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STatiana [176]
3 years ago
10

Producers,____________, and_______________ help to move matter and energy through ecosystems.

Physics
1 answer:
Phoenix [80]3 years ago
3 0
Producers, consumers, and decomposers help to move matter and energy through ecosystems.

Hope this helps! :)
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The lubrication of bone joints is a subject of ongoing medical research. Two bones connected at a joint do not touch. The bones
maks197457 [2]

The question is incomplete. The complete question is :

To measure the effective coefficient of friction in a bone joint, a healthy joint (and its immediate surroundings) can be removed from a fresh cadaver. The joint is inverted, and a weight is used to apply a downward force F⃗ d on the head of the femur into the hip socket. Then, a horizontal force F⃗ h is applied and increased in magnitude until the femur head rotates clockwise in the socket. The joint is mounted in such a way that F⃗ h will cause clockwise rotation, not straight-line motion to the right. The friction force will point in a direction to oppose this rotation.

Draw vectors indicating the normal force n⃗  (magnitude and direction) and the frictional force f⃗ f (direction only) acting on the femur head at point A.

Assume that the weight of the femur is negligible compared to the applied downward force.

Draw the vectors starting at the black dot. The location, orientation and relative length of the vectors will be graded

Solution :

The normal force represented by N is equal to the downward force, $F_d$ which is equal in magnitude but it is opposite in direction.

Also the frictional force acts always to oppose the motion because the bone starts moving in a clockwise direction. The frictional force that will be applied to the right direction so that the movement or the rotation at A is opposed.  

5 0
3 years ago
In order for a ship to stay afloat, its buoyant force must be
Kruka [31]
The buoyant force must be greater than water.
4 0
3 years ago
A jet plane flying 600 m/s experiences an acceleration of 4g when pulling out of the dive. What is the radius of curvature of th
Gnoma [55]

Answer:

91.84 m/s²

Explanation:

velocity, v = 600 m/s

acceleration, a = 4 g = 4 x 9.8 = 39.2 m/s^2

Let the radius of the loop is r.

he experiences a centripetal force.

centripetal acceleration,

a = v² / r

39.2 x r = 600 x 600

r = 3600 / 39.2

r = 91.84 m/s²

Thus, the radius of the loop is 91.84 m/s².

8 0
3 years ago
A 6.0 g marble is fired vertically upward using a spring gun. The spring must be compressed 9.4 cm if the marble is to just reac
RoseWind [281]

Answer:

a) \Delta U_{g} = 12.945\,J, b) \Delta U_{k} = 12.945\,J, c) k = 2930.059\,\frac{N}{m}

Explanation:

a) The change in the gravitational potential energy of the marble-Earth system is:

\Delta U_{g} = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (22\,m)

\Delta U_{g} = 12.945\,J

b) The change in the elastic potential energy of the spring is equal to the change in the gravitational potential energy, then:

\Delta U_{k} = 12.945\,J

c) The spring constant of the gun is:

\Delta U_{k} = \frac{1}{2} \cdot k \cdot x^{2}

k = \frac{2\cdot \Delta U_{k}}{x^{2}}

k = \frac{2\cdot (12.945\,J)}{(0.094\,m)^{2}}

k = 2930.059\,\frac{N}{m}

4 0
3 years ago
Ask Your Teacher Suppose the roller coaster below(h1 = 36 m, h2 = 13 m, h3 = 30) passes point A with a speed of 1.00 m/s. If the
Oliga [24]

Answer:

The answer to the question is

The roller coaster will reach point B with a speed of 14.72 m/s

Explanation:

Considering both kinetic energy KE = 1/2×m×v² and potential energy PE = m×g×h

Where m = mass

g = acceleration due to gravity = 9.81 m/s²

h = starting height of the roller coaster

we have the given variables

h₁ = 36 m,

h₂ = 13 m,

h₃ = 30 m

v₁ = 1.00 m/s

Total energy at point 1 = 0.5·m·v₁² + m·g·h₁

= 0.5 m×1² + m×9.81×36

=353.66·m

Total energy at point 2 = 0.5·m·v₂² + m·g·h₂

= 0.5×m×v₂² + 9.81 × 13 × m = 0.5·m·v₂² + 127.53·m

The total energy at 1 and 2 are not equal due to the frictional force which must be considered

Total energy at point 2 = Total energy at point 1 + work done against friction

Friction work = F×d×cosθ = (\frac{1}{5} × mg)×60×cos 180 = -117.72m

0.5·m·v₂² + 127.53·m = 353.66·m -117.72m

0.5·m·v₂² = 108.41×m

v₂² = 216.82

v₂  =  14.72 m/s

The roller coaster will reach point B with a speed of 14.72 m/s

8 0
3 years ago
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