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yKpoI14uk [10]
3 years ago
5

A relaxed biceps muscle requires a force of 25.0N for an elongation of 3.0 cm; under maximum tension, the same muscle requires a

force 500N for the same elongation. Find the Young's modulus for the muscle tissue under each of these conditions if the muscle can be modeled as a uniform cylinder with an initial length of 0.200 m and a cross-sectional area of 50 cm^2.
Physics
1 answer:
Norma-Jean [14]3 years ago
4 0

3.3 x10^4N/m²

6.7 x105N/m²

Explanation:

Let the young modulus of the relaxed biceps be

Y= F¹Lo/ deta L1 x A

= 25 x0.2/ 0.03* 50cm²(1m²

0.0004cm^-²)

= 3.3x10^4N/m²

But young modules of muscle under maximum tension will be

Y= F"Lo/ deta L" x A

= 500x 0.2/ 0.03* 50cm²(1m²

0.0004cm^-²)

= 6.7 x10^5N/m²

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Mariana [72]

Answer:

The velocity of block = 0.188 \frac{m}{s}

Explanation:

Mass m = 5.6 kg

k = 1040 \frac{N}{m}

\mu = 0.26

x_{1} = 0.035 m  , v_{1} = 0

x_{2} = 0.02 m

From work energy theorem

K_{1} + U_{1} + W_{other} = K_{2} + U_{2}  --------- (1)

Kinetic energy

K = \frac{1}{2} k x^{2}  ------- (1)

Potential energy

U = \frac{1}{2} k x^{2} ------- (2)

Work done

W = F.s ------ (3)

From Newton's second law

R_{N} = mg

R_{N}  = 5.6 × 9.81 = 54.9 N

Friction  force = 0.4 × 54.9 = 21.9 N

Now the work done by the friction

W_{f} = - 21.9 × 0.015

W_{f} = - 0.329 J

Now kinetic energy

At point 1

K_{1} = \frac{1}{2} m v^{2} _{1}

K_{1} = 0

U_{1} = \frac{1}{2} k x^{2}

U_{1} = \frac{1}{2}  (1040) 0.035^{2}

U_{1} = 0.637 J

At point 2

K_{2} = \frac{1}{2} (5.6) v^{2} _{2}

K_{2} = 2.8 v_{2} ^{2}

Potential energy

U_{2} = \frac{1}{2}  k x_2^{2}

U_{2} = \frac{1}{2}  (1040) 0.02^{2}

U_{2} = 0.208 J

From equation (1) we get

0 + 0.637 - 0.329 = 2.8 v_{2} ^{2} + 0.208

2.8 v_{2} ^{2} = 0.1

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Answer:

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Which of the following summarizes cellular respiration?
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An airplane flies eastward and always accelerates at a constant rate. At one position along its path it has a velocity of 34.3 m
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Explanation:

We'll need two equations.

v² = v₀² + 2a(x - x₀)

where v is the final velocity, v₀ is the initial velocity, a is the acceleration, x is the final position, and x₀ is the initial position.

x = x₀ + ½ (v + v₀)t

where t is time.

Given:

v = 47.5 m/s

v₀ = 34.3 m/s

x - x₀ = 40100 m

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a = 0.0135 m/s²

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When a 4.20-kg object is hung vertically on a certain light spring that obeys Hooke's law, the spring stretches 2.00 cm. (a) If
mr_godi [17]

(a) 0.714 cm

First of all, we need to find the spring constant of the spring. This can be done by using Hooke's law:

F=kx

where

F is the force applied on the spring

k is the spring constant

x is the stretching of the spring

At the beginning, the force applied is the weight of the block of m = 4.20 kg hanging on the spring, therefore:

F=mg=(4.20)(9.8)=41.2 N

The stretching of the spring due to this force is

x = 2.00 cm = 0.02 m

Therefore, the spring constant is

k=\frac{F}{x}=\frac{41.2}{0.02}=2060 N/m

Now, a new object of 1.50 kg is hanging on the spring instead of the previous one. So, the weight of this object is

F=mg=(1.50)(9.8)=14.7 N

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(b) 1.65 J

The work done on a spring is given by:

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In this situation,

k = 2060 N/m

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