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san4es73 [151]
3 years ago
13

At exactly 3:14PM, the velocity of a dog running in a park points toward a group of flowers. Which of the following best describ

e the forces acting on the dog at 3:14PM?
There may be many forces acting on the dog, but the forces pointing toward the flowers are larger than the forces pointing away from the flowers.
There may be many forces acting on the dog, but the forces pointing toward the flowers are smaller than the forces pointing away from the flowers.
There may be many forces acting on the dog, but the forces pointing toward the flowers are equal to the forces pointing away from the flowers.
Options A and B are both possible.
Options A and C are both possible.
Options A, B, and C are all possible.
Physics
1 answer:
dmitriy555 [2]3 years ago
4 0

Answer:

Options A, B, and C are all possible.

Explanation:

We know that the instantaneous velocity of the dog at 3:14PM is possitive to toward the flowers. But what about the acceleration to toward the flowers?

If the dog is decreasing speed at 3:14PM, it means that acceleration is negative toward the flowers, hence (since F=ma) the net force points away from the flowers.

If the dog is increasing speed at 3:14PM, it means that acceleration is positive toward the flowers, hence (since F=ma) the net force points toward the flowers.

If the dog is not increasing nor decreasing speed at 3:14PM, it means that acceleration is 0, hence (since F=ma) the net force is null and it does not point neighter to toward the flowers  nor away from the flowers. This happens when the forces acting on the dog are equal to both sides.

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Un alambre de teléfono de 120m de largo y de 2.2mm de diámetro se estira debido a una fuerza de 380 N cual es el esfuerzo longit
sergij07 [2.7K]

Respuesta: verifique amablemente la explicación

Explicación:

Dado lo siguiente:

Longitud (L) del cable = 120 m

Diámetro (d) = 2,2 mm (2,2 / 1000) = 2,2 * 10 ^ -3 m

Fuerza (F) = 380 N

Esfuerzo longitudinal = Fuerza / Área

Área = πd² / 4 = (π * (2.2 * 10 ^ -3) ^ 2) / 4

Área = (3.142 * 4.84 * 10 ^ -6)

Área = 0.00000380132 m²

Estrés = Fuerza / Área

Estrés = 380 / 0.00000380132

Esfuerzo longitudinal = 99952128.12 = 9.9952128 * 10^7 Nm^-2

Deformación longitudinal: extensión / longitud

Extensión = 0.10 m

Longitud = 120 m

Deformación longitudinal = 0,1 m / 120 m

Deformación longitudinal = 0.0008333 = 8.33 × 10 ^ -4

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3 years ago
Us your understanding of asexual reproduction to explain why is it important that organisms reproduce in a variety of ways.
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The genetic material is identical in asexual reproduction- in order for organisms to be strong they need variety so if a disease comes, some of the species may be able to fight it off because of their varied genetics
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Answer:

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Earth is slightly closer to the sun in january than in july. How does the area swept out by earth’s orbit around the sun during
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Two parallel plate capacitors 1 and 2 are identical except that capacitor 1 has charge +q on one plate and charge −q on the othe
Grace [21]

Answer:

a) the capacitance is the same for both capacitors.

b) The potential difference between the plates for the capacitor with charge +2q, is double of the one for the capacitor  with charge +q.

c) The electric field magnitude between the plates for the capacitor with charge +2q, is double of the one for the capacitor with charge +q

d) The energy stored between the plates for the capacitor with charge +2q, is 4 times the value for the one with charge +q  

Explanation:

a) The capacitance of a capacitor, by definition, is as follows:

C = \frac{q}{V}

Appying Gauss' Law to one of plates, it can be showed, that the capacitance (for a parallel plates capacitor) can be  expressed as follows:

C = ε*A / d

As it can be seen, it does not depend on the charge. so we conclude that the capacitance must be the same for both capacitors, due to they are identical except for the value of the charge on the plates.

b) By definition, as we said above, the capacitance is equal to the proportion between the charge of one of the plates, and the potential difference between them.

If this proportion must remain the same, and one of the capacitors has the double of  the charge than the other, the potential difference must be the double also.

c) Applying Gauss' law, to the surface of one of  the plates, and assuming a constant surface charge density σ, it can be  showed that the  electric field can be calculated as follows:

E*A = Q/ε₀ as σ=Q/A

⇒ E = σ/ε₀

As σ is directly proportional to the charge (being the area A the same), we conclude that the electric field for the capacitor with charge +2q must be the double than the one for the capacitior with charge +q.

c) The electric potential energy, stored between plates of a capacitor, can be written as follows:

Ue = \frac{1}{2} *\frac{q^{2}}{C}

If the capacitance remains the same, we can conclude that the electric potential energy for the capacitor with charge +2q, as the charge is raised to the 2nd power, must be 4 times the one for the capacitor with charge +q.

4 0
3 years ago
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