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Komok [63]
3 years ago
10

Which two substances are linked in one recycling process?

Physics
1 answer:
Galina-37 [17]3 years ago
5 0
Carbon and oxygen are two substances that are linked in one recycling process  
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To describe velocity you need to know _______ and _________
Crazy boy [7]
To get the velocity you need to know the displacement and the time.

I hope that helps!
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3 years ago
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Why do sound engineers use spongy materials in the walls? (1 point)
mihalych1998 [28]

Answer:

o to increase the frequency of sound waves. It increases the sound waves to a level of frequency that humans cannot hear so you won't be able to hear many things though the wall other then low noises like pounding.

Explanation:

I am in construction class as well as a student teacher for other construction type programs trust me :D

Brainiest would be appreciated

8 0
3 years ago
Wrist joint is known gliding joint * 1 point true or False ​
zepelin [54]

True

Wrist and ankle joints are also known as the gliding joints : )

it's also called the midcarpal joint

4 0
3 years ago
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To practice Problem-Solving Strategy 21.1 Coulomb's Law. Three charged particles are placed at each of three corners of an equil
Salsk061 [2.6K]

Answer:

The force felt by charge 3 is F=(-5.6*10⁻⁶,3.36⁻⁵)N

Explanation:

As the superposition principle applies to static charges, we can find the net electric force as the sum of the two forces felt by q3.

Looking at the drawing and knowing that they form an equilateral triangle of lenght 4 we can conclude that each internal angle is 60°.

So, the positions in our coordinate system are:

r_1=(0,0)\\r_2=(4\ cos(60\°),4\ sin(60\°))\\r_3=(4,0)\\

Now  using Coulomb's force:

F_{ij}=\frac{-kq_iq_j}{d^2}(\vv{r}_j-\vv{r}_i)

Where d=4, q1 = -7.8*10⁻⁹C, q2 = -15.6 *10⁻⁹C, q3 = 8.0 *10⁻⁹C, k=8.98*10⁹, e0=8.8*10¹⁰:

Replacing we get 2 equations:

F_{13}=\frac{-kq_1q_3}{d^2}(\vv{r}_1-\vv{r}_3)=\frac{-kq_1q_2}{d^2}(-0.04\ cos(60\°),-0.04\ sin(60\°))\\\\F_{23}=\frac{-kq_2q_3}{d^2}(\vv{r}_1-\vv{r}_3)=\frac{-kq_1q_2}{d^2}(0.04-0.04\ cos(60\°),-0.04\ sin(60\°))\\

To work with the sam

F=∑F_i=3.5*10⁻⁴(0.023,0.032)+7*10⁻⁴(-0.016,0.032)=

=((3.5*10⁻⁴-7*10⁻⁴)*0.016,(3.5*10⁻⁴+7*10⁻⁴)*0.032)=

F=(-5.6*10⁻⁶,3.36⁻⁵)N

7 0
3 years ago
If a 60 kg student is standing on the edge of a cliff. Find the students gravitational potential of the cliff is 30 m high
kipiarov [429]

Answer:

17640J

Explanation:

Give mass m= 60kg

Height h =30m

Since the student is standing on the edge of a cliff , acceleration due to gravity g is 9.8m/s^2

The student’s gravitational potential energy can be found by the formula

mgh

That’s 60 x 9.8 x 30

= 17640J

8 0
3 years ago
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