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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 compound microscope is made with an objective lens (f0 = 0.900 cm) and an eyepiece (fe = 1.10 cm). The lenses are separated by
Marizza181 [45]

Answer:

-252.52

Explanation:

L = Distance between lenses = 10 cm

D = Near point = 25 cm

f_o = Focal length of objective = 0.9 cm

f_e = Focal length of eyepiece = 1.1 cm

Magnification of a compound microscope is given by

m=-\frac{L}{f_o}\frac{D}{f_e}\\\Rightarrow m=-\frac{10}{0.9}\times \frac{25}{1.1}\\\Rightarrow m=-252.52

The angular magnification of the compound microscope is -252.52

6 0
3 years ago
Question 30
stellarik [79]

Answer: 0.69\°

Explanation:

The angular diameter \delta of a spherical object is given by the following formula:

\delta=2 sin^{-1}(\frac{d}{2D})

Where:

d=16 m is the actual diameter

D=1338 m is the distance to the spherical object

Hence:

\delta=2 sin^{-1}(\frac{16 m}{2(1338 m)})

\delta=0.685\° \approx 0.69\° This is the angular diameter

3 0
3 years ago
A 0.153 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.700 m/s. It has a head-on col
DedPeter [7]

Answer:

3.1216 m/s.

Explanation:

Given:

M1 = 0.153 kg

v1 = 0.7 m/s

M2 = 0.308 kg

v2 = -2.16 m/s

M1v1 + M2v2 = M1V1 + M2V2

0.153 × 0.7 + 0.308 × -2.16 = 0.153 × V1 + 0.308 × V2

= 0.1071 - 0.66528 = 0.153 × V1 + 0.308 × V2

0.153V1 + 0.308V2 = -0.55818. i

For the velocities,

v1 - v2 = -(V1 - V2)

0.7 - (-2.16) = -(V1 - V2)

-(V1 - V2) = 2.86

V2 - V1 = 2.86. ii

Solving equation i and ii simultaneously,

V1 = 3.1216 m/s

V2 = 0.2616 m/s

8 0
3 years ago
For the circuit shown, calculate
RSB [31]

For the circuit shown

a.the total resistance   = R = 35.9 Ω

b. when total current = 2 A, then total voltage  =  V = 71.8 V

c.the current through resistor of resistance 56Ω = I₁ = 1.28 A  

  the current through resistor of resistance 100Ω = I₂ = 0.72 A

Explanation:

a)

Resistors can be connected in series or in parallel. For series combination Resultant resistance of the circuit is given by

R = R₁ + R₂ + R₃ +...........+Rₙ

But is our case as shown in the picture, Both the resistors are connected in parallel and for parallel combination resultant, resultant resistance of the circuit is given by

(1 / R) = (1 / R₁) + (1 / R₂) + (1 / R₃) +.......+ (1 / Rₙ)

So

1 / R = (1 / R₁) + (1 / R₂)

Let

R₁ = 56Ω and R₂ = 100Ω

1 / R = 1/56 + 1/100

1 / R = 39 / 1400

R = 1400 / 39

R = 35.9 Ω

b)

Part b can be solved by Ohm's Law Which is stated as "The current flowing through a circuit is directly proportional to the potential difference across its ends provided the physical state such as temperature of the conductor (circuit) does not change."

Mathematically

V ∝ I

V = IR

Where V is the potential difference across the ends of the conductor and I is the Current flowing through the circuit. R is the resistance of the circuit.

Given data:

Total current = I = 2 A

Resistance = R = 35.9 Ω

Voltage = ?

By Ohm's Law

V = IR

V = 2*35.9

V = 71.8 V

c)

To solve current through each resistor, We can use current division formula which is given as

I₁ = I[ R₂ / (R₁ + R₂) ]

I₂ = I[ R₁ / (R₁ + R₂) ]

Also we can take Voltage across each resistor equal to V = 71.8 V because Potential difference across each resistors connected in parallel remain same . And by using Ohm's law divide the value of potential difference with the value of respective resistor to find the current through each resistors.

By Ohm's Law

V = IR

I = V / R

I₁ = 71.8 / 56

I₁ = 1.28 A

I₂ = V / R

I₂ = 71.8 / 100

I₂ = 0.72 A

Learn more about Resistors and Ohm's law from

https://brainly.in/question/9744300

#learnwithBrainly

7 0
3 years ago
Bob and John are pulling in different directions. If Bob is pulling to the right with a force of 10N, and John is pulling to the
kykrilka [37]

Answer:

A) -2N

B) Left

C) -0.5

Explanation:

A) -12 + 10

B) More force is acted on in that direction

C) Net force/Mass (-2/4)

5 0
3 years ago
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