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Igoryamba
3 years ago
12

Is it plagiarism/copying to do a science fair project that is similar to one that has been done already, if you modify certain a

spects and you do not know the results?
Physics
2 answers:
Yanka [14]3 years ago
5 0
It is not as plagiarism as it is different project. You are even allowed to repeat past experiments, although you should have a category in your presentation of "background research". Science fair projects are about obtaining research on your own through you personal experiments. It sounds acceptable, but it would be advisable to double check it with your teacher or supervisor.
Ierofanga [76]3 years ago
3 0
If you cite your sources correctly and make sure to credit everything I'm p sure it's cool. You might lose points on originality but it depends on the teacher
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Two point charges of 60.0 C and -12.0 C are separated by a distance of 20.0 cm. A 7.00 C charge is placed midway between these t
iogann1982 [59]

Answer:

4.5\times 10^{14} N

Explanation:

q_1=60 C

q_2=-12 C

q_3=7 C

Distance between q1 and q2,d=20 cm

r=\frac{d}{2}=\frac{20}{2}=10 cm=10\times 10^{-2} m

1m =100 cm

Electric force on charge q3 due to charge q1

F_1=\frac{kq_1q_3}{r^2}=\frac{9\times 10^9\times 60\times 7}{(10\times 10^{-2})^2} (away from q1)

Electric force on charge q3 due to charge q2

F_2=\frac{kq_2q_3}{r^2}=\frac{9\times 10^9\times 12\times 7}{(10\times 10^{-2})^2}(attract toward q2)

Net force=F=F_1+F_2

F=\frac{9\times 10^9\times 60\times 7}{(10\times 10^{-2})^2}+\frac{9\times 10^9\times 12\times 7}{(10\times 10^{-2})^2}

F=\frac{9\times 10^9\times 7}{(10\times 10^{-2})^2}(60+12)

F=4.5\times 10^{14} N

5 0
4 years ago
A 400.0 kg storage box is held 10 m above ground by a forklift. What is its gravitational potential energy? (PE = mgh)
Kobotan [32]

Answer:

The gravitational potential energy (PE) of a particle or a system of particles is a form of energy that exists due to its position. Mathematically, it is expressed as the product of mass, gravitational acceleration, and the vertical displacement.

Explanation:

Given data:

Mass of the storage box, m = 400.0 \ kg

Height, h = 10 \ m

The gravitational potential energy of the storage box can be expressed as,

PE = mgh

PE = 400.0 \times 9.80 \times 10

PE = 39200 \ \rm J.

8 0
4 years ago
Read 2 more answers
2 ways to change frictional force between 2 objects
Zanzabum
In order to change the frictional force between two solid surfaces, it can be changed by shorter distances and by the amount of weight it has or the amount of force that is pushing that object to go however distance it can. 
5 0
3 years ago
Read 2 more answers
Sample Question: 4
Fantom [35]

The gravitational potential energy of the mass in increased approximately by 490 J

Gravitational Potential energy formula is

P=mgh

where mass is given my m= 5 Kg

g is the gravitational force i.e. 9.8 m/s^{2}

h is the height above the ground that is given by 10 m

P = 5*9.8*10

P = 490 J

What is gravitational potential energy?

The potential energy that a huge item has in relation to another massive object due to gravity is known as gravitational energy or gravitational potential energy. When two objects descend toward one another, the potential energy associated with the gravitational field is released (transformed into kinetic energy).

To learn more about gravitational potential energy

brainly.com/question/15978356

#SPJ9

5 0
1 year ago
A horizontal pole is attached to the side of a building. There is a pivot P at the wall and a chain is connected from the end of
PtichkaEL [24]

Answer:

Fscos63

Explanation:

Given that a horizontal pole is attached to the side of a building. There is a pivot P at the wall and a chain is connected from the end of the pole to a point higher up the wall. There is a tension force F in the chain. What is the moment of the force F about the pivot P?

Taking the moment from the pivot point P, that means the moment at point p = 0

Then, if we consider the weight mg of the pole, according to the principle of equilibrium : sum of the upward forces equal to the sum of the downward forces.

Therefore, mg = Fsinø ....... (1)

Also, taking moment at point P

Let the length of the pole = s

The length of the weight of the pole = 1/2 S

Fscosø = mgs/2

The distance s will cancel out

2Fcosø = mg ...... (3)

Substitute mg in equation 1 into equation 3

2fcosø = fsinø

F will cancel out

Tanø = 2

Ø = tan^-1(2)

Ø = 63.4 degree

The moment of force F about pivot point P will be

Moment = force × distance

Moment = Fcos63 × S

Moment = Fscos63

6 0
3 years ago
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