Physics, 20.10.2020 18:01 donttrip18

What is the magnitude of the force exerted by the biceps FbicepsFbicepsF_biceps? What is the magnitude of the force exerted by the elbow FelbowFelbowF_elbow

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Answer from: jfkdhrmd

The force of forearm is 239.9 N

The force of elbow is 215.89 N

Explanation:

Suppose, When you lift an object by moving only your forearm, the main lifting muscle in your arm is the biceps. Suppose the mass of a forearm is 1.50 kg. If the biceps is connected to the forearm a distance $d_{b}$ 2.50 cm from the elbow, how much force $F_{b}$ must the biceps exert to hold a 950 g ball at the end of the forearm at distance dball J 36.0 cm from the elbow, with the forearm parallel to the floor? How much force $F_{l}$ must the elbow exert,

Given that,

Mass of forearm = 1.50 kg

Distance of forearm = 2.50 cm

Mass of ball = 950 g

Distance of ball = 36.0 cm

We need to calculate the force of forearm

Using balancing torque about elbow

$F_{b}\times d_{b}=w_{f}\times\dfrac{d_{f}}{2}+w_{ball}\times d_{ball}$

Put the value into the formula

$F_{b}\times 0.025= 1.50\times9.8\times\dfrac{0.36}{2}+0.95\times9.8\times0.36$

$F_{b}=\dfrac{1.50\times9.8\times\dfrac{0.36}{2}+0.95\times9.8\times0.36}{0.025}$

$F_{b}=239.9\ N$

We need to calculate the force of elbow

Using balancing force

$F_{b}=F_{l}+w_{f}+w_{b}$

$F_{l}=F_{b}-w_{f}-w_{b}$

Put the value into the formula

$F_{l}=239.9-(1.50\times9.8)-(0.95\times9.8)$

$F_{l}=215.89\ N$

Hence, The force of forearm is 239.9 N

The force of elbow is 215.89 N

What is the magnitude of the force exerted by the biceps FbicepsFbicepsF_biceps? What is the magnitu

Answer from: Quest

a.metals can become magnetic if they have many spinning electrons that create magnetic fields.d.only some metals are attracted by magnets; these metals are called ferromagnetic.

d.wrapping an insulated wire around a metal with ferromagnetic properties and applying an electric current.

Answer from: Quest

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

so, refractive index of medium = 300000/200000 = 1.5

1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

∞ refractive index is usually for opaque mediums. so third medium is opaque medium.

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