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RoseWind [281]
3 years ago
12

Suppose that on earth you can throw a ball vertically upward a distance of 1.20 m. Given that the acceleration of gravity on the

Moon is 1.67 m/s2, how high could you throw a ball on the Moon? (Take the y-axis in the vertical direction, and assume that the location of your hand is at y = 0.)
Physics
1 answer:
tatuchka [14]3 years ago
6 0

Answer:

7.04 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity = 0

s = Displacement on Earth = 1.2 m

a = Acceleration due to gravity on Moon = 1.67 m/s²

a = Acceleration due to gravity Earth= 9.81 m/s²

Accelration going up is considered as negetive

Initial Velocity of the ball

v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow -u^2=2\times -9.81\times 1.2-0^2\\\Rightarrow u=\sqrt{2\times 9.81\times 1.2}\\\Rightarrow u=4.85\ m/s

Assuming that the ball is thrown with the same velocity on the Moon, displacement of the ball is

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-4.85^2}{2\times -1.67}\\\Rightarrow s=7.04\ m

The displacement of the ball on the moon is 7.04 m

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Answer:

The stress is  \sigma  =  1.218*10^{6} \  N/m^2

Explanation:

From the question we are told that

   The diameter of the post is  d =  29 \ cm  =  0.29 \  m

   The length is L  =  2.0 \  m

    The weight of the loading mass

Generally  the  radius of the post is mathematically represented as

     r =  \frac{0.29}{2}

=>   r = 0.145 \  m

Generally the area of the post is  

       A =  \pi r^2

=>     A =  3.14 *  0.145 ^2

=>     A =  0.066 \ m^2

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=>      F =  8200  *  9.8

=>      F =  80360 \  N

Generally the stress is mathematically represented as

         \sigma  =  \frac{F}{A}

=>      \sigma  =  \frac{80360 }{0.066}

=>      \sigma  =  1.218*10^{6} \  N/m^2

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