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\times 10^32.00×10 ^3 N with an effective perpendicular lever arm of 3.00 cm, producing an angular acceleration of the forearm of 120 rad/s^2 .
What is the moment of inertia of the boxer's forearm?

Answers

Answer 1

Answer:

Explanation:

From the given information:

The torque produced due to the force can be expressed as:

[tex]\tau = F \times r[/tex]

where;

[tex]\tau[/tex] = torque

F = force exerted

r = lever's arm radius

[tex]\tau[/tex] = [tex]2.00 \times 10^3 \times 0.03 m[/tex]

[tex]\tau[/tex] = 60 N.m

However, equating the torque with the moment of inertia & angular acceleration, we use the equation:

[tex]\tau[/tex] = I∝

60 Nm = I × 120 rad/s²

I = 60 Nm/120 rad/s²

I = 0.5 kg.m²


Related Questions

45. Pressure in air undergoes a decrease when the air
a) rises to higher altitudes.
b) accelerates to higher speed.
c) fills a greater space.
d) All of these.

Answers

D okokokokokokok I’m right promise

A lumberjack is trying to drag a small tree that he cut down. If the static
coefficient of friction of the tree on the ground is 0.5 and the tree weighs 430
N, what is the minimum amount of horizontal force that he will need to apply
so that the tree will start moving?
A. 215 N
B. 430 N
C. 365 N
D. 500 N

Answers

Answer:

A

Explanation:

weight of the tree =normal force

Horizontal force =coefficient of friction x Fnormal

0.5×430=215

Two identical conductors have charge -1.8 C and 5.5 C on them, respectively. They are connected by a conducting wire for a short period of time and then disconnected. What is the net charge on each of the conductors after the interaction? g

Answers

Answer: 1.85 C

Explanation:

Given

charges on the conductors are [tex]-1.8\ C[/tex] and [tex]5.5\ C[/tex]

They are connected by a conducting wire for a short period of time and then disconnected. During this time charge flow from the wire and net charge becomes [tex]5.5-1.8=3.7\ C[/tex]

This charge will be equally distribute among the two conductors i.e. 1.85 C on each conductor.

Which phenomenon occurs when one wave is superimposed on another?
A. Interference
B. Refraction
C. Diffraction
D. Polarization

Answers

Answer:Alternativa A.   Damos o nome de interferência a superposição de efeitos que ocorre ao ser produzido dois pulsos de onda, que serão propagados e acabarão inevitavelmente por se encontrar. No instante em que os pulsos se cruzarem, há então, uma superposição de efeitos individuais de cada um deles. Se durante o cruzamento, houver um reforço das ondas, estará ocorrendo a este fenômeno.

the two factors that affect the amount of heat​

Answers

Answer:

The two important factors that affect heat energy are specific heat and temperature. Specific heat is a heat-constant of a material per unit mass per degree of temperature change (in units of energy per mass and temperature), like Joules/Kg-°C .

Thank you.....

Have a good day.....

If the distance between the center of two objects is quadrupled. The gravitational
force between the two objects will change by a factor of:
1) 16
2) 0.25
3) 4
4) 0.0625

Answers

Answer:

F' = F/16

Explanation:

The gravitational force between masses is given by :

[tex]F=G\dfrac{m_1m_2}{r^2}[/tex]

If the distance between the center of two objects is quadrupled, r' = 4r

New force will be :

[tex]F'=G\dfrac{m_1m_2}{r'^2}\\\\F'=G\dfrac{m_1m_2}{(4r)^2}\\\\F'=\dfrac{Gm_1m_2}{16r^2}\\\\F'=\dfrac{1}{16}\times \dfrac{Gm_1m_2}{r^2}\\\\F'=\dfrac{F}{16}[/tex]

So, the new force will change by a factor of 16.

The equation of damped oscillations is given in the form x=0.05e^-0.25sin½πt (m). Find the velocity of an oscillating point at the moments of time: 0, T, 2T, 3T and 4T.​

Answers

Explanation:

The logarithmic damping decrement of a mathematical pendulum is DeltaT=0.5. How will the amplitude of oscillations decrease during one full oscillation of the pendulum

12. A car is travelling at 30 m/s when the driver sees a red light in the distance and immediately applies the brakes. The car comes to a stop 1.5 s later. How far did the car move from the time the driver applied the brake to when it came to a stop?​

Answers

Answer:

22.5 m

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 30 m/s

Time (t) = 1.5 s

Final velocity (v) = 0 m/s

Distance (s) =?

