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dmitriy555 [2]
3 years ago
5

The tendon from the flexor hallucis longus muscle in the hindlimb of a turkey (Meleagris gallopavo) has a resilience of 0.93. If

9.00 J of work are done on the tendon to stretch it out, how many Joules of work does the tendon do as it is relaxing?
Physics
1 answer:
Vitek1552 [10]3 years ago
8 0

Answer:

8.37 Joules

Explanation:

The amount of energy that a substance, such as animal tissue or rubber can take and yet return to its original condition is known as resilience. When we stress such a substance and then let it restore to its previous state, we are referring to the substance to maintain its elastic area of the stress-strain curve.

From the given information:

the resilence is 0.93

The amount of work done during stretching of the tendon = 9.00 J

Thus,

the work done when relaxing = 0.93 × 9.00 J

= 8.37 Joules

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Answer:

Explanation:

When the central shaft rotates , the seat along with passenger also rotates . Their rotation requires a centripetal force of mw²R where m is mass of the passenger and w is the angular velocity and R is radius of the circle in which the passenger rotates.

This force is provided by a component of   T , the tension in the rope from which the passenger hangs . If θ be the angle the rope makes with horizontal ,

T cos θ will provide the centripetal force . So

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Tsinθ component will balance the weight .

Tsinθ = mg

Dividing the two equation

Tanθ = \frac{g}{\omega^2R}

Hence for a given w , θ depends upon g or weight .

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Astronauts in orbit are not truly weightless. True or False?
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Thermal conductivity of a material is given as 129Btuft–¹ h–¹°F–¹.Calculate this thermal conductivity in Jm–¹s–¹°C–¹(Given: 1Btu
ss7ja [257]

Answer:

223.25 $\text{Jm}^{-1}\text{s}^{-1}^\circ\text{C}^{-1}$

Explanation:

The thermal conductivity of an object is defined as the measure or the ability of the object to transfer heat or conduct heat through its body.

In the context, the thermal conductivity of the material is given as

$=129 \text{ Btu ft}^{-1}\text{h}^{-1}^\circ\text{F}^{-1}$

And it is given that :

1 Btu = 1055 J

1 ft = 0.3048 m

$1^\circ F = \frac{5}{9}^\circ C$

We know that 1 h = 3600 s

So the thermal conductivity of the material in $\text{Jm}^{-1}\text{s}^{-1}^\circ\text{C}^{-1}$ is  :

Thermal conductivity :

$=\frac{129 \text{ Btu}}{1 \text{ ft }\times \text{1 h}\times 1^\circ\text{F}}$

$=\frac{129 \times 1055 \text{ J}}{0.3048 \text{ m} \ \times 3600 \text{ s}\ \times \frac{5}{9}^\circ \text{C}}$

=  223.25 $\text{Jm}^{-1}\text{s}^{-1}^\circ\text{C}^{-1}$

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