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PilotLPTM [1.2K]
3 years ago
7

How far would a cantaloupe fall in 5.0 seconds of true free fall?

Physics
1 answer:
RUDIKE [14]3 years ago
7 0

Answer:

i think the answer is about 50 m/s

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What is the net force experienced by the rope? Include both the magnitude and direction
Lapatulllka [165]

Answer:

All the physical world objects that comers in the contact to exert the force to each other. The contact forces are different from their names and what type of force they exert.

Explanation:

The cables and the ropes are the useful objects that exert the forces that can efficiently transfer the force from a significant distance.

It is noted that tension is a type of force that the rope can not simply push it away effectively. When push happened with rope, the rope goes to slack and lose all the tension that pulls at the first place. Tension only pull objects.

3 0
3 years ago
On which factor potential energy depends?​
Ludmilka [50]

▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂☘️

The potential energy of the object depends on

  • the height of the object with respect to some reference points,
  • the mass of the object,
  • the gravitational field the object is in.

▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂▂☘️

Hope it helps ~

3 0
2 years ago
Read 2 more answers
Which the answer It's science
earnstyle [38]
The answer is A. ive done a 5-k race, so its for sure 3 miles. 
3 0
3 years ago
Everyone can be hypnotized?
MAXImum [283]

Answer:

false

Explanation:

8 0
3 years ago
A 41-turn square coil of area 0.074 m2 and a 123-turn circular coil are both placed perpendicular to the same changing magnetic
vesna_86 [32]

Answer:

<h3>The area of second coil is ≅ 0.025 m^{2}</h3>

Explanation:

Given :

No. of turns in the first coil N_{1} = 41

No. of turns in the second coil N_{2}  = 123

Area of first coil A_{1} = 0.074 m^{2}

According to the law of electromagnetic induction,

Induced emf = -N \frac{d \phi}{dt}

Where \phi = magnetic flux.

Since given in question emf of both coil is same so we compare above equation.

    -\frac{N_{1} d\phi _{1}   }{dt_{1} }  = -\frac{N_{2} d\phi _{2}   }{dt_{2} }

   \frac{N_{1} A_{1}   dB_{1}  }{dt_{1} }  = \frac{N_{2} A_{2} dB_{2}     }{dt_{2} }

        A_{2} = \frac{N_{1} A_{1}  }{N _{2}  }

        A_{2} = \frac{41 \times 0.074 }{123  }

        A_{2} = 0.0246 = 0.025 m^{2}

Therefore, the area of second coil is ≅ 0.025 m^{2}

4 0
3 years ago
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