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givi [52]
3 years ago
14

The specialty of an athlete on the women's track team is the pole vault. She has a mass of 48 kg and her approach speed is 8.9 m

/s. When she is directly above the bar, her speed is 1.6 m/s. Neglecting air resistance and any energy absorbed by the pole, determine the amount(in m) she has raised herself as she crosses the bar.
Physics
1 answer:
Lyrx [107]3 years ago
4 0

Answer:

H = 3.9 m

Explanation:

mass (m) = 48 kg

initial velocity (initial speed) (U) = 8.9 m/s

final velocity (V) = 1.6 m/s

acceleration due to gravity (g) = 9.8 m/s^{2}

find the height she raised her self to as she crosses the bar (H)

from energy conservation, the change in kinetic energy = change in potential energy

0.5m(V^{2} - [test]U^{2}[/tex]) = mg(H-h)

where h = initial height = 0 since she was on the ground

the equation becomes

0.5m(V^{2} - [test]U^{2}[/tex]) = mgH

0.5 x 48 x (1.6^{2} - [test]8.9^{2}[/tex]) = 48 x 9.8 x H

-1839.6 = 470.4 H  (the negative sign indicates a decrease in kinetic energy so we would not be making use of it further)

H = 3.9 m

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Lemur [1.5K]

Answer:

The magnitude of the acceleration of the tip of the minute hand of the clock 1.675\times10^{-6}\ m/s^2.

Explanation:

Given that,

Length of minute hand = 0.55 m

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T= 3600\ sec

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Using formula of angular frequency

\omega=\dfrac{2\pi}{T}

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\omega=\dfrac{2\pi}{3600}

\omega=0.001745\ rad/s

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a=r\omega^2

Put the value into the formula

a=0.55\times(0.001745)^2

a=1.675\times10^{-6}\ m/s^2

Hence, The magnitude of the acceleration of the tip of the minute hand of the clock 1.675\times10^{-6}\ m/s^2.

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2 years ago
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Mashutka [201]

Answer:

4v/3

Explanation:

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Substitute m_2 = m_1/2 \& v_1 = v/3

m_1v = \frac{m_1v}{3} + \frac{m_1v_2}{2}

Divide both side by m_1, then multiply by 6 we have

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3 years ago
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Which statement correctly describes the current in a circuit that is made up of any two resistors connected in parallel with a b
AysviL [449]
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3 years ago
15 points. give me the method.
AveGali [126]

Answer:

\boxed{{160 \:  m(s)}^{ - 1} }

Explanation:

if \:the \:  frequencies \: are \to \\   f_{1} =  640Hz  \\ and \\f_{2}   = 480Hz \:  \\ but \:  \boxed{v = f \gamma }:   f =  \frac{v}{ \gamma } \\ if \:  \gamma_{1}  -  \gamma _{2}  = 1 =  \gamma  \\ f_{1}  - f_{2}  = 640 - 480 = \boxed{ 160Hz} = f \\ v = f \gamma = 160 \times 1 =  \boxed{{160 \:  m(s)}^{ - 1} }

5 0
2 years ago
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