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Aleonysh [2.5K]
3 years ago
12

romeo threw a small bundle toward Juliet on a balcony 10.0 m above him with a velocity of 5.0 m/s. The bundle stopped rising aft

er 1.5 seconds. How high did the bundle travel? Was that high enough for her to catch it?
Physics
1 answer:
choli [55]3 years ago
4 0

I can tell that you've never read much Shakespeare.

If the bundle is pitched with a velocity of 5.0 m/s, it must follow
as the night follows the day that it cannot then continue to rise
for more than barely 1/2 second before it reacheth the zenith
of its arc, reverseth its course, and plungeth Earthward to its
ultimate demise.

But hey !  To thine own self be true, eh ?

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A vw beetle goes from 0 to 60 mi/h with an acceleration of 2.35 m/s^2.
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Part a.
u = 0, the initial velocity
v = 60 mi/h, the final velocity
a = 2.35 m/s², the acceleration.

Note that
1 m = 1609.34 m.
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Use the formula
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(26.822 m/s) = (2.35 m/s²)*(t s)
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Answer: 11.4 s

Part b.
We already determined that v = 60 mi/h = 26.822 m/s.
t = 0.6 s
Therefore
(26.822 m/s) = (a m/s²)*(0.6 s)
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Answer:  44.7 m/s²
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111,000 Pa

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A bicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of 3.6 m/s . Two seconds later t
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Answer:

A) t = 7.0 s    

B) x = 25 m  

C) v = 10 m/s

Explanation:

The equations for the position and velocity of an object traveling in a straight line is given by the following expressions:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

A)When both friends meet, their position is the same:

x bicyclist = x friend

x0 + v0 · t + 1/2 · a · t² = x0 + v · t

If we place the center of the frame of reference at the point when the bicyclist starts following his friend, the initial position of the bicyclist will be 0, and the initial position of the friend will be his position after 2 s:

Position of the friend after 2 s:

x = v · t

x = 3.6 m/s · 2 s = 7.2 m

Then:

1/2 · a · t² = x0 + v · t       v0 of the bicyclist is 0 because he starts from rest.

1/2 · 2.0 m/s² · t² = 7.2 m + 3.6 m/s · t

1  m/s² · t² - 3.6 m/s · t - 7.2 m = 0

Solving the quadratic equation:

t = 5.0 s

It takes the bicyclist (5.0 s + 2.0 s) 7.0 s to catch his friend after he passes him.

B) Using the equation for the position, we can calculate the traveled distance. We can use the equation for the position of the friend, who traveled over 7.0 s.

x = v · t

x = 3.6 m/s · 7.0 s = 25 m

(we would have obtained the same result if we would have used the equation for the position of the bicyclist)

C) Using the equation of velocity:

v = a · t

v = 2.0 m/s² · 5.0 s = 10 m/s

8 0
3 years ago
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