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Karolina [17]
3 years ago
13

A population is comprised of all one species, whereas a ___________ is made up of many of these.

Physics
1 answer:
miv72 [106K]3 years ago
5 0

Answer: Ecosystem

Explained: There's a minimum word count I'm filling up, don't mind me.

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The triceps muscle in the back of the upper arm extends the forearm. This muscle in a professional boxer exerts a force of 2615
mart [117]

Answer:

0.68 kg-m²

Explanation:

F = Force applied by the muscle = 2615 N

r = effective perpendicular lever arm = 2.85 cm = 0.0285 m

α = Angular acceleration of the forearm = 110.0 rad/s²

I  = moment of inertia of the boxer's forearm = ?

Torque is given as

τ = I α                                   eq-1

Torque is also given as

τ = r F                                   eq-2

using eq-1 and eq-2

r F =  I α

(0.0285)(2615) = (110.0) I

I = 0.68 kg-m²

7 0
3 years ago
Match the bones with their common name.
mrs_skeptik [129]

1. clavicle = collarbone

2. vertebrae = backbone

3. scapula = shoulder blade

4. femur = thigh

5. humerus = upper arm

6. patella = kneecap

7. cranium = skull

8. tibia = lower leg

9. radius/ulna = forearm

10. phalanges = fingers/toes

7 0
3 years ago
FIll in the blanks.: It's a beautiful day and you decide to go to the beach with some friends. The sun is ________
gregori [183]
Converting
releases
absorbs
cooling
transfer
liberating
4 0
3 years ago
Read 2 more answers
) determine the density of a 32.5 g metal sample that displaces 8.39 ml of water.
sweet-ann [11.9K]
Density is the ratio of a substance's mass to its volume. On the other hand, according to Archimedes' principle, the volume of water displaced is equal to the volume of the object placed on the water. Thus, the density of the metal is equal to 8.39 mL. So, the density would be

Density = 32.5 g/8.39 mL = 3.87 g/mL
3 0
4 years ago
Read 2 more answers
Use the diagram below to answer the following question:
d1i1m1o1n [39]

Answer:

3.0 cm

Explanation:

We can solve this problem by using the mirror equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

f is the focal length of the mirror

p is the distance of the object from the mirror

q is the distance of the image from the mirror

In this problem we have:

f = 1.5 cm is the focal length of the mirror (positive for a concave mirror)

p = 3.0 cm is the distance of the object from the mirror

Therefore, the distance of the image is:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{1.5}-\frac{1}{3.0}=\frac{1}{3.0}\\\rightarrow q=3.0 cm

And the positive sign means that the image is real.

(The second part of the exercise is just the description of the image of the first exercise).

5 0
3 years ago
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