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oksian1 [2.3K]
3 years ago
13

You make an observation that two objects attract. Which object below indicates the claim(s) that could be supported by this obse

rvation?
1.Both objects have a net charge.

2.Neither of the objects have a net charge.

3.One object has a net charge and the other object is neutral.

4.All of the above.

5.and C.
Physics
1 answer:
wel3 years ago
4 0
4. All of the above.





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How can you use distance and displacement to describe an object's motion​
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What graph shape is this what does the snap tell you
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What do you think would happen to the force of attraction of two interacting charges if their distance apart is halved?
sweet [91]

Answer:

The new force becomes 4 times the initial force.

Explanation:

The force of attraction or repulsion is given by the relation as follows :

F=k\dfrac{q_1q_2}{d^2}

Where

d is the distance between the interacting charges

F is inversely proportional to the distance between charges.

If the distance is halved, d'=(d/2), new force is given by :

F'=k\dfrac{q_1q_2}{d'^2}\\\\=k\dfrac{q_1q_2}{(\dfrac{d}{2})^2}\\\\=k\dfrac{q_1q_2}{\dfrac{d^2}{4}}\\\\=4\times \dfrac{kq_1q_2}{d^2}\\\\F'=4F

So, the new force becomes 4 times the initial force.

4 0
3 years ago
A submarine emerges 1/9 of its volume when it partially floats on the sea surface. For make it completely submerge it is necessa
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Answer:

4,524,660 N

Explanation:

Assuming the submarine's density is uniform, 1/9th of the submarine's mass is equal to the mass of the displaced water.

m/9 = (1026 kg/m³) (50 m³)

m = 461,700 kg

mg = 4,524,660 N

3 0
3 years ago
A stone is dropped from the
ICE Princess25 [194]
  • Height=h=500m
  • Acceleration=g=10m/s^2
  • Initial velocity=u=0
  • Speed of sound=c=340m/s
  • TIME TAKEN BY STONE TO HIT WATER=t
  • Time taken by sound to hear back=T

Now

\\ \sf\longmapsto h=ut+\dfrac{1}{2}gt^2

\\ \sf\longmapsto h=0t+\dfrac{1}{2}10t^2

\\ \sf\longmapsto 500=5t^2

\\ \sf\longmapsto t^2=100

\\ \sf\longmapsto t=10s

Now

\\ \sf\longmapsto h=cT

\\ \sf\longmapsto T=\dfrac{h}{c}

\\ \sf\longmapsto T=\dfrac{500}{340}

\\ \sf\longmapsto T=1.47\approx 1.5s

Total time:-

\\ \sf\longmapsto T_{net}=t+T=10+1.5=11.5s

8 0
2 years ago
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