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NISA [10]
3 years ago
14

Why is it important for a muscle to be attached to a fixed origin at one end and a moving insertion at the other? Discuss how th

is promotes movement, using one of the muscles you observed as an example.
Physics
1 answer:
vlabodo [156]3 years ago
6 0
Muscles function only by contracting. This makes it necessary for one end of the muscle to be fixed and the other mobile.
Take the bicep for example.
Its origin is at the shoulder and its two heads connect to the bones of the forearm, the radius and ulna.
Now, had the muscle not been fixed at one end, and contracted, it would pull both our shoulder and forearm together resulting in an ineffective movement. The desired motion is to lift the forearm (proximal and distal movement) which can only be achieved if the bicep is fixed at the shoulder and allowed to move at the forearm.
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Compare the benefits of wildfires to grasslands, northern forests, and deciduous forests.
Dimas [21]
Wildfires benefit grasslands, northern forests, and deciduous forests. Grasslands are benefited by improved soil quality and control of tree cover. Invertebrate species diversity is maintained through wildfire as well. Northern forests, like grasslands, experience increased production and nutritional quality of food as a result of wildfires. Deciduous forests experience an increase in the nutritional quality of food as well, but the effects are more temporary. The amount of shrubs in deciduous forests is reduced as a result of wildfires, allowing more herbaceous plants such as mosses and lichens to grow.
6 0
3 years ago
Read 2 more answers
The bar graph shows energy data taken from a roller coaster at a theme park. Analyze the data and assess its validity. In 3–5 se
alexdok [17]

The data given in the bar graph is valid because it follows the law of conservation of energy, since the GPE at top of 2nd hill plus KE at top of 2nd hill equals KE at bottom of 1st hill.

<h3>What is law of conservation of energy?</h3>

The law of conservation of energy states that energy can neither be created nor destroyed but can be transformed from one form to another.

Based on the law of conservation of energy, kinetic energy of a roller coaster can be converted into potential energy of the roller coaster and vice versa.

ΔK.E = ΔP.E

where;

  • ΔK.E is change in kinetic energy
  • ΔP.E is change in potential energy

The kinetic energy of the coaster is greatest at the bottom of the hill, as the coaster moves upward, the kinetic energy decreases and will be converted into potential energy. The potential energy of the coaster increases as the coaster moves up the hill and will become maximum at the highest point of the hill.

From the given data;

GPE at top of 2nd hill + KE at top of 2nd hill = KE at bottom of 1st hill

Learn more about conservation of energy here: brainly.com/question/166559

#SPJ1

6 0
1 year ago
What is the unique geological feature found on Mercury surface?
Cloud [144]

Answer:

The surface of Mercury has landforms that indicate its crust may have contracted. They are long, sinuous cliffs called lobate scarps. These scarps appear to be the surface expression of thrust faults, where the crust is broken along an inclined plane and pushed upward.

Explanation:

I hope this helps a little bit.

3 0
3 years ago
A spherical asteroid of average density would have a mass of 8.7×1013kg if its radius were 2.0 km. 1. If you and your spacesuit
Law Incorporation [45]

1. 0.16 N

The weight of a man on the surface of asteroid is equal to the gravitational force exerted on the man:

F=G\frac{Mm}{r^2}

where

G is the gravitational constant

M=8.7\cdot 10^{13}kg is the mass of the asteroid

m = 100 kg is the mass of the man

r = 2.0 km = 2000 m is the distance of the man from the centre of the asteroid

Substituting, we find

F=(6.67\cdot 10^{-11}m^3 kg^{-1} s^{-2})\frac{(8.7\cdot 10^{13} kg)(110 kg)}{(2000 m)^2}=0.16 N

2. 1.7 m/s

In order to stay in orbit just above the surface of the asteroid (so, at a distance r=2000 m from its centre), the gravitational force must be equal to the centripetal force

m\frac{v^2}{r}=G\frac{Mm}{r^2}

where v is the minimum speed required to stay in orbit.

Re-arranging the equation and solving for v, we find:

v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{(6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2})(8.7\cdot 10^{13} kg)}{2000 m}}=1.7 m/s

3 0
3 years ago
A 1000-kilogram car traveling due east at 15 meters per second hit from behind and receives a forward impulse of 6000 newton-sec
strojnjashka [21]

The change in momentum of the car is 6000 kg m/s

Explanation:

According to the impulse theorem, the change in momentum of an object is equal to the impulse exerted on the object, therefore:

\Delta p = I

where

\Delta p is the change in momentum

I is the impulse exerted

For the car in this problem, the impulse received is

I = 6000 kg m/s (in the forward direction)

Therefore, the change in momentum of the car is equal to this value:

\Delta p = I = 6000 kg m/s (in the forward direction)

We can also calculate what is the new momentum of the car. In fact, the initial momentum is

p_i = mu = (1000 kg)(15 m/s)=15,000 kg m/s

And  so, the new momentum is

p_f = p_i + \Delta p = 15,000 + 6,000 = 21,000 kg m/s

Learn more about impulse and momentum:

brainly.com/question/9484203

#LearnwithBrainly

8 0
3 years ago
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