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telo118 [61]
3 years ago
8

discuss four contributing factors that may lead to an increase of learners abusing substance in scools​

Physics
1 answer:
Oksana_A [137]3 years ago
3 0

Answer:

What are the four contributing factors that may lead to an increase of learners abusing substances in school?

Peer pressure. ...

Socializing.

Community.

Socioeconomic status.

Stress.What are the four contributing factors that may lead to an increase of learners abusing substances in school?

Peer pressure. ...

Socializing.

Community.

Socioeconomic status.

Stress.What are the four contributing factors that may lead to an increase of learners abusing substances in school?

Peer pressure. ...

Explanation:

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What kind of motion for a star does not produce a Doppler effect? Explain.
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Answer:

 Transverse motion does not produced a doppler effect for a star. As, the doppler effect are occurred when the object are emitting light either in the direction towards the observer or moving away. The various astronomical objects are moving and the observer observed the blue wavelength of the light.

Doppler effect in the star is the process when the observer moves towards the star then it seems like the star move towards the observer.

6 0
3 years ago
Yung hangs a mass from a rod, sets the pendulum in simple harmonic motion, and determines the period of the pendulum. How could
kow [346]
He could use a shorter rod.
8 0
3 years ago
Three +3.0-μC point charges are at the three corners of a square of side 0.50 m. The last corner is occupied by a −3.0-μC charge
Naddik [55]

Answer:

4.30 x 10⁵ N/C

Explanation:

Two positive +3.0-μC point charges at opposite corners of the square creates equal and opposite electric field at the center. hence the electric field by these two positive charges at opposite corners becomes zero.

a = length of the square of side = 0.50 m

r = distance of the center from each corner = \frac{a}{sqrt(2)} = \frac{0.50}{sqrt(2)} = 0.354 m

Magnitude of net electric field at the center is given as

E = \frac{2kq}{r^{2}} \\E = \frac{2(8.99\times10^{9})(3\times10^{-6})}{(0.354)^{2}} \\E = 4.30\times10^{5} NC^{-1}

6 0
3 years ago
50pts: Fill this out.
Bogdan [553]

Push is in, pull is out

----------------------------------------------------------------------------------------------------

Contact Forces  

Frictional Force  

Tension Force  

Normal Force  

Air Resistance Force  

Applied Force

----------------------------------------------------------------------------------------------------

Action-at-a-Distance Forces  

Gravitational Force  

Electrical Force  

Magnetic forces

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Very bottom, last part

First drawing: Arrow down above box

Second drawing: Arrow up from below box    

 


4 0
3 years ago
Read 2 more answers
On the International Space Station an object with mass m = 280 g is attached to a massless string L = 0.97 m. The string can han
katovenus [111]

Answer:

5.16 m/s

Explanation:

The tension in the string provides the centripetal force that keeps the object spinning in circular motion, therefore we can write:

T=m\frac{v^2}{r}

where

T is the tension

m is the mass of the object

v is the speed

r is the radius of the circle

Here we have:

m = 280 g = 0.280 kg

r = L = 0.97 m (the radius of the circle is the length of the string)

The maximum tension allowed is

T = 7.7 N

Therefore, by solving for v, we find the maximum speed allowed before the string breaks:

v=\sqrt{\frac{Tr}{m}}=\sqrt{\frac{(7.7)(0.97)}{.280}}=5.16 m/s

6 0
3 years ago
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