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Setler [38]
2 years ago
15

What size force does the femur exerts on the kneecap if the tendons are oriented as in the figure and the tension in each tendon

is 60 N?

Physics
1 answer:
sesenic [268]2 years ago
5 0
 Sum of horizontal forces<span> = 0 
F(horizontal) = 60 cos 42
                     = 44.58 

</span>Sum of vertical forces = 0 <span>
F(vertical) = 60 - 60 sin 42
                 = 19.85 

F = sqrt (Fh² + Fv²)
   = sqrt (44.58² + 19.85²)
   = 48.8 N</span>
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Find the object's speeds v1, v2, and v3 at times t1=2.0s, t2=4.0s, and t3=13s.
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Since this is a distance/time graph, the speed at any time is the slope
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At time  t1 = 2.0 s
That's in the middle of the first segment of the graph,
that extends from zero to 3 seconds.
Its slope is  7/3 .              v1 = 7/3 m/s .

At time  t2 = 4.0 s
That's in the middle of the horizontal part of the graph
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Its slope is zero.
                                     v2 = zero .

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You are given a material which produces no initial magnetic field when in free space. when it is placed in a region of uniform m
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A curtain hangs straight down in front of an open window. A sudden gust of wind blows past the window; and the curtain is pulled
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option B.

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The correct answer is option B.

The phenomenon of the curtains to pull out of the window can be explained using Bernoulli's equation.

According to Bernoulli's Principle when the speed of the moving fluid increases the pressure within the fluid decrease.

When wind flows in the outside window the pressure outside window decreases and pressure inside the room is more so, the curtain moves outside because of low pressure.

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The transmutation of a radioactive uranium isotope, 234/92 U, into a radon isotope, 222/86 Rn, involves a series of three nuclea
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You are called as an expert witness to analyze the following auto accident: Car B, of mass 2000 kg, was stopped at a red light w
OLga [1]

Answer:

a). va=17.23 \frac{m}{s} or 38.54 mph

b). v=38.54 mph and limit is 35 mph

c). Completely inelastic

d). Eka=192.967 kJ

Ekt=76.071 kJ

Explanation:

m_{a}=1300kg\\m_{b}=2000kg\\x_{f}=7.25m\\u_{k}=0.65

The motion is an inelastic collision so

m_{a}*v_{a}+m_{b}*v_{b}=(m_{a}+m_{b})*v_{f}

The force of the motion is contrarest by the force of friction so

F-F_{uk} =0\\F=F_{uk}\\F_{uk}=u_{k}*m*g\\F=m*a\\a=\frac{F}{m}\\ a=\frac{F_{uk}}{m}\\a=\frac{u_{k}*m*g}{m}\\a=u_{k}*g\\a=0.65*9.8\frac{m}{s^{2}} \\a=6.39\frac{m}{s^{2}}

Now with the acceleration can find the time and the velocity final that make the distance 7.25m being united

x_{f}=x_{o}+v_{o}*t+2*a*t^{2}\\x_{o}=0\\v_{o}=0\\x_{f}=2*a*t^{2}\\t^{2}=\frac{x_{f}}{2*a}\\t=\sqrt{\frac{7.25m}{6.37\frac{m}{s^{2} } } } \\t=1.06s

So the velocity final can be find using this time

v_{f}=v_{o}+a*t\\v_{o}=0\\v_{f}=6.37\frac{m}{s^{2} } *1.06s\\v_{f}=6.79 \frac{m}{s}

a).

Replacing in the first equation the final velocity can find the initial velocity

m_{a}*v_{a}+m_{b}*v_{b}=(m_{a}+m_{b})*v_{f}

v_{b}=0

v_{a}= \frac{(m_{a}+m_{b)*v_{f}}}{m_{a}}\\v_{a}= \frac{(1300+2000)*6.37}{1300}\\v_{a}=17.23 \frac{m}{s}

b).

35mph*\frac{1m}{0.000621371mi} *\frac{1h}{3600s}=15.646\frac{m}{s}

Velocity limit in m/s is 15.646 m/s and the initial velocity is 17.23 m/s

so is exceeding the speed limit in about 1.58 m/s

or in miles per hour

3.5 mph

c).

The collision is complete inelastic because any mass can be returned to the original mass, so even they are no the same mass however in the moment they move the distance 7.25m as a same mass the motion is considered completely inelastic

d).

Ek=\frac{1}{2}*m*(v)^{2}\\  Eka=\frac{1}{2}*1300kg*(17.23\frac{m}{s})^{2}\\Eka=192.967 kJ\\Ekt=\frac{1}{2}*m*(v)^{2}\\Ekt=\frac{1}{2}*3300kg*(6.79\frac{m}{s})^{2}\\Ekt=76.071 kJ

8 0
2 years ago
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