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liubo4ka [24]
4 years ago
3

What movement occurs when one moves the foot from the anatomical position to point the toes laterally, with the foot flat on the

floor?
Physics
1 answer:
Oksana_A [137]4 years ago
8 0

Answer:

Literal Rotation (It is sometimes called outward or external rotation)

Explanation:

Rotation of the neck or body is the twisting movement produced by the summation of the small rotational movements available between adjacent vertebrae.Rotation can occur within the vertebral column, at a pivot joint, or at a ball-and-socket joint.  At a pivot joint, one bone rotates in relation to another bone.

Lateral rotation is a rotating movement of the limb (in this case foot) so that the anterior surface moves away from the midline (center of the body).

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Motion can be detected by using background objects that are not moving called_______________ points
gtnhenbr [62]

Answer: A reference point

Explanation: If an object is in motion then its distance from another object is changing.  Whether an object is moving or not depends on your point of view. For example, a woman riding on a bus is not moving in relation to the seat she is sitting on, but she is moving in relation to the buildings the bus passes. A reference point is a place or object used for comparison to determine if something is in motion. An object is in motion if it changes position relative to a reference point. You assume that the reference point is stationary, or not moving.

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3 years ago
5. Identify the reactants in this chemical equation:<br> 2H3PO4 → HAP20, +H2O
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Answer:

H=Hydrogen, PO4=Phosphate

Explanation:

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3 years ago
A 5.50 kg sled is initially at rest on a frictionless horizontal road. The sled is pulled a distance of 3.20 m by a force of 25.
kiruha [24]

(a) 69.3 J

The work done by the applied force is given by:

W=Fd cos \theta

where:

F = 25.0 N is the magnitude of the applied force

d = 3.20 m is the displacement of the sled

\theta=30^{\circ} is the angle between the direction of the force and the displacement of the sled

Substituting numbers into the formula, we find

W=(25.0 N)(3.20 m)(cos 30^{\circ})=69.3 J

(b) 0

The problem says that the surface is frictionless: this means that no friction is acting on the sled, therefore the energy dissipated by friction must be zero.

(c) 69.3 J

According to the work-energy theorem, the work done by the applied force is equal to the change in kinetic energy of the sled:

\Delta K = W

where

\Delta K is the change in kinetic energy

W is the work done

Since we already calculated W in part (a):

W = 69.3 J

We therefore know that the change in kinetic energy of the sled is equal to this value:

\Delta K=69.3 J

(d) 4.9 m/s

The change in kinetic energy of the sled can be rewritten as:

\Delta K=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2 (1)

where

Kf is the final kinetic energy

Ki is the initial kinetic energy

m = 5.50 kg is the mass of the sled

u = 0 is the initial speed of the sled

v = ? is the final speed of the sled

We can calculate the variation of kinetic energy of the sled, \Delta K, after it has travelled for d=3 m. Using the work-energy theorem again, we find

\Delta K= W = Fd cos \theta =(25.0 N)(3.0 m)(cos 30^{\circ})=65.0 J

And substituting into (1) and re-arrangin the equation, we find

v=\sqrt{\frac{2 \Delta K}{m}}=\sqrt{\frac{2(65.0 J)}{5.50 kg}}=4.9 m/s

6 0
4 years ago
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There will be a high current drop a cross it
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3 years ago
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a

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