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Kruka [31]
3 years ago
5

Two leopards are fighting over a piece of meat they caught while hunting. The leopards pull on the meat muscle with a force of 1

00. N, stretching the 0.10-m-long tendon by 0.0080 m. If the cross-sectional area of the tendon is 1.0 105 m2, what is its stretch modulus?
Physics
1 answer:
kirza4 [7]3 years ago
3 0

Answer:

Stretch Modulus of the given is

Modulus = 1.25 \times 10^8 N/m^2

Explanation:

By the law of elasticity we know that the ratio of stress and strain is known as modulus of elasticity

So here we have

Modulus = \frac{stress}{strain}

here we have

stress = \frac{Force}{Area}

stress = \frac{100}{1.0 \times 10^{-5}}

stress = 10^7 m^2

now similarly we have

strain = \frac{\Delta L}{L}

strain = \frac{0.0080}{0.10}

strain = 0.08

Now we have

Modulus = \frac{10^7}{0.08}

Modulus = 1.25 \times 10^8 N/m^2

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C. How bright it seems to us

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4 years ago
As the temperature rises the solubility of all gases in water also ______
Anastasy [175]

Answer:

As the temperature rises the solubility of all gases in water also will decrease.  

Explanation:

The solubility of gases in water increase with the decrease of the temperature, and vice-versa.  

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I hope it helps you!                        

4 0
4 years ago
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two identical steel balls mounted on wooden posts initially have different amounts of charge, one with -14uC and the other with
GrogVix [38]

According to the Law of Conservation of Charge, the net charge remains constant. If both things have different charges, upon contact, they would share the charge equally. In this case, the total charge is -16μC. The final charge for each ball would be -8 μC.

7 0
4 years ago
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A car is traveling at 120 km/h (75 mph). When applied the braking system can stop the car with a deceleration rate of 9.0 m/s2.
Bumek [7]

Answer:

the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15

Explanation:

Given that;

speed of car V  = 120 km/h = 33.3333 m/s

Reaction time of an alert driver = 0.8 sec

Reaction time of an alert driver = 3 sec

extra time taken by sleepy driver over an alert driver = 3 - 0.8 = 2.2 sec

now, extra distance that car will travel in case of sleepy driver  will be'

S_d = V × 2.2 sec

S_d = 33.3333 m/s × 2.2 sec

S_d = 73.3333 m

hence, number of car of additional car length  n will be;

n = S_n / car length

n = 73.3333 m / 5m

n = 14.666 ≈ 15

Therefore, the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15

8 0
4 years ago
What is the rotational inertia of an object that rolls without slipping down a 2.00-m-high incline starting from rest, and has a
krek1111 [17]

We will apply the concepts of energy conservation to solve the problem. We know that gravitational potential energy is equivalent to the sum of translational kinetic energy and rotational kinetic energy. Additionally, the relation of the angular velocity with the tangential velocity will be determined to eliminate the angular term and obtain the expression of the Inertia in terms of the data given, therefore, we know that

PE_{g} = KE_{trans} +KE_{rot}

mgh = \frac{1}{2} mv^2+\frac{1}{2}I\omega^2

The angular velocity in terms of tangential velocity and radius is defined as,

\omega = \frac{v}{R}

Replacing,

mgh = \frac{1}{2} mv^2+\frac{1}{2} I(\frac{v}{R})^2

Multiplying by R^2,

gh(mR^2) = \frac{v^2}{2} (mR^2)+\frac{v^2}{2}I

\frac{v^2}{2}I = (gh-\frac{v^2}{2})mR^2

I = (\frac{2gh}{v^2}-1)mR^2

Replacing with our values we have,

I = (\frac{2(9.8)(2)}{6^2}-1)mR^2

I = 0.089mR^2

6 0
3 years ago
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