The distance to which the car move before stopping from the time the driver applied the brake can be obtained as follow:

s = (u + v)t/2

s = (30 + 0)1.5 / 2

s = (30 × 1.5) / 2

s = 45 / 2

s = 22.5 m

Thus, the car will move to a distance of 22.5 m before stopping from the time the driver applied the brake.

A sack of groceries with a mas of 22 kg is lifted off the floor with a velocity of 6 m/s. What is the kinetic energy of the sack
of groceries?

Answers

the answer is 396 joules :D

Use the simulation to compare the masses of the three colored and unlabeled weights of different sizes. To do so, set the spring constant of both springs to the same value. Hang known weights on the left spring and an unknown weight on the right spring, and compare the two. Use as many known weights as necessary to determine the unknown masses, and then place each into the appropriate mass bins in the ranking task below. A. M<50 g
B. M = 50 g
C. 50 g D. M = 100 g
E. 100 g F. M = 250 g
G. M> 250 g
1. Blue medium sized weight
2. Magenta small sized weight
3. Gold large sized weight

Answers

Answer:

The answer is given as follows,

Explanation:

Gold large-sized weight 100 g < M < 250g

50 g < Magenta small-sized weight < 100g

100g < Blue medium-sized weight < 250g

Hence,

100g < Blue medium-sized weight < 250g

50 g < Magenta small-sized weight < 100g

100 g < Gold large-sized weight < 250g.

A canoe has a velocity of 0.330 m/s southeast relative to the earth. The canoe is on a river that is flowing at 0.540 m/s east relative to the earth. Find the velocity (magnitude and direction) of the canoe relative to the river.

Answers

Answer:

The velocity of the canoe relative to the river is 0.385 m/s, S37.26⁰W

Explanation:

Given;

velocity of the canoe relative to the earth, [tex]V_{r/e} = 0.33 \ m/s[/tex]

velocity of the river relative to the earth, [tex]V_{r/e} = 0.54 \ m/s[/tex]

The velocity of the canoe relative to the river is calculated as;

[tex]V_{(c/r)x} = V_{(c/e)x}- V_{(r/e)x} \ \ ----(1)\\\\V_{(c/r)y} = V_{(c/e)y}- V_{(r/e)y} \ \ ----(2)[/tex]

The x - component of the velocity of the canoe relative to the earth;

[tex]V_{(c/e)x} = 0.33 \times cos \ 45^0\\\\V_{(c/e)x} = 0.2333 \ m/s[/tex]

The y-component of the velocity of the canoe relative to the earth;

[tex]V_{(c/e)y} = 0.33 \times sin \ 45^0\\\\V_{(c/e)y} = 0.2333 \ m/s[/tex]

Note: velocity of the river relative to the earth has only x-component = 0.54 m/s

Apply equation (1) and (2) to calculate the velocity of the canoe relative to the river;

[tex]V_{(c/r)}x = 0.2333 - 0.54 = -0.3067 \ m/s\\\\V_{(c/r)}y = 0.2333 - 0 = 0.2333 \ m/s\\\\The \ resultant \ velocity;\\\\V_{c/r} = \sqrt{(-0.3067)^2 + (0.2333)^2} \\\\V_{c/r} = 0.385 \ ms/\\\\The \ direction:\\\\\theta = tan^{-1} (\frac{0.2333}{0.3067} ) = 37.26^0 \ south \ west \ of \ the \ river[/tex]

In a football game a kicker attempts a field goal. The ball remains in contact with the kicker's foot for 0.0342 s, during which time it experiences an acceleration of 186 m/s2. The ball is launched at an angle of 45.9 ° above the ground. Determine the (a) horizontal and (b) vertical components of the launch velocity.

