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2 years ago

Need help with #6. The digestive system Section Review.

Physics

1 answer:

uysha [10]2 years ago

5 0

1)Food is crushed and ground in the mouth by teeth
2)Chewed food is pushed into the pharynx
3)Food enters the esophagus
4)Food passes through a valve and into the stomach
5)Food is broken down by pancreatic enzymes and absorbed by the small intestine
6)Undigested material is pushed into the large intestine
7)Waste material is compressed in the rectum and eliminated through the anus

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An object with a mass of 2.0 kg accelerates 2.0 m/s2 when an unknown force is applied to it. What is the amount of the force?

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Answer:

4 Newton’s

Explanation:

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Sophie [7]

Decompose the forces acting on the block into components that are parallel and perpendicular to the ramp. (See attached free body diagram. Forces are not drawn to scale)

• The net force in the parallel direction is

∑ F (para) = -mg sin(21°) - f = ma

• The net force in the perpendicular direction is

∑ F (perp) = n - mg cos(21°) = 0

Solving the second equation for n gives

n = mg cos(21°)

n = (0.200 kg) (9.80 m/s²) cos(21°)

n ≈ 1.83 N

Then the magnitude of friction is

f = µn

f = 0.25 (1.83 N)

f ≈ 0.457 N

Solve for the acceleration a :

-mg sin(21°) - f = ma

a = (-0.457N - (0.200 kg) (9.80 m/s²) sin(21°))/(0.200 kg)

a ≈ -5.80 m/s²

so the block is decelerating with magnitude

a = 5.80 m/s²

down the ramp.

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2 years ago

If you hold your arm outstretched with palm upward, the force to keep your arm from falling comes from your deltoid muscle. assu

Maurinko [17]

Your question seems to be incorrect. Please check below:

What force must the deltoid muscle provide to keep the arm in this position? By what factor does this force exceed the weight of the arm?If you hold your arm outstretched with palm upward, as in (Figure 1) , the force to keep your arm from falling comes from your deltoid muscle. Assume that the arm has mass 4 kg and the distances and angles shown in (Figure 1) .

F=?
F/w= ?

The answer is 339 N

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2 years ago

Two objects, T and B, have identical size and shape and have uniform density. They are carefully placed in a container filled wi

Korvikt [17]

Complete Question:

Two objects, T and B, have identical size and shape and have uniform density. They are carefully placed in a container filled with a liquid. Both objects float in equilibrium. Less of object T is submerged than of object B, which floats, fully submerged, closer to the bottom of the container. Which of the following statements is true?

Object T has a greater density than object B.
Object B has a greater density than object T.
Both objects have the same density.

Answer:

Object B has a greater density than object T

Explanation:

Any object partially or completely submerged in a liquid, experiments an upward force, equal to the weight of the volume displaced by the liquid. This force is called the buoyant force, and can be expressed as follows:

Fb = ρl * Vs*g

where ρl is the density of the liquid, and Vs is the submerged volume.

This force must be compared with the weight of the object, which is always downward, and can be expressed as follows:

Fg = ρb* Vb * g

where ρb, is the density of the object, and Vb is the total volume of the object, regardless which portion is submerged.

For object B, as it floats fully submerged, this means that both forces are equal in magnitude:

Fg = Fb⇒ ρb* Vb * g = ρl * Vs*g

As Vb = Vs (the object is fully submerged) this means that ρb =ρl.

For object T, as it floats partially submerged, this means that Fg < Fb:

Fg= ρt* Vt * g < Fb = ρl * Vs*g.

Now, we know that ρb =ρl, so we can replace in the equation above:

ρT* Vt * g < ρb*Vs*g

Simplifying common terms, and replacing Vs by KVt (where K is the fraction of the total volume which is submerged, i.e. K<1), we have:

ρt*Vt < ρb*K*Vt ⇒ ρt / ρb < K < 1 ⇒ ρt < ρb ⇒ ρb > ρt

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2 years ago