

<u>(a) What is the magnitude of F when the crate is in this final position</u>

Let us first determine vertical angle as follows

=>Sin \theta = \frac{d }{L}

=> \theta = Sin^{-1} \frac{d}{L} =

Now substituting thje values

=> \theta = Sin^{-1} \frac{4}{12} =

=> \theta = Sin^{-1} \frac{1}{3}

=> \theta = Sin^{-1}(0.333)

=> \theta = 19.5^{\circ}

Now the tension in the string resolve into components

The vertical component supports the weight

=>Tcos\theta = mg

=>T = \frac{mg}{cos\theta}

=>T = \frac{230 \times 9.8 }{cos(19.5)}

=>T = \frac{2254 }{cos(19.5)}

=>T = \frac{2254 }{0.9426}

=>T =2391N

Therefore the horizontal force

F = TSin(19.5)

F = 797 N

b) The total work done on it

As there is no change in Kinetic energy

The total work done W = 0

<u>c) The work done by the gravitational force on the crate</u>

The work done by gravity

Wg = Fs.d = - mgh

Wg = - mgL ( 1 - Cosθ )

Substituting the values                                                            

= -230 \times 9.8\times 12 ( 1 - cos(19.5) )

= -230 \times 9.8\times 12 ( 1 - 0.9426) )

= -230 \times 9.8\times 12 (0.0574)

= -230 \times 9.8\times 0.6888

=  -230 \times 6.750

= -1552.55 J

The work done by gravity = -1.55 kJ

<u>d) the work done by the pull on the crate from the rope</u>

Since the pull  is perpendicular to the direction of motion,

The work done = 0

e)Find the work your force F does on the crate.

Work done by the Force on the crate

WF = - Wg  

WF = -(-1.55)

WF = 1.55 kJ

<u>(f) Why is the work of your force not equal to the product of the horizontal displacement and the answer to (a)</u>

Here the work done by force is not equal to F*d  

and it is equal to product of the cos angle and F*d

So, it is not equal to the product of the horizontal displacement and the answer to (a)      

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