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Ne4ueva [31]
3 years ago
14

A 75.0-kg man steps off a platform 3.10 m above the ground. he keeps his legs straight as he falls, but his knees begin to bend

at the moment his feet touch the ground; treated as a particle, he moves an additional 0.60 m before coming to rest. (a) what is his speed at the instant his feet touch the ground? (b) if we treat the man as a particle, what is his acceleration (magnitude and direction) as he slows down, if the acceleration is assumed to be constant?

Physics
1 answer:
sasho [114]3 years ago
8 0
Refer to the diagram shown below.

u = 0, the initial vertical velocity
Assume g = 9.8 m/s² and ignore air resistance.

At the first stage of landing on the ground, the distance traveled is
h = 3.1 - 0.6 = 2.5 m.
If v =  the vertical velocity at this stage, then
v² = u² + 2gh
v² = 2*(9.8 m/s²)*(2.5 m) = 49 (m/s)²
v = 7 m/s

At the second stage of landing on the ground, let a =  the acceleration (actually deceleration) that his body provides to come to rest.
The distance traveled is 0.6 m.
Therefore
0 = (7 m/s)² + 2(a m/s²)*(0.6 m)
a = - 49/1.2 = - 40.833 m/s²

Answers:
(a) The velocity when the man first touches the ground is 7.0 m/s.
(b) The acceleration is -40.83 m/s² (deceleration of 40.83 m/s²) to come to rest within 0.6 m.

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harina [27]

Answer:

See step by step sexplanation

Explanation:

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P₁ * V₁  = P₂ * V₂

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3.-No se puede estar de acuerdo con el criterio del plomero. En su solución no plantea el aumento de la altura del tanque,  para el logro del aumento de la presión que es realmente lo que hay que hacer

4 0
3 years ago
What formula is used to calculate resistance using the color code?
jolli1 [7]
<span><em>The answer is </em><em>A</em><em> :</em><em>" R = (First digit * 10 + second digit) * multiplier. "

</em>
<em>yw peasant XD</em><em>
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5 0
3 years ago
In a series RLC resonance circuit, the resonance frequency f0 = 700 kHz. The resistor R = 10 Ohm. The specified bandwidth (BW) s
sladkih [1.3K]

Answer:

  • quality factor (Q) = 69.99
  • inductor = 1.591 x 10⁻⁴ H
  • capacitor = 3.248 x 10⁻¹⁰ F

Explanation:

Given;

resonance frequency (F₀) = 700 kHz

resistor, R =  10 Ohm

bandwidth (BW) = 10 kHz

bandwidth (BW)  = \frac{R}{2\pi L}

BW = \frac{R}{2\pi L}

make L (inductor) the subject of the formula

L = \frac{R}{2\pi *BW}  =  \frac{10}{2\pi *10,000} =1.591 *10^{-4} \ H = \ 0.1591\ mH

F_o =\frac{1}{2\pi\sqrt{LC} } \\\\\sqrt{LC} = \frac{1}{2\pi F_o} \\\\LC = \frac{1}{4\pi ^2F_o^2}= \frac{1}{4\pi ^2(700,000)^2} = 5.168*10^{-14}

make C (capacitor)  the subject of the formula

C = \frac{5.168*10^{-14}}{1.591*10^{-4}} = 3.248*10^{-10} \ F = \ 3.248*10^{-4} \ \mu F

quality factor (Q) = \frac{1}{R} \sqrt{\frac{L}{C}} \ = \frac{1}{10} \sqrt{\frac{1.591*10^{-4}}{3.248*10^{-10}}}=69.99

quality factor (Q) =  69.99

5 0
3 years ago
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