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3 years ago

13

The triceps muscle in the back of the upper arm is primarily used to extend the forearm. Suppose this muscle in a professional b

oxer exerts a force of 1.95 × 103 N with an effective perpendicular lever arm of 3.1 cm, producing an angular acceleration of the forearm of 125 rad/s2. What is the moment of inertia of the boxer's forearm?

1 answer:

taurus [48]3 years ago

8 0

Answer:

I = 0.483 kgm^2

Explanation:

To know what is the moment of inertia I of the boxer's forearm you use the following formula:

$\tau=I\alpha$ (1)

τ: torque exerted by the forearm

I: moment of inertia

α: angular acceleration = 125 rad/s^2

You calculate the torque by using the information about the force (1.95*10^3 N) and the lever arm (3.1 cm = 0.031m)

$\tau=Fr=(1.95*10^3N)(0.031m)=60.45J$

Next, you replace this value of τ in the equation (1) and solve for I:

$I=\frac{\tau}{\alpha}=\frac{60.45Kgm^2/s ^2}{125rad/s^2}=0.483 kgm^2$

hence, the moment of inertia of the forearm is 0.483 kgm^2

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A copper rod of cross-sectional area 11.6 cm2 has one end immersed in boiling water and the other in an ice-water mixture, which

julia-pushkina [17]

Answer:

0.686 g of ice melts each second.

Solution:

As per the question:

Cross-sectional Area of the Copper Rod, A = $11.6\ cm^{2} = 11.6\times 10^{- 4}\ m^{2}$

Length of the rod, L = 19.6 cm = 0.196 m

Thermal conductivity of Copper, K = $390\ W/m.^{\circ}C$

Conduction of heat from the rod per second is given by:

$q = \frac{KA\Delta T}{L}$

where

$\Delta T = 100^{\circ} - 0^{\circ} = 100^{\circ}C$ = temperature difference between the two ends of the rod.

Thus

$q = \frac{390\times 11.6\times 10^{- 4}\times 100}{0.196} = 228.48\ J/s$

Now,

To calculate the mass, M of the ice melted per sec:

$M = \frac{q}{L_{w}}$

where

$L_{w}$ = Latent heat of fusion of water = 333 kJ/kg

$M = \frac{228.48}{333\times 10^{3}} = 6.86\times 10^{- 4}\ kg = 0.686\ g$

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3 years ago

A rocket travels in the x-direction at speed 0.70c with respect to the earth. An experimenter on the rocket observes a collision

marishachu [46]

Answer:

A) The space time coordinate x of the collision in Earth's reference frame is

$x \approx 103,46x10^{9}m$ .

B) The space time coordinate t of the collision in Earth's reference frame is

$t=377,29s$

Explanation:

We are told a rocket travels in the x-direction at speed v=0,70 c (c=299792458 m/s is the exact value of the speed of light) with respect to the Earth. A collision between two comets is observed from the rocket and it is determined that the space time coordinates of the collision are (x',t') = (3.4 x 10¹⁰ m, 190 s).

An event indicates something that occurs at a given location in space and time, in this case the event is the collision between the two comets. We know the space time coordinates of the collision seen from the reference frame of the rocket and we want to find out the space time coordinates in Earth's reference frame.

<em>Lorentz transformation</em>

The Lorentz transformation relates things between two reference frames when one of them is moving with constant velocity with respect to the other. In this case the two reference frames are the Earth and the rocket that is moving with speed v=0,70 c in the x axis.

