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3 years ago

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approximately what force,FM must the extensor muscle in the upper arm exert on the lower arm to hold a 7.6kg shot put? assume th

e lower arm has a mass of 2.8kg and it's CG is 11.6cm froom the pivot point.

1 answer:

MAXImum [283]3 years ago

7 0

Lets do the sum of the forces about the elbow joint.

Fm = Force of Muscle; Fe = Force Elbow; Fb = Force Ball

Sum Force about Joint = (-2.5)Fm + 12.5Fe + 30Fb = 0

(-2.5)Fm + 12.5(2.8) + 30(6.9) = 0

Fm = 96.8kg

Fm = 96.8 * 9.8 = 948.6N

Do you understand why the -2.5 is negative?
<span> Because I put the origin at the joint. So when you go left it is negative and when you go right it is positive. </span>

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What is the original source of energy for gasoline?

Yuki888 [10]

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3 years ago

Read 2 more answers

Continuous sinusoidal perturbation Assume that the string is at rest and perfectly horizontal again, and we will restart the clo

Elena-2011 [213]

a) $3.14 \cdot 10^{-4} s$

b) See plot attached

c) 10.0 m

d) 0.500 cm

Explanation:

a)

The position of the tip of the lever at time t is described by the equation:

$y(t)=(0.500 cm) sin[(2.00\cdot 10^4 s^{-1})t]$ (1)

The generic equation that describes a wave is

$y(t)=A sin (\frac{2\pi}{T} t)$ (2)

where

A is the amplitude of the wave

T is the period of the wave

t is the time

By comparing (1) and (2), we see that for the wave in this problem we have

$\frac{2\pi}{T}=2.00\cdot 10^4 s^{-1}$

Therefore, the period is

$T=\frac{2\pi}{2.00\cdot 10^4}=3.14 \cdot 10^{-4} s$

b)

The sketch of the profile of the wave until t = 4T is shown in attachment.

A wave is described by a sinusoidal function: in this problem, the wave is described by a sine, therefore at t = 0 the displacement is zero, y = 0.

The wave than periodically repeats itself every period. In this sketch, we draw the wave over 4 periods, so until t = 4T.

The maximum displacement of the wave is given by the value of y when $sin(...)=1$ , and from eq(1), we see that this is equal to

y = 0.500 cm

So, this is the maximum displacement represented in the sketch.

c)

When standing waves are produced in a string, the ends of the string act as they are nodes (points with zero displacement): therefore, the wavelength of a wave in a string is equal to twice the length of the string itself:

$\lambda=2L$

where

$\lambda$ is the wavelength of the wave

L is the length of the string

In this problem,

L = 5.00 m is the length of the string

Therefore, the wavelength is

$\lambda =2(5.00)=10.0 m$

d)

The amplitude of a wave is the magnitude of the maximum displacement of the wave, measured relative to the equilibrium position.

In this problem, we can easily infer the amplitude of this wave by looking at eq.(1).

$y(t)=(0.500 cm) sin[(2.00\cdot 10^4 s^{-1})t]$

And by comparing it with the general equation of a wave:

$y(t)=A sin (\frac{2\pi}{T} t)$

In fact, the maximum displacement occurs when the sine part is equal to 1, so when

$sin(\frac{2\pi}{T}t)=1$

which means that

$y(t)=A$

And therefore in this case,

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6 0

3 years ago

if escape velocity and orbital velocity of a satellite for Orbit close to the Earth's surface then these are related by

n200080 [17]

Answer:

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The formula of the escape velocity is

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The formula of the orbital velocity is

$v=\sqrt\frac{GM}{R}$

The ratio of the escape velocity to the orbital velocity is

$\frac{v_e}{v_o}=\sqrt 2$

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3 years ago

A coin is rolling across a table at 0.23 m/s. It rolls off the table and lands 0.15 meters away from the edge of the table. How

timurjin [86]

Answer:

<em>The table is 2.08 m high</em>

Explanation:

<u>Horizontal Motion</u>

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$\displaystyle d=v\cdot\sqrt{\frac {2h}{g}}$

If we know the range d = 0.15 m and the speed v = 0.23 m/s, we can solve the above equation for h:

$\displaystyle h=\frac{d^2\cdot g}{2v^2}$

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3 years ago

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Andrei [34K]

Answer:

$v_{y}=35.21m/s$

Explanation:

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$v_{o}=51.6m/s\\\beta =37.0º\\y_{o}=7m$

If we want to know whats the <u>y-component of velocity</u> we need to use the following formula:

$v_{y}^2=v_{oy}^2+ag(y-y_{o})$

Knowing that $g=-9.8m/s^2$

$v_{y}=\sqrt{((51.6m/s)sin(37))^2-2(9.8m/s^2)(0m-7m)}=35.21m/s$

So, the cannonball's y-component of velocity is $v_{y}=35.21m/s$

6 0

3 years ago