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Eduardwww [97]
4 years ago
12

A ball is thrown straight up with enough speed so that it is in the air for several seconds. Assume the positive direction is up

wards. Part APart complete What is the velocity of the ball when it reaches its highest point? Express your answer to two significant figures and include the appropriate units. v = 0 ms Previous Answers Correct Part B What is its velocity 0.70 s before it reaches its highest point?
Physics
1 answer:
Andrew [12]4 years ago
6 0

Answer:

a) v= 0 m/s b) v= 6.86 m/s

Explanation:

a) When the ball reaches to its highest point, under the influence of gravity, before starting to fall down, it momentarily comes to an stop (this is needed prior to change direction in any movement), so, applying the definition of acceleration, and replacing the acceleration a by g, we have:

vf = v₀ - g*t (1)

The minus sign means that the acceleration due to gravity is always downward, so if we assume that the positive direction is upwards it must be negative.

At the highest point, vf= 0.

b) Prior to solve this point, we need to know which is the time when the ball reaches to its highest point.

As we know vf=0, we can solve (1) for t, as follows:

th = v₀/g

Now, for a time that is 0.7 s before this time, applying the acceleration definition and solving for v again, we have:

v = v₀ -(g *(th-0.7 s)), but th= v₀/g, so we get:

v= v₀ -g((v₀/g)-0.7 s) = v₀ - v₀ + g*0.7 s

⇒ v=g*0.7 s = 9.8 m/s²*0.7 s

⇒ v = 6.86 m/s

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A 2.2-m-long steel rod must not stretch more than 1.2 mm when it is subjected to a 8.5-kN tension force. Knowing that E = 200 GP
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Answer:

a) 0.00996 m

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Explanation:

Unit conversions:

E = 200GPa = 200\times10^9 Pa

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If the 2.2m rod cannot stretch more than 0.0012 m, its maximum strain is

\epsilon = \frac{\Delta L}{L} = \frac{0.0012}{2.2} = 0.000545455

With elastic modulus being E = 200 GPa, then its maximum stress must be

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Knowing the tension force being F = 8500 N, we can calculate the appropriate cross section area

A = \frac{F}{\sigma} = \frac{8500}{109090909} = 7.79\times10^{-5}m^2

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A = \pi d^2/4

7.79\times10^{-5} = \pi d^2/4

d^2 = \frac{4*7.79\times10^{-5}}{\pi} = 9.92\times10^{-5}

d = \sqrt{9.92\times10^{-5}} = 0.00996 m \approx 1 cm

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The Nardo ring is a circular test track for cars. It has a circumference of 12.5 km. Cars travel around the track at a constant
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Given

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Answer:

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