PHYSICS COLLEGE

A relaxed biceps muscle requires a force of 25.4 N for an elongation of 3.20 cm; under maximum tension, the same muscle requires a force of 520 N for the same elongation. Find Young's modulus (Pa) 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 50.0 cm2.

Answers

Answer 1
Answer:

Answer:

Y = 31750 Pa = 31.75 KPa (For 24.5 N force)

Y = 312500 Pa = 312.5 KPa (For 250 N force)

Explanation:

Since the elongation is constant. Therefore, the strain will remain the same in both cases:

Strain = (Elongation)/(Original\ Length)\n\nStrain = (0.032\ m)/(0.2\ m)\n\nStrain = 0.16

FOR A FORCE OF 25.4 N:

Stress = (Force)/(Area)\n\nStress = (25.4\ N)/(0.005\ m^2)\n\nSress = 5080\ Pa = 5.08\ KPa

Now, for Young's Modulus:

Y = (Stress)/(Strain)\n\nY = (5080\ Pa)/(0.16)

Y = 31750 Pa = 31.75 KPa

FOR A FORCE OF 520 N:

Stress = (Force)/(Area)\n\nStress = (250\ N)/(0.005\ m^2)\n\nSress = 50000\ Pa = 50\ KPa

Now, for Young's Modulus:

Y = (Stress)/(Strain)\n\nY = (50000\ Pa)/(0.16)

Y = 312500 Pa = 312.5 KPa

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How fast can the car take this curve this curve without skidding to the outside of the curve?

On average, both arms and hands together account for 13 % of a person's mass, while the head is 7.0% and the trunk and legs account for 80 % . We can model a spinning skater with her arms outstretched as a vertical cylinder (head, trunk, and legs) with two solid uniform rods (arms and hands) extended horizontally. Suppose a 61.0-kg skater is 1.80 m tall, has arms that are each 70.0 cm long (including the hands), and a trunk that can be modeled as being 35.0 cm in diameter. The skater is initially spinning at 70.0 rpm with his arms outstretched.Required:
What will his angular velocity be (in rpm) when he pulls in his arms until they are at his sides parallel to his trunk?

Answers

Final answer:

To find the final angular velocity when the skater pulls in his arms, we use the conservation of angular momentum.

Explanation:

To find the final angular velocity when the skater pulls in his arms, we can make use of the conservation of angular momentum. Initially, the skater's arms are outstretched, and the moment of inertia can be calculated using the parallel axis theorem. After the skater pulls in his arms, we can calculate the new moment of inertia using the same theorem. Equating the initial and final angular momentum values, we can solve for the final angular velocity.

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Final answer:

The problem involves the concept of conservation of angular momentum. The skater's spinning speed will increase when they pull their arms in. For a precise value of the final velocity, a complex calculation taking into account body mass distribution is needed.

Explanation:

This question involves the principle of conservation of angular momentum, which states that the angular momentum of an object remains constant as long as no external torques act on it. The total initial angular momentum of the skater spinning with outstretched arms is equal to his final angular momentum when he pulls his arms in.

Calculating the skater's initial and final angular momentum, you can then solve for his final velocity.

However, note that the calculation needs to take into account the skater's mass distribution. Specifically, we need to consider the percentage distributions for the arms/hands (13%), head (7%) and trunk/legs (80%), and integrate these over the skater's body.

This can result in a significantly complex calculation if done accurately, involving calculus level mathematics. However, using the qualitative knowledge that the skater's spinning speed will increase when they pull their arms in, it's reasonable to estimate, considering the mass distribution, the final velocity will be somewhere near 2 to 3 times the original rpm. But for an exact value, a detailed and complex calculation is needed.

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A magnet of mass 0.20 kg is dropped from rest and falls vertically through a 35.0 cm copper tube. Eddy currents are induced, causing the copper to warm up. The speed of the magnet as it emerges from the tube is 1.10 m/s. How much heat energy is dissipated to the environment?

Answers

Answer:

0.58 J

Explanation:

We know that Total energy is conserved.

Initial Kinetic energy + Initial potential energy = final kinetic energy+ final potential energy + dissipated heat energy

Initial kinetic energy = 0 ( magnet is at rest initially)

Initial Potential energy = m g h = (0.20 kg)(9.81 m/s²)(0.35 m) = 0.69 J

Final kinetic energy = 0.5 m v² = 0.5 ×0.20 kg × 1.10 m/s = 0.11 J

Final potential energy = 0

∴ Dissipated heat energy = (0.69 -0.11) J = 0.58 J

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Answers

The dependent variable in this scenario is the outcome or result that you are trying to measure or analyze based on the student's skateboarding activity.

Since the student goes skateboarding a few times a week, the dependent variable could be any aspect related to their skateboarding experience or its effects.

