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3 years ago

7

Is the direction of current through a battery always from the negative terminal to the positive one?

2 answers:

rodikova [14]3 years ago

8 0

Answer:

Yes. The current is actually flow of charge and in actual situation electrons flow from negative terminal to positive terminal. But in physics the convention of positive charge flow is used i.e. we take the direction of current from positive to negative terminal.

Explanation:

diamong [38]3 years ago

3 0

Poorly written question.

The direction of current through a CIRCUIT is always from the negative terminal of the battery to the positive one.

But the battery develops it's voltage and supplies its current and energy by chemical processes going on everywhere inside the battery, so it's not applicable to talk about any particular current "through" a battery.

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Explanation:

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3 years ago

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Consider two massless springs connected in series. Spring 1 has a spring constant k1, and spring 2 has a spring constant k2. A c

Andru [333]

Answer:

a. k = (1/k₁ + 1/k₂)⁻¹ b. k = (1/k₁ + 1/k₂ + 1/k₃)⁻¹

Explanation:

Since only one force F acts, the force on spring with spring constant k₁ is F = k₁x₁ where x₁ is its extension

the force on spring with spring constant k₂ is F = k₂x₂ where x₁ is its extension

Let F = kx be the force on the equivalent spring with spring constant k and extension x.

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x = F/k = F/k₁ + F/k₂

1/k = 1/k₁ + 1/k₂

k = (1/k₁ + 1/k₂)⁻¹

B

The force on spring with spring constant k₃ is F = k₃x₃ where x₃ is its extension

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x = F/k = F/k₁ + F/k₂ + F/k₃

1/k = 1/k₁ + 1/k₂ + 1/k₃

k = (1/k₁ + 1/k₂ + 1/k₃)⁻¹

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3 years ago

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A student uses a spring loaded launcher to launch a marble vertically in the air. The mass of the marble is

GarryVolchara [31]

Answer:

Part a)

When spring compressed by 2 cm

H = 1.47 m

Part b)

When spring is compressed by 4 cm

H = 5.94 m

Explanation:

Part a)

As we know that the spring is compressed and released

so here spring potential energy is converted into gravitational potential energy at its maximum height

So we will have

$\frac{1}{2}kx^2 = mg(H + x)$

$0.5(220)(0.02)^2 = 0.003(9.81)(H + 0.02)$

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$H = 1.47 m$

Part b)

Similarly when spring is compressed by 4 cm

then we have

$\frac{1}{2}kx^2 = mg(H + x)$

$0.5(220)(0.04)^2 = 0.003(9.81)(H + 0.04)$

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$H = 5.94 m$

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3 years ago