Answers

Answer:

b)      v_y = 4.57 m / s

a)      vₓ = 4.43 m / s

Explanation:

This is an exercise in kinematics, where we assume that the acceleration is in the direction of the force and the initial body with zero velocity

          v = v₀ + a t

          v = 0 + a t

          v = 186  0.0342

          v = 6.36 m / s

let's use trigonometry to decompose this velocity

           sin 45.9 = v_y / v

           cos 45.9 = vₓ / v

           v_y = v sin 45.9

           vₓ = v cos 45.9

           v_y = 6.36 sin 45.9

           vₓ = 6.36 cos 45.9

           v_y = 4.57 m / s

           vₓ = 4.43 m / s

An astronaut throws a wrench in interstellar space. How much force is required to keep the wrench moving continuously with constant velocity?
A.
a force equal to its weight on Earth
B.
a force equal to zero
C.
a force equal to half of its weight on Earth
D.
a force equal to double its weight on Earth

Answers

Answer:

0 N

Explanation:

This is a trick question, the mass of the wrench would be 0 due to it being in space and has no gravitational pull to weight it down. And since acceleration is defined as the rate and change of velocity with no respect of time and the wrench is moving at a constant velocity, that means the velocity is 0. and since F = m*a it would be F = 0 * 0 = 0 N

Which term describes friction that acts on a stationary object?
O A. Static friction
B. Sliding friction
C. Kinetic friction
D. Resistance friction

Answers

Answer:

Static friction

Explanation:

Answer:

Static friction

Explanation:

What is the source of almost all energy on Earth?

Earth’s hot core
the Sun
stored carbon
moving water

Answers

Answer:

the sun

Explanation:

The sun is the source of almost all energy on Earth beacuse both plants and animal on Earth derive their energy from the sun.

Source of energy on Earth

The Earth is one of the planets that make up the solar system. The Sun is the center of the universe and the Earth revolves round the sun.

The source of almost all energy on Earth is from the sun. The energy from the sun is callled solar energy.

This energy from the sun can be used by the planet to manufatcure its own food. The plants are consumed by animals to provide energy their metabolic activities.

Thus, we can conclude that the sun is the source of almost all energy on Earth beacuse both plants and animal on Earth derive their energy from the sun.

Learn more about solar energy here: https://brainly.com/question/17711999

Dr. John Paul Stapp was a U.S. Air Force officer who studied the effects of extreme acceleration on the human body. On December 10, 1954, Stapp rode a rocket sled, accelerating from rest to a top speed of 282 m/s (1015 km/h) in 5.2 s and was brought jarringly back to rest in only 1 s. Calculate his (a) magnitude of acceleration in his direction of motion and (b) magnitude of acceleration opposite to his direction of motion. Express each in multiples of g (9.80 m/s2) by taking its ratio to the acceleration of gravity. g g

Answers

Answer:

    a = 5.53 g ,   a = -15g

Explanation:

This is an exercise in kinematics.

a) Let's look for the acceleration

         as part of rest v₀ = 0

          v = v₀ + a t

           a = v / t

           a = 282 / 5.2

          a = 54.23 m / s²

in relation to the acceleration of gravity

          a / g = 54.23 / 9.8

          a = 5.53 g

b) let's look at the acceleration to stop

         va = 0

         0 = v₀ -2 a y

         a = vi / y

         a = 282/2 1

         a = 141 m /s²

         a / G = 141 / 9.8

          a = -15g

A laser emits a single 3.0-ms pulse of light that has a frequency of 2.83E11 Hz and a total power of 65000 W. How many photons are in the pulse? Please provide all equations and work.

6.0E23
1.0E24
2.4E25
3.6E25
4.8E26

Answers

Answer:

The number of photons in the pulse is 1.04 x 10²⁴

Explanation:

Given;

frequency of the emitted photons, f = 2.83 x 10¹¹ Hz

duration of the incident light, t = 3 ms = 3 x 10⁻³ s

power of the incident light, P = 65,000 W

The energy of each photon emitted is calculated as;

E = hf

where;

h is Planck's constant, = 6.626 x 10⁻³⁴ Js

E =  6.626 x 10⁻³⁴ x  2.83 x 10¹¹

E = 1.875 x 10⁻²² J

let the number of photons in the pulse = n

n(E)= Power x time

[tex]n = \frac{Pt}{E} \\\\n = \frac{65,000 \times 3\times 10^{-3}}{1.875 \times 10^{-22}} \\\\n = 1.04 \times 10^{24} \ photons[/tex]

What is the energy equivalent of an object with a mass of 1.05g?​

Answers

Answer:

The equivalent energy of an object given its mass is calculated through the equation,

                             E = mc²

where c is the speed of light (3 x 10^8 m/s)

Substituting the known values,

                            E = (1.05 g/ 1000) (3 x 10^8 m/s)²

                               E = 9.45x10^13 J

Explanation:

The correct formula for finding the relative velocity of an object is:

WILL MARK BRAINLIEST TO THE CORRECT ANSWER!!

