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uranmaximum [27]
3 years ago
10

In some recent studies it has been shown that women are men when competing in similar sports (most notably in soccer and basketb

all). Select the statement that explains why this disparity might exist. a. The cross-sectional area of the ACL is typically larger in men, and therefore experiences less strain for the sam tensile force and Young's modulus. b. The Young's modulus of women's ACLS is typically smaller than that of men's, resulting in more stress for the same amount of strain. c. The cross-sectional area of the ACL is typically smaller in women, and therefore experiences less stress for the same tensile force. d. The ACL of women is more elastic than the ACL of men.
Physics
1 answer:
Elza [17]3 years ago
7 0

Answer: The correct option is B (The Young's modulus of women's ACLS is typically smaller than that of men's, resulting in more stress for the same amount of strain)

Explanation:

Anterior cruciate ligament (ACL) is one of the important ligaments found at the knee joint which helps to stabilise the joint. It connects the femur to the tibia bone at the knee joint.

Anterior cruciate ligament tear is one of the common knee joint injury which is seen in individuals( especially females) involved in sports( example soccer and basketball which involves sudden change in direction causing the knee to rotate inwards)

ACL tear occurs through both contact and non contact mechanisms. The contact mechanism of ACL injury occurs when force is directly applied at the lateral part of the knee while in non contact mechanism,tear occurs when the tibia is externally rotated on the planted foot.

Research has proven that women are prone to have ACL tear than men when competing in similar sports. This disparity exists due to structural differences that pose as risk factors. These includes

- the female ACL size is smaller than the male.

- the ACL of female has a lower modulus if elasticity( that is, less stiff) than in males leading to greater joint mobility than in the male.. therefore the option, (The Young's modulus of women's ACLS is typically smaller than that of men's, resulting in more stress for the same amount of strain) is correct.

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A 4kg watermelon is dropped from a height of 45m. What is the velocity of the watermelon just before it hits the ground?
Makovka662 [10]

Answer:

v=30 m/s

Explanation:

h - height

g - acceleration due to gravity=10

t - time

v- velocity

h =  \frac{1}{2}  \times g \times t {}^{2}

45 = 5t²

t² = 9

t=3 seconds

v=g×t

v=10×3

v=30 m/s

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What force arises from the physical contact of two objects?
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Which of the following terms is used for the chemical composition of a compound?
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Chemical formula is the term used for the chemical composition of a compound.
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3 years ago
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Six artificial satellites complete one circular orbit around a space station in the same amount of time. Each satellite has mass
oee [108]

Answer:

The ranking of the net force acting on different satellite from largest to smallest is {F_E} > {F_F} > {F_A} = {F_B} = {F_D} > {F_C}

Explanation:

In order to get a good understanding of this solution we need to understand that the main concepts used to solve this problem are centripetal force and velocity of satellite.

Initially, use the expression of the velocity of satellite and find out its dependence on the radius of orbit. Use the dependency in the centripetal force expression.

Finally, we find out the velocity of the six satellites and use that expression to find out the force experienced by the satellite. Find out the force in terms of mass (m) and radius of orbit (L) and at last compare the values of force experienced by six satellites.

Fundamentals

The centripetal force is necessary for the satellite to remain in an orbit. The centripetal force is the force that is directed towards the center of the curvature of the curved path. When a body moves in a circular path then the centripetal force acts on the body.

The expression of the centripetal force experienced by the satellite is given as follows:

                    {F_{\rm{c}}} = \frac{{m{v^2}}}{L}

Here, m is the mass of satellite, v is the velocity, and L is the radius of orbit.

The velocity of the satellite with which the satellite is orbiting in circular path is given as follows:

                        v = \frac{{2\pi L}}{T}

Here, T is the time taken by the satellite.

The velocity of the satellite with which the satellite is orbiting in circular path is given as follows;

                    v = \frac{{2\pi L}}{T}

Since, all the satellites complete the circular orbit in the same amount of time. The factor of   \frac{{2\pi }}{T}   is not affected the velocity value for the six satellites. Therefore, we can write the expression of v given as follows:

Substitute  v = \frac{{2\pi L}}{T} in the force expression {F_{\rm{c}}} = \frac{{m{v^2}}}{L}   as follows:

                              \begin{array}{c}\\{F_c} = \frac{{m{{\left( {\frac{{2\pi L}}{T}} \right)}^2}}}{L}\\\\ = \frac{{4{\pi ^2}}}{{{T^2}}}mL\\\end{array}

Since, all the satellites complete the circular orbit in the same amount of time. The factor of \frac{{4{\pi ^2}}}{{{T^2}}}  not affect the force value for six satellites.Therefore, we can write the expression of {F_c}  given as follows:

        {F_c} = kmL

Here, k refers to constant value and equal to  \frac{{4{\pi ^2}}}{{{T^2}}}

    {F_A} = k{m_A}{L_A}

Substitute 200 kg for {m_A}   and 5000 m for LA in the expression                                  {F_A} = k{m_A}{L_A}

\begin{array}{c}\\{F_A} = k\left( {200{\rm{ kg}}} \right)\left( {5000{\rm{ m}}} \right)\\\\ = {10^6}k{\rm{ N}}\\\end{array}

