What is the number of significant digits in the measurement 34.000 m?

15. A volleyball player who weighs 650 Newtons jumps 0.500 meters vertically off the floor. What is her kinetic energy just befo

allochka39001 [22]

<h2>Kinetic energy just before hitting the floor is 324.57 J</h2>

Explanation:

Weight of volleyball player = 650 N

That is

Mass x Acceleration due to gravity = 650

Mass x 9.81 = 650

Mass = 66.26 kg

We also have equation of motion v² = u² + 2as

Initial velocity, u = 0 m/s

Acceleration, a = 9.81 m/s²

Final velocity, v = ?

Displacement, s = 0.5 m

Substituting

v² = u² + 2as

v² = 0² + 2 x 9.81 x 0.5

v = 3.13 m/s

Velocity with which he lands on ground is 3.13 m/s

We have kinetic energy = 0.5 x Mass x Velocity²

Substituting

Kinetic energy = 0.5 x 66.26 x 3.13²

Kinetic energy = 324.57 J

Kinetic energy just before hitting the floor is 324.57 J

3 0

3 years ago

What effect will a turning point have on an individuals life

ankoles [38]

Answer:

The turning points are those instants, moments or situations that happen in an absolutely unexpected way, as a result of which your life changes ... and nothing is the same as before.

4 0

3 years ago

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A string with a mass density of 3 * 10^-3 kg/m is under a tension of 380 N and is fixed at both ends. One of its resonance frequ

Delvig [45]

Answer:

(a) the fundamental frequency of this string is 65 Hz

(b) the harmonics of the given frequencies are third and fourth respectively.

(c) the length of the string is 2.74 m

Explanation:

Given;

mass density of the string, μ = 3 x 10⁻³ kg/m

tension of the string, T = 380 N

resonating frequencies, 195 Hz and 260 N

For the given resonant frequencies;

$195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } ---(1)\\\\260 = \frac{n+1}{2l} \sqrt{\frac{T}{\mu} } ---(2)\\\\divide \ (2) \ by (1)\\\\\frac{260}{195} = \frac{n+1 }{n} \\\\260n = 195(n+1)\\\\260 n = 195 n + 195\\\\260n - 195n = 195\\\\65n = 195\\\\n = \frac{195}{65} \\\\n = 3$

(c) From any of the equations, solve for Length of the string (L);

$195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } \\\\195 = \frac{3}{2l}\sqrt{\frac{380}{3\times 10^{-3}} } \\\\l = \frac{3}{2\times 195}\sqrt{\frac{380}{3\times 10^{-3}} }\\\\l = 2.74 \ m$