PHYSICS COLLEGE

The triceps muscle in the back of the upper arm extends the forearm. This muscle in a professional boxer exerts a force of 2.00 x 103N with an effective perpendicular arm of 3.0 cm, producing an angular acceleration of the forearm of 130 rad/s2

What is the moment of inertia of the boxer's forearm?

Answers

Answer 1
Answer:

To solve this problem it is necessary to apply the concepts related to Torque as a function of Force and distance. Basically the torque is located in the forearm and would be determined by the effective perpendicular lever arm and force, that is

\tau = F * r

Where,

F = Force

r = Distance

Replacing,

\tau = 2*10^3*0.03

\tau = 60N\cdot m

The moment of inertia of the boxer's forearm can be calculated from the relation between torque and moment of inertia and angular acceleration

\tau = I \alpha

I = Moment of inertia

\alpha = Angular acceleration

Replacing with our values we have that

I = (\tau)/(\alpha)

I = (60)/(120)

I = 0.5kg\cdot m^2

Therefore the value of moment of inertia is 0.5kg\cdot m^2

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Answers

Answer:

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In Europe, gasoline efficiency is measured in km/L. If your car's gas mileage is 28.0 mi/gal , how many liters of gasoline would you need to buy to complete a 142-km trip in Europe

Answers

Answer:

The volume required is  V  = 11.91 \ liters

Explanation:

From the question we are told that

  The cars mileage is  v =  28.0 mi/gal

   The  distance is  d  =  142 km

Converting the distance from km to  miles

        d = 142 * 0.6214 = 88.24 \ miles

Generally the volume of gasoline needed is mathematically represented as  

       V =  (d)/( v)

=>     V =  (88.24)/( 28.0 )

=>     V = 3.151 \ gal

Converting to  Liters

    =>   V = 3.151  *   3.78

     =>   V  = 11.91 \ liters

A 2.00 kg cart on a frictionless track is pulled by force of 3.00 N. What is the acceleration of the cart?

Answers

1.5 will be its acceleration

We had a homework problem in which the Arrhenius equation was applied to the blinking of fireflies. Several other natural phenomena also obey that equation, including the temperature dependent chirping of crickets. A particular species, the snowy tree cricket, has been widely studied. These crickets chirp at a rate of 178 times per minute at 25.0°C, and the activation energy for the chirping process is 53.9 kJ/mol. What is the temperature if the crickets chirp at a rate of 126 times per minute?

Answers

Answer:

Temperature = 20.35°C

Explanation:

Arrhenius equation is as follows:

k = A*exp(-Ea/(R*T)), where

k = rate of chirps

Ea = Activation Energy

R = Universal Gas Constant

T = Temperature (in Kelvin)

A = Constant

Given Data

Ea = 53.9*10^3 J/mol

R = 8.3145  J/(mol.K)

T = 273.15 + 25  K

k = 178  chirps per minutes

Calculation

Using the Arrhenius equation, we can find A,

A= 4.935x10^11

Now we can apply the same equation with the data below to find T at k=126,

k = A*exp(-Ea/(R*T))

Ea = 53.9*10^3

R = 8.3145

k = 126

T = 20.35°C

A trombone can produce pitches ranging from 85 Hz to 660 Hz approximately. When the trombone is producing a 357 Hz tone, what is the wavelength of that tone in air at standard conditions?

Answers

Answer:

The wavelength of that tone in air at standard condition is 0.96 m.

Explanation:

Given that, a trombone can produce pitches ranging from 85 Hz to 660 Hz approximately. We need to find the wavelength of that tone in air when the trombone is producing a 357 Hz tone.

We know that the speed of sound in air is approximately 343 m/s. Speed of a wave is given by :

v=f\lambda\n\n\lambda=(v)/(f)\n\n\lambda=(343\ m/s)/(357\ Hz)\n\n\lambda=0.96\ m

So, the wavelength of that tone in air at standard condition is 0.96 m. Hence, this is the required solution.

An iceskater is turning at a PERIOD of (1/3) second with his arms outstretched. a) What is his ANGULAR VELOCITY w? b) If he pulls his arms towards his body to reduce his MOMENT OF INTERTIA by 1/2, what is his ANGULAR VELOCITY w? c) How much does his ROTATIONAL KINETIC ENERGY change? That is, if the initial Kinetic Energy is (KE)initial, what is the final KE? d) Where did that ENERGY come from, or go to?

Answers

Answer:

Explanation:

a )

Time period T = 1/3 s

angular velocity = 2π / T

= 2 x 3.14 x 3

ω = 18.84 radian / s

b )

Applying conservation of angular momentum

I₁ ω₁ = I₂ ω₂

I₁ / I₂ = ω₂ / ω₁

2 = ω₂ / ω

ω₂ = 2 ω

c )

(KE)initial = 1/2 I₁ ω²

(KE)final =  1/2 I₂ ω₂²

= 1/2 (I₁ / 2)  (2ω)²

=  I₁ ω²

c )

Change in rotational kinetic energy

=  I₁ ω² -  1/2 I₁ ω²

=  +  1/2 I₁ ω²

d )

This energy comes from the work done by centripetal force which is increased to increase the speed of rotation.

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