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harina [27]
3 years ago
12

What are the body parts to this figure? Any of the body parts. (Please)

Physics
1 answer:
vlabodo [156]3 years ago
3 0
1 - Skull
2 - Mandible
3 - Scapula
4 - Sternum
5 - Ulna
6 - Radius
7 - Pelvis
8 - Femur
9 - Patella
10 - Tibia
11 - Fibula
12 - Metatarsals
13 - Clavicle
14 - Ribs (rib cage)
15 - Humerus
16 - Spinal column
17 - Carpals
18 - Metacarpals
19 - Phalanges
20 - Tarsals
21 - Phalanges
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The molecule that traps the sun's energy is ?.
kakasveta [241]
The molecule that trap's the sun's energy is chlorophyll.
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2 years ago
A charge of +2.00 x 10^-9 C is placed at the origin, and another charge of +4.50 x 10^-9 C is placed at x = 1.6 m. The Coulomb c
igor_vitrenko [27]

Answer:

0.64 m from the first charge

Explanation:

Force is given by

F_1=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F_1=\dfrac{k2\times 10^{-9}\times 3.7\times 10^{-9}}{x^2}

F_2=\dfrac{kq_2q_3}{r^2}\\\Rightarrow F_1=\dfrac{k4.5\times 10^{-9}\times 3.7\times 10^{-9}}{(1.6-x)^2}

These forces are equal

\dfrac{k2\times 10^{-9}\times 3.7\times 10^{-9}}{x^2}=\dfrac{k4.5\times 10^{-9}\times 3.7\times 10^{-9}}{(1.6-x)^2}\\\Rightarrow \dfrac{2}{x^2}=\dfrac{4.5}{(1.6-x)^2}\\\Rightarrow \dfrac{2}{4.5}=\dfrac{x^2}{(1.6-x)^2}\\\Rightarrow \dfrac{4.5}{2}=\dfrac{(1.6-x)^2}{x^2}\\\Rightarrow \sqrt{\dfrac{4.5}{2}}=\dfrac{1.6-x}{x}\\\Rightarrow 1.5=\dfrac{1.6-x}{x}\\\Rightarrow 1.5x+x=1.6\\\Rightarrow x=\dfrac{1.6}{2.5}\\\Rightarrow x=0.64\ m

The distance that charge should be placed is 0.64 m from the first charge

3 0
3 years ago
A merry-go-round at a playground is a circular platform that is mounted parallel to the ground and can rotate about an axis that
julsineya [31]

Since angular speed is persistent, v/r is constant.
Where:

 v is tangential speed; and

 r is distance from axis.

 

Then equate v/r in both cases to get v in the second case.

Hence, speed = 2.2 x 2.1 / 1.4 meters

= 3.3 meters / seconds

 

Alternative solution would be:

w = 2.2 / 1.4 = 1.57

v = rw = 2.1 x 1.57 = 3.3 meters / seconds 

3 0
2 years ago
When doing a squat, how do you do it without getting hurt?
Aleksandr [31]
I do weight lifting so you need feet flat on floor but out back straight and also a spotter
7 0
3 years ago
Read 2 more answers
A car is traveling at a speed of 38.0 m/s on an interstate highway where the speed limit is 75.0 mi/h. Is the driver exceeding t
lidiya [134]

Answer:

Yes, the car driver is exceeding the given limit.

Explanation:

<u>Given:</u>

  • Speed of the car, v = 38.0 m/s.
  • Speed limit of the highway, \rm v_o=75.0\ mi/h.

<h2><u>Converting the speed limit from mi/h to m/s:</u></h2>

We know,

1 mi = 1.60934 km.

1 km = 1000 m.

Therefore, 1 mi = 1.60934 × 1000 m = 1609.34 m.

1 hour = 60 minutes.

1 minute = 60 seconds.

Therefore, 1 hour = 60 × 60 seconds = 3600 seconds.

Using these values,

\rm 1\ \dfrac{mi}{h}=\dfrac{1609.34\ m}{3600\ s}=0.447\ m/s.

Therefore,

\rm v_o = 75.0\ mi/h=75.0\times 0.447=33.52\ m/s.

Clearly,

\rm v_o

which means, the car driver is exceeding the given speed limit.

6 0
3 years ago
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