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ss7ja [257]
3 years ago
12

A weight lifter is trying to do a bicep curl with a weight of 300 N. At the "sticking point", the moment arm of this weight is 3

0 cm about the elbow axis of rotation. The maximum force that the elbow flexor muscles can create for this situation is 4000 N and the moment arm of the elbow flexors is 2 cm. Can the lifter get past this sticking point?
Physics
1 answer:
lesantik [10]3 years ago
7 0

Answer:

The weight lifter would not get past this sticking point.

Explanation:

Generally torque applied on the weight is mathematically represented as

             T =  F z

To obtain Elbow torque we substitute 4000 N for F (the force ) and 2cm = \frac{2}{100} = 0.02m for z the perpendicular distance

So Elbow Torque is   T_e= 4000 * 0.02

                                   = 80Nm

 To obtain the torque required we substitute 300 N for F and 30cm =\frac{30}{100} = 0.3 m

  So the Required Torque is T_R = 300 *0.3

                                                     =90Nm

Now since   T_e < T_R it mean that the weight lifter would not get past this sticking point

                                   

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Explanation:

The partial negative charge on oxygen and partial positive charge on hydrogen enables them to make hydrogen bond and also makes it to dissolve the the other substances having partial charges.

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A long, thin rod parallel to the y-axis is located at x = - 1 cm and carries a uniform positive charge density λ = 1 nC/m . A se
zheka24 [161]

Answer:

The electric field at origin is 3600 N/C

Solution:

As per the question:

Charge density of rod 1, \lambda = 1\ nC = 1\times 10^{- 9}\ C

Charge density of rod 2, \lambda = - 1\ nC = - 1\times 10^{- 9}\ C

Now,

To calculate the electric field at origin:

We know that the electric field due to a long rod is given by:

\vec{E} = \frac{\lambda }{2\pi \epsilon_{o}{R}

Also,

\vec{E} = \frac{2K\lambda }{R}                  (1)

where

K = electrostatic constant = \frac{1}{4\pi \epsilon_{o} R}

R = Distance

\lambda = linear charge density

Now,

In case, the charge is positive, the electric field is away from the rod and towards it if the charge is negative.

At x = - 1 cm = - 0.01 m:

Using eqn (1):

\vec{E} = \frac{2\times 9\times 10^{9}\times 1\times 10^{- 9}}{0.01} = 1800\ N/C

\vec{E} = 1800\ N/C     (towards)

Now, at x = 1 cm = 0.01 m :

Using eqn (1):

\vec{E'} = \frac{2\times 9\times 10^{9}\times - 1\times 10^{- 9}}{0.01} = - 1800\ N/C

\vec{E'} = 1800\ N/C     (towards)

Now, the total field at the origin is the sum of both the fields:

\vec{E_{net}} = 1800 + 1800 = 3600\ N/C

7 0
3 years ago
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