Answers

Answer:

[tex]V_{a/c} = V_{a/b} + V_{b/c}[/tex]

Explanation:

The relative velocity of an object is the velocity of the object relative to the observer or frame of reference.

The velocity of particle "A" with respect to particle "B" is written as [tex]V_{A/B} = V_{A} - V_{B}[/tex]

From the given options, the second option is the correct answer.

[tex]V_{a/c} = V_{a/b} + V_{b/c}\\\\Re-arrange \ the \ above \ equation;\\\\V_{a/c} - V_{b/c}= V_{a/b}\\\\or\\\\V_{a/b}= V_{a/c} - V_{b/c}[/tex]

A 10 kg block rests on a 30o inclined plane. The block is attached to a bucket by pulley system depicted below. The mass in the bucket is gradually increased by the addition of sand. At some point, the bucket will accumulate enough sand to set the block in motion. The coefficients of static and kinetic friction are 0.60 and 0.50 respectively.

Required:
a. Determine the mass of sand in [kg], including the bucket, needed to start the block moving.
b. Find the blocks acceleration, in [m/s^2] up the plane?

Answers

Answer:

a). M = 20.392 kg

b). am = 0.56 [tex]m/s^2[/tex] (block),  aM = 0.28 [tex]m/s^2[/tex] (bucket)

Explanation:

a). We got  N = mg cos θ,

                  f = [tex]$\mu_s N$[/tex]

                    = [tex]$\mu_s mg \cos \theta$[/tex]

If the block is ready to slide,

T = mg sin θ + f

T = mg sin θ + [tex]$\mu_s mg \cos \theta$[/tex]   .....(i)

2T = Mg ..........(ii)

Putting (ii) in (i), we get

[tex]$\frac{Mg}{2}=mg \sin \theta + \mu_s mg \sin \theta$[/tex]

[tex]$M=2(m \sin \theta + \mu_s mg \cos \theta)$[/tex]

[tex]$M=2 \times 10 \times (\sin 30^\circ+0.6 \cos 30^\circ)$[/tex]

M = 20.392 kg

b). [tex]$(h-x_m)+(h-x_M)+(h'+x_M)=l$[/tex]  .............(iii)

   Here, l = total string length

Differentiating equation (iii) double time w.r.t t, l, h and h' are constants, so

[tex]$-\ddot{x}-2\ddot x_M=0$[/tex]

[tex]$\ddot x_M=\frac{\ddot x_m}{2}$[/tex]

[tex]$a_M=\frac{a_m}{2}$[/tex]   .....................(iv)

We got,   N = mg cos  θ

                [tex]$f_K=\mu_K mg \cos \theta$[/tex]

∴ [tex]$T-(mg \sin \theta + f_K) = ma_m$[/tex]

  [tex]$T-(mg \sin \theta + \mu_K mg \cos \theta) = ma_m$[/tex]  ................(v)

Mg - 2T = M[tex]a_M[/tex]

[tex]$Mg-Ma_M=2T$[/tex]

[tex]$Mg-\frac{Ma_M}{2} = 2T$[/tex]    (from equation (iv))

[tex]$\frac{Mg}{2}-\frac{Ma_M}{4}=T$[/tex]   .....................(vi)

Putting (vi) in equation (v),

[tex]$\frac{Mg}{2}-\frac{Ma_M}{4}-mf \sin \theta-\mu_K mg \cos \theta = ma_m$[/tex]

[tex]$\frac{g\left[\frac{M}{2}-m \sin \theta-\mu_K m \cos \theta\right]}{(\frac{M}{4}+m)}=a_m$[/tex]