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2. Physical fitness (e.g., measured by changes in endurance, strength, or flexibility).

3. Time spent skateboarding per session.

4. The number of skateboarding injuries or accidents.

5. Overall enjoyment or satisfaction with skateboarding.

6. Changes in stress levels or mood before and after skateboarding sessions.

7. Social interactions and friendships formed through skateboarding.

The specific dependent variable would depend on the research question or hypothesis you are investigating in relation to the student's skateboarding activity.

The tension in a string is 15 N, and its linear density is 0.85 kg/m. A wave on the string travels toward the -x direction; it has an amplitude of 3.6 cm and a frequency of 12 Hz. What are the: a. Speed
b. Wavelength of the wave?
c. Write down a mathematical expression for the wave, substituting numbers for the variables

Answers

Answer:

(a) The speed of the wave, v is 4.2 m/s

(b) Wavelength of the wave, λ is 0.35 m

(c) mathematical expression of the wave, Y = 0.036sin(5.71πx - 24πt)

Explanation:

Given;

tension on the string, T = 15 N

Linear density, μ = 0.85 kg/m

amplitude of the wave, A = 3.6 cm = 0.036 m

frequency of the wave, f = 12 Hz

(a) The speed of the wave, v is calculated as;

v = \sqrt{(T)/(\mu) } \n\nv = \sqrt{(15)/(0.85) }\n\nv = 4.2 \ m/s

(b) Wavelength of the wave, λ

v = fλ

λ = v / f

λ = 4.2 / 12

λ = 0.35 m

(c) mathematical expression of the wave;

Y = Asin((2\pi x)/(\lambda) -\omega t)\n\nY = Asin((2\pi x)/(\lambda) -2\pi f t)\n\nY = 0.036sin ((2\pi )/(0.35)x -2\pi *12 t)\n\nY = 0.036sin (5.71 \pi x - 24 \pi t)

Calculate the slope of the 25-coil line and the 50-coil line to determine the average number of paper clips that a 1 V battery would pick up.

Answers

Answer:

For 25-turn electromagnet, Number of clips = 4.1

For 50-turn electromagnet number of clips = 9.6

Explanation:

To calculate the slope of the 25-coil line and the 50-coil line to determine the average number of paper clips that a 1 V battery would pick up.

Hence;

Using the equations gotten from the graph in the previous question and 1.0 V as the value for x, we get

For 25-turn electromagnet y = 3.663x * 0.5

(rounded to one decimal place) Number of clips = 4.1

For 50-turn electromagnet y = 7.133x 2.5

(rounded to one decimal place) Number of clips = 9.6

An unruly student with a spitwad (a lump of wet paper) of mass 20 g in his pocket finds himself in the school library where there is a ceiling fan overhead. He relieves his boredom by throwing the spitwad up at the ceiling fan where it collides with, and sticks to, the end of one of the blades of the stationary ceiling fan. Its horizontal velocity vector is perpendicular to the long axis of the blade. If the fan is free to rotate (no friction at all) and has moment of inertia I=1.4kgm2 , if the spitwad has horizontal velocity 4 m/s, and if the spitwad sticks to the fan blade at a distance of 0.6 m from the rotation axis of the fan, how much time will it take the fan to move through one complete revolution after the spitwad hits it (closest answer)?a. 1min
b. 2min
c. 3min
d. 4min
e. 5min
f. 6min

Answers

Answer:

T = 188.5 s, correct is  C

Explanation:

This problem must be worked on using conservation of angular momentum. We define the system as formed by the fan and the paper, as the system is isolated, the moment is conserved

         

initial instant. Before the crash

        L₀ = r m v₀ + I₀ w₀

the angular speed of the fan is zero w₀ = 0

final instant. After the crash

        L_f = I₀ w + m r v

        L₀ = L_f

        m r v₀ = I₀ w + m r v

angular and linear velocity are related

        v = r w

        w = v / r

        m r v₀ = I₀ v / r + m r v

         m r v₀ = (I₀ / r + mr) v

       v = (m)/((I_o)/(r) +mr) \ r v_o

let's calculate

       v = (0.020)/((1.4)/(0.6 ) + 0.020 \ 0.6 ) \ 0.6 \ 4

       v = (0.020)/(2.345) \ 2.4

       v = 0.02 m / s

         

To calculate the time of a complete revolution we can use the kinematics relations of uniform motion

        v = x / T

         T = x / v

the distance of a circle with radius r = 0.6 m

         x = 2π r

we substitute

         T = 2π r / v

let's calculate

         T = 2π 0.6/0.02

         T = 188.5 s

reduce

         t = 188.5 s ( 1 min/60 s) = 3.13 min

correct is  C
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