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zaharov [31]
3 years ago
15

A pebble is thrown into a calm lake, ripples are formed from the center and move outward. The water particles in the lake travel

in a circular pattern that moves up and down on the surface of a lake, and the energy travels A) diagonally. B) downward. C) horizontally. D) upward.
Physics
1 answer:
Misha Larkins [42]3 years ago
7 0

Answer: The option (C) is correct. The energy travels horizontally.

Explanation: When the pebble is thrown into a calm lake, there will be disturbance in the lake. The ripples are formed from the center and move outward because the energy is carried out from particle to particle.

The distribution of the energy among the particles starts from center and it will carry out outward. This disturbance will occur horizontally.

Therefore, the energy travels horizontally.

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Please help.. urgent Which statement is equivalent to Newton's first law? a. 15,300 N b. 1.20*10^3 N c. 2,030 N d. 1,560 N
taurus [48]
According to Newton laws of motion, 
F = m*a
Here, m = 1,560 Kg
a = 1.30 m/s²

Substitute their values, 
F = 1,560 * 1.30
F = 2028 N ~ 2030 N  [ Closest value ]

In short, Your Answer would be Option C

Hope this helps!
6 0
3 years ago
How are domains arranged in materials that are magnetic and in ones that are not
Valentin [98]

Answer:

In most materials, atoms are arranged in such a way that the magnetic orientation of one electron cancels out the orientation of another

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
In the diagram, q1=+6.25 * 10^ -8 C. What is the potential difference when you go from point A to point B? Include the correct s
Nimfa-mama [501]

Answer:

Moving a unit "positive" test charge from A to B will result in a reduction in potential

V = K Q / R      potential at a point

V2 - V1 = K Q (1 / .4 - 1 / .15) = = k Q (.15 - .4) / .06 = -4.17 K Q

V2 - V1 = -4.17 * 9 & 10E9 * 6.25 E-8

V2 - V1 = -4.17 * 562.5 J/C

V = - 2346 Volts

7 0
2 years ago
Pablo preto lift a barbell. During which stage of the lift does Pablo do work on the barbell
Arlecino [84]
While he is lifting the dumbell as the definition of work done = moving a mass through a distance = F x d
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2 years ago
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