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liq [111]
2 years ago
11

Se lanza una piedra de 3.00 N verticalmente hacia arriba desde el suelo. Se observa que, cuando está 15.0 m sobre el suelo, viaj

a a 25.0 m/s hacia arriba. Use el teorema trabajo-energía para determinar a) su rapidez en el momento de ser lanzada. b) su altura máxima
Physics
1 answer:
BARSIC [14]2 years ago
6 0

Answer:

(a). The speed at the moment of being thrown is 30.41 m/s.

(b). The maximum height is 47.18 m.

Explanation:

Given that,

Weight of stone = 3.00 N

Height = 15 m

Speed = 25.0 m/s

(a). We need to calculate the speed at the moment of being thrown

Using work energy theorem

W=\dfrac{1}{2}m(v_{2}^2-v_{1}^2)

-mg\times d=\dfrac{1}{2}m(v_{2}^2-v_{1}^2)

Put the value into the formula

-9.8\times15=\dfrac{1}{2}\times(v_{2}^2-v_{1}^2)

-2\times9.8\times15=25^2-v_{1}^2

-v_{1}^2=-300-25^2

v_{1}=\sqrt{925}

v_{1}=30.41\ m/s

(b). We need to calculate the maximum height

Using work energy theorem

[tex]W=\dfrac{1}{2}mv_{2}^2-\dfrac{1}{2}mv_{1}^2

mg\times d=\dfrac{1}{2}mv_{2}^2-\dfrac{1}{2}mv_{1}^2

Here, \dfrac{1}{2}mv_{2}^2=0

-(mg)\times d=\dfrac{1}{2}mv_{1}^2

d=\dfrac{v_{1}^2}{2g}

Put the value into the formula

d=\dfrac{30.41^2}{2\times9.8}

d=47.18\ m

Hence, (a). The speed at the moment of being thrown is 30.41 m/s.

(b). The maximum height is 47.18 m.

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This question is incomplete, the complete question is;

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- Consider a cylindrical segment of a blood vessel 2.70 cm long and 3.10 mm in diameter. What additional outward force would such a vessel need to withstand in the person's feet compared to a similar vessel in her head

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