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yaroslaw [1]
3 years ago
8

A weightlifter is attempting a biceps curl with a 200 n barbell. the moment arm of the barbell about the elbow joint is 40 cm. t

he moment arm of the elbow flexor muscles is 4 cm. how much force must be produced by these muscles to hold the 200 n barbell in static equilibrium?
Physics
1 answer:
CaHeK987 [17]3 years ago
4 0
Weight of the barbell W = 200 Ndistance of the joint is r = 40 cm = 0.4 mtorque created by the weight at the joint is                  τ = F*r                     = 200 N*0.4 m                     = 80 N.mat equilibrium condition ,    Στ = force*distance - 80 N.m = 0             F'*0.4 - 80 N.m = 0             F'*0.4 = 80          force F' = 200 N
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A 95kg fullback (football player for those not into sports) moving south with a speed of 5.0 m/s has a perfectly inelastic colli
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Answer:

a.  v=3.11mls, 29.4^{0}

b.   K.E =-697.8J

Explanation:

To calculate the values in the  question, a deep understanding of perfect inelastic collision is important.

When two bodies undergo inelastic collision, two important parameters must be well understood i.e

Momentum: the momentum is always conserved in perfectly inelastic collision. i.e the total momentum after collision is the sum of the individual momentum before collision

Kinetic energy: Kinetic energy is not conserved due to dissipative force.

a.To calculate the velocity, we first find the total momentum before collision

Momentum of player 1 p_{1} =mv=95kg*5m/s\\p_{1} =475kgm/s\\

Momentum of player 2 p_{2} =mv=90kg*3m/s\\p_{1} =270kgm/s\\

Hence the total momentum p_{12}=p_{1}+p_{2}\\

Note, since the direction of movement before collision is due south and  due north respectively we have to represent the velocity using the rectangular coordinate

Hence  p_{12}=(m_{1}+m_{2})v=p_{1}i+p_{2}j\\

(95+90)v=475i+270j\\

v=2.57i+1.45j\\

solving for the resultant velocity, we have

v=\sqrt{2.75^{2} +1.45^{2}}\\ v=3.11mls

To calculate the direction of movement, we have

\alpha =tan^{-1}=\frac{v_{j} }{v_{i}}\\  \alpha =tan^{-1}=\frac{1.45}{2.57}\\\alpha =29.4^{0}

b. to calculate the decrease in total kinetic energy, before collision, the total kinetic was

K.E_{initial} =\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}.\\K.E_{initial} =((1/2)*95*5^{2})+((1/2)*90*3^{2})\\K.E_{initial} =1187.5+405\\K.E_{initial} =1592.5J\\

And the final kinetic energy after collision is

K.E_{final} =\frac{1}{2}(m_{1}+m_{2} )v^{2}\\  K.E_{final} =\frac{1}{2}(95+90)* 3.11^{2}\\ K.E_{final} =894.7J

The decrease in Kinetic energy is

K.E =K.E_{final}- K.E_{initial}=894.7-1592.5

K.E =-697.8J

The negative sign indicate a decrease in Kinetic energy

4 0
3 years ago
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pentagon [3]

Answer:

B) x^2+6x+8

Explanation:

x-4 | x^3+2x^2-16x-32

    -  x^3-4x^2             <-- (x-4)(x^2)

_________________

              6x^2-16x-32

           -  6x^2-24x     <-- (x-4)(6x)

_________________

                          8x-32

                       -  8x-32 <- (x-4)(8)

___________________________

                                 0 | x^2+6x+8

This means the answer is B) x^2+6x+8

             

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Answer:

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