20 points + Brainliest. Help please!

To develop muscle tone, a woman lifts a 2.50 kg weight held in her hand. She uses her biceps muscle to flex the lower arm throug

Romashka [77]

To solve this problem we will use the concepts related to Torque as a function of the Force in proportion to the radius to which it is applied. In turn, we will use the concepts of energy expressed as Work, and which is described as the Torque's rate of change in proportion to angular displacement:

$\tau = Fr$

Where,

F = Force

r = Radius

Replacing we have that,

$\tau = Fr$

$\tau = 21cm (\frac{1m}{100cm})* 550N$

$\tau = 11.55Nm$

The moment of inertia is given by 2.5kg of the weight in hand by the distance squared to the joint of the body of 24 cm, therefore

$I = 0.25Kg\cdot m^2 +(2.5kg)(0.24m)^2$

$I = 0.394kg\cdot m^2$

Finally, angular acceleration is a result of the expression of torque by inertia, therefore

$\tau = I\alpha \rightarrow \alpha = \frac{\tau}{I}$

$\alpha = \frac{11.55}{0.394}$

$\alpha = 29.3 rad/s^2$

PART B)

The work done is equivalent to the torque applied by the distance traveled by 60 °° in radians $(\pi / 3)$ , therefore

$W = \tau \theta$

$W = 11.5* \frac{\pi}{3}$

$W = 12.09J$

4 0

3 years ago

The dot or scalar product of two (3d) vec- tors ⃗a = ⟨a1,a2,a3⟩ and ⃗b = ⟨b1,b2,b3⟩ is defined as

Neko [114]

Yes, yes, we know all of that. It certainly took you long enough to
get around to asking your question.

If
   a = (14, 10.5, 0)
and
   b = (4.62, 9.45, 0) ,

then, to begin with, neither vector has a z-component, and they
both lie in the x-y plane.

Their dot-product a · b = (14 x 4.62) + (10.5 x 9.45) =

         (64.68)   +   (99.225) = 163.905 (scalar)

I feel I earned your generous 5 points just reading your treatise and
finding your question (in the last line). I shall cherish every one of them.

7 0

3 years ago

Circular Motion A 650-kg car moving at 8.5 m/s takes a turn around a circle with a radius of 48.0 m. Determine the acceleration

ollegr [7]

Answer:

Explanation:

The mass of the car doesn't matter because On a flat curve the mass of the car does not affect the speed at which it can stay on the curve. You would need the mass if you were solving the the centripetal force acting on the car, but not the acceleration.

$a=\frac{v^2}{r}$ and filling in