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Helga [31]
3 years ago
15

The chewing muscle, the masseter, is one of the strongest in the human body. It is attached to the mandible (lower jawbone) as s

hown in figure
(A) The jawbone is pivoted about a socket just in front of the auditory canal. The forces acting on the jawbone are equivalent to those acting on the curved bar in figure
(B) In figure a, a skull is shown. The lower jaw is labeled "Mandible".
A muscle labeled "Masseter" connects the back of the lower jaw to the upper part of the skull. In figure b, The jawbone is modeled by a curved bar. The left portion of the bar is horizontal, the right portion bends diagonally up and to the right. Three forces act on the bar: Vector T points directly upward at the point where the bar bends. Vector FC acts directly downward at the left end of the bar, a horizontal distance of 7.50 cm from vector T. Vector T points directly upward at the point where the bar bends. Vector R acts directly downward at the right end of the bar, a horizontal distance of 3.50 cm from vector T. F C is the force exerted by the food being chewed against the jawbone, T is the force of tension in the masseter, and R is the force exerted by the socket on the mandible. Find T and R (in N) for a person who bites down on a piece of steak with a force of 56.6 N. magnitude of R N

Physics
1 answer:
kifflom [539]3 years ago
6 0

Answer:

Explanation:

Check attachment

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A horizontal 810-N merry-go-round of radius 1.60 m is started from rest by a constant horizontal force of 55 N applied tangentia
Sloan [31]

Answer:

576 joules

Explanation:

From the question we are given the following:

weight = 810 N

radius (r) = 1.6 m

horizontal force (F) = 55 N

time (t) = 4 s

acceleration due to gravity (g) = 9.8 m/s^{2}

K.E = 0.5 x MI x ω^{2}

where MI is the moment of inertia and ω is the angular velocity

MI = 0.5 x m x r^2

mass = weight ÷ g = 810 ÷ 9.8 = 82.65 kg

MI = 0.5 x 82.65 x 1.6^{2}

MI = 105.8 kg.m^{2}

angular velocity (ω) = a x t

angular acceleration (a) = torque ÷ MI

where torque = F x r = 55 x 1.6 = 88 N.m

a= 88 ÷ 105.8 = 0.83 rad /s^{2}

therefore

angular velocity (ω) = a x t = 0.83 x 4 = 3.33 rad/s

K.E = 0.5 x MI x ω^{2}

K.E = 0.5 x 105.8 x 3.33^{2} = 576 joules

6 0
3 years ago
Rex decides to launch a priceless vase into the air at a 90-degree angle. The initial velocity vase is +7.0 m/s. Dylan claims th
Sergeu [11.5K]

Answer:

D. Dylan is incorrect because a 90-degree launch angle results in the largest vertical range​

Explanation:

Projectile is the motion of an object thrown into space. When an object is thrown into space, the only force which acts on it is the acceleration due to gravity.

An object thrown into space would reach maximum height (vertical range) if it is launched at an angle of 90 degrees. For maximum horizontal range, the object needs to be launched at an angle of 45 degrees.

Therefore Dylan is incorrect because a 90-degree launch angle results in the largest vertical range​

6 0
2 years ago
Determine the magnitude of the effective value of g⃗ at a latitude of 60 ∘ on the earth. assume the earth is a rotating sphere.
dezoksy [38]
In addition to acceleration of gravity we experience centrifugal acceleration away from the axis of rotation of the earth. this additional acceleration has value ac = r w^2 where w = angular velocity and r is distance from your spot on earth to the earth's axis of rotation so r = R cos(l) where l = 60 deg is the lattitude and R the earth's radius and w = 1 / (24hr x 3600sec/hr) 
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3 years ago
Molecules are made up of: A. elements B. compounds C. protons
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Atoms make up molecules
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A seagull flying horizontally over the ocean at a constant speed of 2.60 m/s carries a small fish in its mouth. It accidentally
Ivenika [448]

(a) +2.60 m/s

The motion of the fish dropped by the seagul is a projectile motion, which consists of two independent motions:

- a horizontal uniform motion, at constant speed

- a vertical motion, at constant acceleration (acceleration of gravity, g=-9.8 m/s^2, downward)

In this part we are only interested in the horizontal motion. As we said the horizontal component of the fish's velocity does not change, therefore its value when the fish reaches the ocean is equal to its initial value, which is the speed at which the seagull was flying (because it was flying horizontally):

v_x = +2.60 m/s

(b) -17.2 m/s

The vertical component of the fish's velocity instead follows the equation:

v_y = u_y +gt

where

u_y = 0 is the initial vertical velocity, which is zero

g=-9.8 m/s^2 is the acceleration of gravity

t is the time

Since the fish reaches the ocean at t = 1.75 s, we can substitute this time into the formula to find the final vertical velocity:

v_y = 0+(-9.8)(1.75)=-17.2 m/s

where the negative sign indicates the direction (downward).

(c)

The horizontal component of the fish's velocity would increase

The vertical component of the fish's velocity would stay the same.

As we said from part (a) and (b):

- The horizontal component of the fish's velocity is constant during the motion and it is equal to the initial velocity of the seagull -> so if the seagull's initial speed increases, the horizontal velocity of the fish will increase too

- The vertical component of the fish's velocity does not depend on the original speed of the seagull, therefore it is not affected.

4 0
3 years ago
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