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3 years ago

14

The triceps muscle in the back of the upper arm extends the forearm. This muscle in a professional boxer exerts a force of 1783

N with an effective perpendicular lever arm of 2.80 cm , producing an angular acceleration of the forearm of 120.0 rad / s 2 . What is the moment of inertia of the boxer's forearm

1 answer:

denis23 [38]3 years ago

6 0

Answer:I=0.416kgm^2

Explanation:

Torque is the rotational equivalence of Force and mathematicallly, it is given by

T = r X F

Where

T=Torque

r= perpendicular distance between the applied force and the axis

F =force

T= 0.028 X 1783 =49.924 Nm

T=αI

Where α=angular acceleration produ ed

I=moment of inertia

I= T/α

I= 49.924/120

I= 0.416kgm^2

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A copper cylinder is initially at 21.1 ∘C . At what temperature will its volume be 0.163 % larger than it is at 21.1 ∘C?

fomenos

To solve this problem we will apply the concepts related to the final volume of a body after undergoing a thermal expansion. To determine the temperature, we will use the given relationship as well as the theoretical value of the volumetric coefficient of thermal expansion of copper. This is, for example to the initial volume defined as $V_1$ , the relation with the final volume as

$V_2 = V_1 +0.163\% V_1$

$V_2 = V_1 +0.00163V_1$

$V_2 = 1.00163V_1$

Initial temperature = $21.1\°C$

Let T be the temperature after expanding by the formula of volume expansion

we have,

$V_2 = V_1 (1+\gamma \Delta t)$

Where $\gamma$ is the volume coefficient of copper $5.1*10^{-5}/C$

$1.00163V_1 = V_1(1+\gamma(T-21.1\°))$

$1.00163 = 1+5.1*10^{-5}(T-21.1\°)$

$0.00163 = 0.000051T-0.0010761$

$T = 53.0608\°C$

Therefore the temperature is 53.06°C

7 0

3 years ago

Runner A is initially 5.7 km west of a flagpole and is running with a constant velocity of 8.9 km/h due east. Runner B is initia

Oksanka [162]

Answer:

B meet A 0.01 km east of flagpole

Explanation:

given data

distance A = 5.7 km west

velocity V1 = 8.9 km/h

distance B = 4.5 km east

velocity V2 = 7 km/h

to find out

How far runners from the flagpole, when paths cross

solution

we know A and B are 5.7 + 4.5 = 10.2 km apart

and we consider here B will run distance x km for meet

so time will be for B is

time B = distance / velocity

time B = x / 7 ...................1

and

for A distance for meet = ( 10.2 - x ) km

so time A = distance / velocity

time A = ( 10.2 - x ) / 8.9 .............2

now equating equation 1 and 2

time A = time B

x / 7 = ( 10.2 - x ) / 8.9

x = 4.490

so distance of B run for meet is 4.490 km

so distance from the flagpole when their paths cross is 4.5 - 4.490 = 0.01 km

so B meet A 0.01 km east of flagpole

4 0

2 years ago

A charge is divided q1 and (q-q1)what will be the ratio of q/q1 so that force between the two parts placed at a given distance i

Arturiano [62]

Answer:

$q / q_{1} = 2$ , assuming that $q_{1}$ and $(q - q_{1})$ are point charges.

Explanation:

Let $k$ denote the coulomb constant. Let $r$ denote the distance between the two point charges. In this question, neither $k$ and $r$ depend on the value of $q_{1}$ .

By Coulomb's Law, the magnitude of electrostatic force between $q_{1}$ and $(q - q_{1})$ would be:

$\begin{aligned}F &= \frac{k\, q_{1}\, (q - q_{1})}{r^{2}} \\ &= \frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\end{aligned}$ .

Find the first and second derivative of $F$ with respect to $q_{1}$ . (Note that $0 < q_{1} < q$ .)

First derivative:

$\begin{aligned}\frac{d}{d q_{1}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\right] \\ &= \frac{k}{r^{2}}\, \left[\frac{d}{d q_{1}} [q\, q_{1}] - \frac{d}{d q_{1}}[{q_{1}}^{2}]\right]\\ &= \frac{k}{r^{2}}\, (q - 2\, q_{1})\end{aligned}$ .

Second derivative:

$\begin{aligned}\frac{d^{2}}{{d q_{1}}^{2}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q - 2\, q_{1})\right] \\ &= \frac{(-2)\, k}{r^{2}}\end{aligned}$ .

The value of the coulomb constant $k$ is greater than $0$ . Thus, the value of the second derivative of $F$ with respect to $q_{1}$ would be negative for all real $r$ . $F\!$ would be convex over all $q_{1}$ .

By the convexity of $\! F$ with respect to $\! q_{1} \!$ , there would be a unique $q_{1}$ that globally maximizes $F$ . The first derivative of $F\!$ with respect to $q_{1}\!$ should be $0$ for that particular $\! q_{1}$ . In other words:

$\displaystyle \frac{k}{r^{2}}\, (q - 2\, q_{1}) = 0$ <em>.</em>

$2\, q_{1} = q$ .

$q_{1} = q / 2$ .

In other words, the force between the two point charges would be maximized when the charge is evenly split:

$\begin{aligned} \frac{q}{q_{1}} &= \frac{q}{q / 2} = 2\end{aligned}$ .

3 0

2 years ago

What is electrical energy

creativ13 [48]

Explanation:

it is energy that flows of electric charge it has the ability to do work or apply force to move an object.

8 0

2 years ago

The half-life of the radioactive isotope Carbon-14 is about 5730 years. Find N in terms of t. The amount of a radioactive elemen

Fudgin [204]

The complete queston is The amount of a radioactive element A at time t is given by the formula

A(t) = A₀e^kt

Answer: A(t) =N e^( -1.2 X 10^-4t)

Explanation:

Given

Half life = 5730 years.

A(t) =A₀e ^kt

such that

A₀/ 2 =A₀e ^kt

Dividing both sides by A₀

1/2 = e ^kt

1/2 = e ^k(5730)

1/2 = e^5730K

In 1/2 = 5730K

k = 1n1/2 / 5730

k = 1n0.5 / 5730

K= -0.00012 = 1.2 X 10^-4

So that expressing N in terms of t, we have

A(t) =A₀e ^kt

A₀ = N

A(t) =N e^ -1.2 X 10^-4t

7 0

2 years ago