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damaskus [11]
3 years ago
12

A 5​-gram ​[g] rubber ball is released from a height of 8 meters​ [m] above a flat surface on the moon. Gravitational accelerati

on on the moon is 1.62 meters per second squared ​[m/s2​]. Assume that no energy is lost from frictional drag. What is the​ velocity, in units of meters per second​ [m/s], of the rubber ball the instant before it strikes the flat​ surface?

Physics
2 answers:
Neko [114]3 years ago
6 0

Answer:

v = 5.0911 m/s

Explanation:

See it in the pic.

melamori03 [73]3 years ago
4 0

Answer:

The​ velocity is 5.09 m/s

Explanation:

the solution is in the attached Word file

Download docx
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Answer:

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solution

we know function

T(x, y) = 54 /( 7 + x² + y²)

so derivative it x and y direction i.e

dt/dx = -54× 2x / (7 +x² + y²)²    .........................1

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now put the value point (1,3) as x = 1 and y = 3 in equation 1 and 2

dt/dx = -54× 2(1) / (7 +(1)² + (3)²)²  

dt/dx = -0.373702

and

dt/dy =  -54× 2(3) / (7 + (1)² + (3)²)²

dt/dy =  -1.121107

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Explanation:

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