The force acting on satellite B from their rocket is given as follows:{F_B} = k{m_B}{L_B}

Substitute 400 kg for {m_B} and 2500 m for in the expression {F_B} = k{m_B}{L_B}

\begin{array}{c}\\{F_B} = k\left( {400{\rm{ kg}}} \right)\left( {2500{\rm{ m}}} \right)\\\\ = {10^6}k{\rm{ N}}\\\end{array}

The force acting on satellite C from their rocket is given as follows:{F_C} = k{m_C}{L_C}

Substitute 100 kg for {m_C}and 2500 m for in the above expression  {F_C} = k{m_C}{L_C}

\begin{array}{c}\\{F_C} = k\left( {100{\rm{ kg}}} \right)\left( {2500{\rm{ m}}} \right)\\\\ = 0.25 \times {10^6}k{\rm{ N}}\\\end{array}

The force acting on satellite D from their rocket is given as follows:{F_D} = k{m_D}{L_D}

Substitute 100 kg for {m_D} and 10000 m for {L_D} in the expression{F_D} = k{m_D}{L_D}

\begin{array}{c}\\{F_D} = k\left( {100{\rm{ kg}}} \right)\left( {10000{\rm{ m}}} \right)\\\\ = {10^6}k{\rm{ N}}\\\end{array}

The force acting on satellite E from their rocket is given as follows:{F_E} = k{m_E}{L_E}

Substitute 800 kg for {m_E}  and 5000 m for  {L_E} in the expression {F_E} = k{m_E}{L_E}

\begin{array}{c}\\{F_E} = k\left( {800{\rm{ kg}}} \right)\left( {5000{\rm{ m}}} \right)\\\\ = 4.0 \times {10^6}k{\rm{ N}}\\\end{array}

The force acting on satellite F from their rocket is given as follows:{F_F} = k{m_F}{L_F}

Substitute 300 kg for {m_F} and 7500 m for {L_F} in the expression {F_F} = k{m_F}{L_F}

\begin{array}{c}\\{F_F} = k\left( {300{\rm{ kg}}} \right)\left( {7500{\rm{ m}}} \right)\\\\ = 2.25 \times {10^6}k{\rm{ N}}\\\end{array}

The value of forces obtained for the six-different satellite are as follows.

\begin{array}{l}\\{F_A} = {10^6}k{\rm{ N}}\\\\{F_B} = {10^6}k{\rm{ N}}\\\\{F_C} = 0.25 \times {10^6}k{\rm{ N}}\\\\{F_D} = {10^6}k{\rm{ N}}\\\\{F_E} = 4.0 \times {10^6}k{\rm{ N}}\\\\{F_F} = 2.25 \times {10^6}k{\rm{ N}}\\\end{array}

     The ranking of the net force acting on different satellite from largest to smallest is {F_E} > {F_F} > {F_A} = {F_B} = {F_D} > {F_C}

7 0
4 years ago
Dennis throws a volleyball up in the air. It reaches its maximum height 1.1\, \text s1.1s1, point, 1, start text, s, end text la
rewona [7]

Answer:

If max height = 1.1 meters, then initial velocity is 3.28 m/s

If max height is 1.1 feet, then the initial velocity is 5.93  ft/s

Explanation:

Recall the formulas for vertical motion under the acceleration of gravity;

for the vertical velocity of the object we have

v=v_0-g \,t

for the object's vertical displacement we have

y-y_0=v_0\,t - \frac{g}{2} \,t^2

If the maximum height reached by the object is given in meters, we use the value for g in m/s^2 which is: 9.8\,\,m/s^2

If the maximum height of the object is given in feet, we use the value for g in  ft/s^2  which is : 32\,\,ft/s^2

Now, when the ball reaches its maximum height, the ball's velocity is zero, so that allows us to solve for the time (t) the process of reaching the max height takes:

v=v_0-g \,t\\0=v_0-g \,t\\g\,\,t=v_0\\t=\frac{v_0}{g}

and now we use this to express the maximum height in the second equation we typed:

y-y_0=v_0\,t - \frac{g}{2} \,t^2\\max\,height=v_0\,(\frac{v_0}{g})  - \frac{g}{2} \,(\frac{v_0}{g})^2\\max\,height= \frac{v_0^2}{2\,g}

Then if the max height is 1.1 meters, we use the following formula to solve for v_0:

1.1= \frac{v_0^2}{2\,9.8}\\(9.8)\,(1.1)=v_0^2\\v_0=10.78\\v_0=\sqrt{10.78} \\v_0=3.28\,\,m/s

If the max height is 1.1 feet, we use the following formula to solve for v_0:

1.1= \frac{v_0^2}{2\,32}\\(32)\,(1.1)=v_0^2\\v_0=35.2\\v_0=\sqrt{35.2} \\v_0=5.93\,\,ft/s

5 0
3 years ago
Read 2 more answers
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