Ask question

Ask question

All categories

1 year 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?

1 answer:

kirza4 [7]1 year 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

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

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

You might be interested in

A student measures the pH of a solution to be 6.8. Which should the student add if she wants to decrease the pH of the solution?

zloy xaker [14]

The neutral pH is 7. Less than 7 indicates an acid and more than 7 indicates a base (up to 14).
<span> NaCl - it's a salt (we can't measure the pH)
H2O - it can be an acid but also a base (the pH it is almost neutral,meaning close to 7 )
HF - it is a strong acid
</span><span> KOH - it is a strong base (pH=14)
</span>
↓

He needs to use HF (Hydrogen fluoride) to decrease the pH.

7 0

1 year ago

Read 2 more answers

An ideal gas is contained in a vessel at 300 K. The temperature of the gas is then increased to 900 K. (i) By what factor does t

Dahasolnce [82]

The question is missing some parts. Here is the complete question.

An ideal gas is contained in a vessel at 300K. The temperature of the gas is then increased to 900K.

(i) By what factor does the average kinetic energy of the molecules change, (a) a factor of 9, (b) a factor of 3, (c) a factor of $\sqrt{3}$ , (d) a factor of 1, or (e) a factor of $\frac{1}{3}$ ?

Using the same choices in part (i), by what factor does each of the following change: (ii) the rms molecular speed of the molecules, (iii) the average momentum change that one molecule undergoes in a colision with one particular wall, (iv) the rate of collisions of molecules with walls, and (v) the pressure of the gas.

Answer: (i) (b) a factor of 3;

(ii) (c) a factor of $\sqrt{3}$ ;

(iii) (c) a factor of $\sqrt{3}$ ;

(iv) (c) a factor of $\sqrt{3}$ ;

(v) (e) a factor of 3;

Explanation: (i) Kinetic energy for ideal gas is calculated as:

$KE=\frac{3}{2}nRT$

where

n is mols

R is constant of gas

T is temperature in Kelvin

As you can see, kinetic energy and temperature are directly proportional: when tem perature increases, so does energy.

So, as temperature of an ideal gas increased 3 times, kinetic energy will increase 3 times.

For temperature and energy, the factor of change is 3.

(ii) Rms is root mean square velocity and is defined as

$V_{rms}=\sqrt{\frac{3k_{B}T}{m} }$

Calculating velocity for each temperature:

For 300K:

$V_{rms1}=\sqrt{\frac{3k_{B}300}{m} }$

$V_{rms1}=30\sqrt{\frac{k_{B}}{m} }$

For 900K:

$V_{rms2}=\sqrt{\frac{3k_{B}900}{m} }$

$V_{rms2}=30\sqrt{3}\sqrt{\frac{k_{B}}{m} }$

Comparing both veolcities:

$\frac{V_{rms2}}{V_{rms1}}= (30\sqrt{3}\sqrt{\frac{k_{B}}{m} }) .\frac{1}{30} \sqrt{\frac{m}{k_{B}} }$

$\frac{V_{rms2}}{V_{rms1}}=\sqrt{3}$

For rms, factor of change is $\sqrt{3}$

(iii) Average momentum change of molecule depends upon velocity:

q = m.v

Since velocity has a factor of $\sqrt{3}$ and velocity and momentum are proportional, average momentum change increase by a factor of

(iv) Collisions increase with increase in velocity, which increases with increase of temperature. So, rate of collisions also increase by a factor of $\sqrt{3}$ .

(v) According to the Pressure-Temperature Law, also known as Gay-Lussac's Law, when the volume of an ideal gas is kept constant, pressure and temperature are directly proportional. So, when temperature increases by a factor of 3, Pressure also increases by a factor of 3.

4 0

1 year ago

(a) when rebuilding her car's engine, a physics major must exert 300 n of force to insert a dry steel piston into a steel cylind

Vilka [71]

There are some missing data in the text of the problem. I've found them online:
a) coefficient of friction dry steel piston - steel cilinder: 0.3
b) coefficient of friction with oil in between the surfaces: 0.03

Solution:
a) The force F applied by the person (300 N) must be at least equal to the frictional force, given by:
$F_f = \mu N$
where $\mu$ is the coefficient of friction, while N is the normal force. So we have:
$F=\mu N$
since we know that F=300 N and $\mu=0.3$ , we can find N, the magnitude of the normal force:
$N= \frac{F}{\mu}= \frac{300 N}{0.3}=1000 N$

b) The problem is identical to that of the first part; however, this time the coefficienct of friction is $\mu=0.03$ due to the presence of the oil. Therefore, we have:
$N= \frac{F}{\mu}= \frac{300 N}{0.03}=10000 N$

8 0

1 year ago

How many significant figures do each of the following numbers have: (a) 214, (b) 81.60, (c) 7.03, (d) 0.03, (e) 0.0086, (f) 3236

Korolek [52]

In determining the number of significant figures in a given number, there are three rules to always remember / follow:

First: All integers except zero are always significant.

<span>Second: Any zeros located between non zeroes are always significant.</span>

Third: A zero located after a non zero in a decimal is always significant whether it is before or after the decimal

Therefore using this rule, the number of significant digits in the given numbers are:

(a) 214 = 3

(b) 81.60 = 4

(c) 7.03 = 3

(d) 0.03 = 1

(e) 0.0086 = 2

(f) 3236 = 4

(g) 8700 = 2

4 0

2 years ago

Passengers on a carnival ride move at constant speed in a horizontal circle of radius 5.0 m, making a complete circle in 4.0 s.

Nataliya [291]

Answer:

bonita sisisisiisisisisiisissiissiisiiss

7 0

1 year ago