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4 years ago

Energy

Physics

2 answers:

Ahat [919]4 years ago

7 0

I think the answer may be D- mechanical

Setler79 [48]4 years ago

3 0

The answer is OD) Mechanical

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What is newtons first law of motion

Brums [2.3K]

An object in motion will remain in motion unless an outside force stops is.

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3 years ago

A teacher applles a force to a wall and becomes axhausted.<br> Explanation:

Finger [1]

Because they are using all of there energy to apply the force

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3 years ago

3. If I run 150m in 30 seconds, what speed will I have been running at?

Radda [10]

Answer:

speed = distance/time

Explanation:

speed = 150/30

speed =5m/s

you were running fast .....5m/s is a good speed

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3 years ago

Una fuerza F de 200 lb actúa a lo largo de AB, sobre la rampa mostrada. fuerza de F respecto del eje OC. Calcule el momento de f

ruslelena [56]

Answer:

Moc = -613.25 [lb*in]

Explanation:

Este problema se puede resolver mediante la mecánica vectorial, es decir se realizara un analisis de vectores.

Primero se calculara el momento de la fuerza F_AB con respecto al punto O, debemos recordar que el momento con respecto a un punto se define como el producto cruz de la distancia por la fuerza.

$M_{o}=r_{A/O} * F_{AB}$ (producto cruz)

Necesitamos identificar los puntos:

O (0,0,0) [in]

A (12,0,0) [in]

B (0, 24,8) [in]

C (12,24,0) [in]

$r_{A/O}=(12,0,0) - (0,0,0)\\r_{A/O} = 12 i + 0j+0k [in]\\AB = (0,24,8) - (12,0,0)\\AB = -12i+24j+8k [in]\\[LAB]=\frac{-12i+24j+8k}{\sqrt{(12)^{2} +(24)^{2} +(8)^{2} } }\\ LAB=-\frac{3}{7} i+\frac{6}{7}j+\frac{2}{7}k$

El ultimo vector calculado corresponde al vector unitario (magnitud = 1) de AB. El vector fuerza corresponderá al producto del vector unitario por la magnitud de la fuerza = 200 [lb].

$F_{AB}=-\frac{600}{7} i +\frac{1200}{7}j+\frac{400}{7} k [Lb]$

De esta manera realizando el producto cruz tenemos

$M_{O}=r_{A/O} * F_{AB}$

$M_{O}=0i-685.7j+2057.1k [Lb*in]$

Para calcular el momento con respecto a la diagonal OC, necesitamos el vector unitario de esta diagonal.

$OC = (12,24,0)-(0,0,0)\\OC= 12i+24j+0k[Lb]\\LOC = \frac{12i+24j+0k}{\sqrt{(12)^{2} +(24)^{2} +(0)^{2} } } \\LOC=\frac{12}{\sqrt{720}}i+\frac{24}{\sqrt{720}}j +0k$

El vector con respecto al eje OC, es igual al producto punto del momento en el punto O por el vector unitario LOC

$M_{OC}=L_{OC}*M_{O}\\M_{OC}=(\frac{12}{\sqrt{720}}i +\frac{24}{\sqrt{720}} j+0k )* (0i-685.7j+2057.1k)\\M_{OC}= -613.32[Lb*in]$

7 0

3 years ago

The Achilles tendon connects the muscles in your calf to the back of your foot. When you are sprinting, your Achilles tendon alt

lesantik [10]

Answer:

(a) $\triangle l=5 mm$

(b) 0.033

Explanation:

(a)

Force F=mg where m is mass and g is acceleration due to gravity whose value is taken as $9.81 m/s^{2}$

However, for this case, the maximum force is 8 times the weight of runner hence F=8mg

Assume Young's modulus for tendon is $0.15*10^{10} N/m^{2}$

Young's modulus is given by

$E=\frac {Fl}{A\triangle l}$ and $\triangle l=\frac {Fl}{EA}$ and substituting F with 8mg we obtain $\triangle l=\frac {8mgl}{EA}$

Where E is young's modulus, l is stretched length and \triangle l is change in length

Substituting m as 70 kg, g as $9.81 m/s^{2}$ , l as 15cm=0.15 m, E as $0.15*10^{10} N/m^{2}$ and A as $110 m^{2}=0.000110 m^{2}$

$\triangle l=\frac {8*70 Kg*9.81 m/s^{2}*0.15m}{0.15*10^{10} N/m^{2} *0.00011 m^{2}}=0.004994182 m$

$\triangle l=5 mm$

(b)

Strain, $\epsilon=\frac {\triangle l}{l}$ and the fraction of tendon’s length is the ratio of change in length to the stretched length

The fraction of tendon, f is given by

$f=\frac {\triangle l}{l}$ . Substituting $\triangle l$ with 0.005m and l with 0.15m we obtain

$\epsilon=f=\frac {0.005}{0.15}=\frac {1}{30}=0.033$

Therefore, fraction of the tendon’s length is 0.033

5 0

4 years ago