[tex]$\frac{9.8\left[\frac{20.392}{2}-10(\sin 30+0.5 \cos 30)\right]}{(\frac{20.392}{4}+10)}=a_m$[/tex]

[tex]$a_m= 0.56 \ m/s^2$[/tex]

Using equation (iv), we get,

[tex]a_M= 0.28 \ m/s^2[/tex]

A positively charged plastic ruler is brought close to a piece paper resting on the desk. The piece of paper was initially neutral. When the ruler was brought closer, the paper is attracted to the ruler. The surface of the paper became charged through:_________

Answers

Answer: static electricity

Explanation:

When the plastic ruler is rubbed, friction opposes the motion and causes the transfer of electron from one surface to another such that plastic becomes negatively charged. When ruler is brought nearer to the paper, it induces the   positive charge in the piece of paper.

If a sprinter ran a distance of 100 meters starting at his top speed of 11 m/s and running with constant spreed throughout. How long would it take him to cover the distance?

Answers

Answer:

9.09 s

Explanation:

If the sprinter ran the 100 meters at the constant speed of 11 m/s it would take him 9.09 s to cover the full distance.

We can find this number by dividing 100 meters (the distance covered) by 11 meters per second (the speed)

[tex]\frac{100}{11} =9.09[/tex]

define a system who's momentum is observed​

Answers

product of force and perpendicular distance

Suppose that 2 J of work is needed to stretch a spring from its natural length of 32 cm to a length of 46 cm. (a) How much work (in J) is needed to stretch the spring from 37 cm to 41 cm

Answers

Answer:

the work required is 0.163 J

Explanation:

Given;

Energy applied to the spring, E = 2 J

initial length of the spring, x₀ = 32 cm

final length of the spring, x₁ = 46 cm

Extension of the spring, Δx = x₁ - x₀ = 46 cm - 32 cm = 14 cm = 0.14 m

The spring constant is calculated as follows;

E = ¹/₂kΔx²

[tex]k = \frac{2E}{(\Delta x)^2} = \frac{2\times 2}{(0.14)^2} = 204.1 \ N/m^2[/tex]

The extension of the spring when it is stretched from 37 cm + 41 cm:

Δx =  41 cm - 37 cm = 4 cm = 0.04 m

The work required:

W = ¹/₂kΔx²

W = ¹/₂ x (204.1) x (0.04)²

W = 0.163 J

Therefore, the work required is 0.163 J

URGENTT

Which statement best defines the term "superconductivity"?

Answers

Answer:

the ability of certain substances at very low temperatures to conduct electricity with no resistance

A spring of spring constant k=8.25N/m is displaced from equilibrium by a distance of 0.150m. What is the stored energy in the form of spring potential energy? Show your work.

Answers

Answer:

0.0928J

Explanation:

the pulling force of spring F=-kx

where x is the displacement from equilibrium position.

energy stored:

[tex]\int\limits^x_0 {-F} \, dx \\=\int\limits^x_0 {kx} \, dx \\\\=\frac{kx^{2} }{2}[/tex]

*** Its fine if you know nothing about calculus. Just apply the equation

    [tex]U=\frac{kx^{2} }{2}[/tex]

  where U is the potential energy of the spring***

  put x=0.150, [tex]U=\frac{8.25}{2}[/tex]×[tex]0.150^{2}[/tex] = 0.0928J (corr. to 3 sig. fig.)


List the 5 theoretical perspectives that underlie much of the research on human development. Also, name an individual strongly associated with each perspective.

Answers

Answer:

25

Explanation:

5 theroical name indvivdual perspective asssssoitive each persp

Your hand and wrist curl in toward the center of your body (chest and stomach) to prepare to throw the frisbee.
O True
O False

Answers

True

Hope this helps! :)

Answer:

true because when trow the frisbee gives u level

semiconductor have negative temperature coefficient of resistance why​

Answers

Answer:

As the number of free electrons increases, the resistance of this type of non-metallic material decreases with increasing temperature.

Explanation:

The energy wasted in using a machine is 600j. if the machine is 70% efficient. calculate the volume of water pumb by the machine through a height of 15m.​

Answers

Answer:

yhgigy6ftu5cg l8vbbnnbbgtccccvhklhaywje nc 62bbnzmakbdbvfvdbf93bdldmffmfkqhdv

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