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bearhunter [10]
3 years ago
14

Tendons are strong elastic fibers that attach muscles to bones. To a reasonable approximation, they obey Hooke's law. In laborat

ory tests on a particular tendon, it was found that, when a 251 g object was hung from it, the tendon stretched 1.23 cm .Find the force constant of this tendon.
Physics
1 answer:
meriva3 years ago
5 0

Answer:

k = 200 N/m

Explanation:

given,

mass of the object  = 251 g

                                = 251 x 10⁻³ Kg = 0.251 Kg

distance of tendon stretch = x = 1.23 cm

                                            = 1.23 x 10⁻² = 0.0123 m

using the formula

F = k  x

where

k is the force constant of the tendon

F = m g

F = 0.251 x 9.8 = 2.4598 N

2.4598 = k x 0.0123

k = 199.98 N/m

k = 200 N/m

hence, force constant of the tendon is approximately equal to 200 N/m

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Answer:

S = V0 t + 1/2 a t^2

S = 5 m/s * 300 s + 1/2 * 1.2 m/s * (300 s^2)

S = 1500 m + .6 * 90000 m = 55,500 m

Check:     V0 = 5 m/s

                V2 = V0 + a t  = 5 + 1.2 * 300 = 365 m/s

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We calculated V2 above at 365 m/s  the speed after 300 sec

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3 years ago
While standing on a bridge 40.0 m above the ground, you drop a stone from rest. when the stone has fallen 3.80 m, you throw a se
Norma-Jean [14]
Given: distance 1 d₁ = 40 m;  distance 2 d₂ = 3.8 m   g = -9.8 m/s²

 Initial Velocity Vi = 0  Final Velocity of stone 2 is unknown = ?

Total distance dₓ = d₁ - d₂  = 40 m - 3.8 m = 36.2 m

Formula: a = Vf² -  Vi²/2d   derive for Final Velocity Vf
 
acceleration is now due to gravity, therefore a = g

Vf = √2gd   Vf = √2(9.8 m/s²)(36.2 m)

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Reason: The second stone will still start from rest.


3 0
3 years ago
Two objects are moving in the xy-plane. Object A has a mass of 3.2 kg and has a velocity = (2.3 m/s)i+ (4.2 m/s)j and object B h
prisoha [69]

The total momentum of the system is 2.14i + 21.27j.

A vector quantity with both direction and magnitude is momentum. Kg m/s (kilogram meter per second) or N s serve as its units (newton second).

The total starting momentum of a system must match the entire final momentum of the system since momentum is a conserved quantity. The overall momentum does not change.

The total momentum of the system is defined as follows:

As momentum is vector quantity and vectors can be added, so, the momentum of a system is given by

P = Pₓ + P'

where Pₓ is the x-component of momentum

P' is the y-component of the momentum

Also, we know that

P=mv

where m is mass

v is velocity

Thus,

P = Pₓ + P'

P = m₁vₓ + m₂v'

vₓ is the x-component of the velocity

v' is the y- component of the velocity

Given, m₁= 3.2kg

m₂ = 2.9kg

Now,

P = 3.2 (2.3i + 4.2j) + 2.9 (-1.8i +2.7j)

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Thus, the total momentum of the given system is 2.14i + 21.27j.

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8 0
1 year ago
A stubborn, 100 kgkg mule sits down and refuses to move. To drag the mule to the barn, the exasperated farmer ties a rope around
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Answer:

No, the farmer is not able to move the mule.

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The coefficient between the mule and the ground=\mu_s=0.8

\mu_k=0.5

Static friction force,f=\mu N

Normal force=N=mg

Static friction force,f=\mu_s mg=0.8\times 100\times 9.8=784 N

Using g=9.8m/s^2

F<f

Static friction force is greater than applied force.

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