The Lorentz transformation is

$x'=\frac{x-vt}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$y'=y$

$z'=z$

$t'=\frac{t-\frac{v}{c^{2}}x}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

prime coordinates are the ones from the rocket reference frame and unprimed variables are from the Earth's reference frame. Since we want position x and time t in the Earth's frame we need the inverse Lorentz transformation. This can be obtained by replacing v by -v and swapping primed an unprimed variables in the first set of equations

$x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$y=y'$

$z=z'$

$t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

First we calculate the expression in the denominator

$\frac{v^{2}}{c^{2}}=\frac{(0,70)^{2}c^{2}}{c^{2}} =(0,70)^{2}$

$\sqrt{1-\frac{v^{2}}{c^{2}}} =0,714$

then we calculate t

$t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$t=\frac{190s+\frac{0,70c}{c^{2}}.3,4x10^{10}m}{0,714}$

$t=\frac{190s+\frac{0,70c .3,4x10^{10}m}{299792458\frac{m}{s}}}{0,714}$

$t=\frac{190s+79,388s}{0,714}$

finally we get that

$t=377,29s$

then we calculate x

$x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$x=\frac{3,4x10^{10}m+0,70c.190s}{0,714}}$

$x=\frac{3,4x10^{10}m+0,70.299792458\frac{m}{s}.190s}{0,714}}$

$x=\frac{3,4x10^{10}m+39872396914m}{0,714}}$

$x=\frac{73872396914m}{0,714}}$

$x=103462740775,91m$

finally we get that

$x \approx 103,46x10^{9} m$

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3 years ago

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makkiz [27]

<u>Answer:</u>

<h2>All the waves are pertubations that propagate (transport) energy.</h2><h2></h2>

Nevertheless, they have some differences:

1. Light waves are<u> electromagnetic waves</u>, while sound and water waves are <u>mechanical waves</u>, this is the first and principal difference.

2. Electromagnetic waves can<u> propagate in vacuum</u> (they do not need a medium or material), but mechanical waves obligatory need a material to propagate

3. Light waves are always <u>transversal waves</u>, this means <u>the oscillatory movement is in a direction that is perpendicular to the propagation</u>; but mechanical waves may be both: <u>longitudinal waves</u> (the oscillation occurs in the same direction as the propagation) or transversal waves.

4. Electromagnetic waves propagates at a <u>constant velocity</u> (Light velocity) while the velocity of mechanical waves will depend on the type of wave and the <u>density</u> of the medium or material.

5. <u>Mechanical waves</u> are characterized by the regular variation of a single magnitude, while <u>electromagnetic waves</u> are characterized by the variation of two magnitudes: the electric field and the magnetic field

6. <u>Water waves</u> are 2-dimensional waves, while the <u>light and the sound</u> are tridimensional spherical waves

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3 years ago

Describe how one plays Dr.Dogeball

JulsSmile [24]

•To play Dr. Dodgeball you need to have 2 teams to verse each other.
•Next, select one person from each team to be the doctor (depending on the size of the teams you can have varying amounts of doctors)
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•When a player gets hit and is “out” they have to sit on the ground and wait for the doctor to “revive them” (this usually requires the doctor dragging,touching, or moving the player that is out to a “revival place” which is usually decided on by the advisor or person in charge.
•Finally, try to get all the doctors and players out from the other team. Get the doctors first, for they cannot revive themselves. Which means the other players are out after they get hit with a ball since the doctors are out. (Some games are played where if all doctors are out the game ends)
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4 0

3 years ago

Motor oil , with a viscosity of 0 . 250 Ns / m2 , is flowing through a tube that has a radius of 5 . 00 mm and is 25 . 0 cm long

Thepotemich [5.8K]

Answer:

1.1775 x 10^-3 m^3 /s

Explanation:

viscosity, η = 0.250 Ns/m^2

radius, r = 5 mm = 5 x 10^-3 m

length, l = 25 cm = 0.25 m

Pressure, P = 300 kPa = 300000 Pa

According to the Poisuellie's formula

Volume flow per unit time is

$V=\frac{\pi \times Pr^{4}}{8\eta l}$

$V=\frac{3.14 \times 300000\times \left ( 5\times 10^{-3} \right )^{4}}{8\times 0.250\times 0.25}$

V = 1.1775 x 10^-3 m^3 /s

Thus, the volume of oil flowing per second is 1.1775 x 10^-3 m^3 /s.

3 0

3 years ago