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2 years ago

Which circuit has the largest equivalent resistance?.

1 answer:

6 0

The series circuit has the higher equivalent resistance, but to find the larger equivalent resistance you have to use Ohm's law (V2=I2R2), the power dissipated by the resistor can also be found using P2=I22R2=V22R2. To find the equivalent resistance of the circuit, notice that the parallel connection of R2 and R3 is in series with R1, so the equivalent resistance is Req=R1+(1R2+1R3)−1=1.00Ω+(16.00Ω+113.00Ω)−1=5.10Ω.

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The impact of the type of material of which the slope is made on acceleration

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3 years ago

The first person to map the Milky Way Galaxy using radio waves was

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Answer:B

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3 years ago

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g A box having mass 0,5kg is placed in front of a 20 cm compressed spring. When the spring released, box having mass m1, collide

vagabundo [1.1K]

Answer:Expression given below

Explanation:

Given mass of spring $\left ( m_1\right )=0.5 kg$

Compression in the spring $\left ( x\right )=20 cm$

Let the spring constant be K

Using Energy conservation

potential energy stored in spring =Kinetic energy of Block $\left ( m_1\right )$

$\frac{1}{2}Kx^2=\frac{1}{2}m_1v^2$

$v=x\sqrt{\frac{k}{m_1}}$

now conserving momentum

$m_1v=\left ( m_1+m_2\right )v_0$

$v_0=\frac{m_1}{m_1+m_2}v$

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4 years ago

when a metal ball is heated through 30°c,it volume becomes 1.0018cm^3 if the linear expansivity of the material of the ball is 2

Vlad1618 [11]

Answer:

The original volume of the metal sphere is approximately 1 cm³

Explanation:

The given parameters are;

The temperature change of the metal ball, ΔT = 30°C = 30 K

The new volume of the metal ball, V₂ = 1.0018 cm³

The linear expansivity of the material ball, α = 2.0 × 10⁻⁵ K⁻¹

We have;

d₂ = d₁·(1 + α·ΔT)

Where;

d₁ = The original diameter of the metal ball

d₂ = The new diameter of the ball

From the volume of the ball, V₂, we have;

V₂ = 1.0018 cm³ = (4/3)×π×r₂³

Where;

r₂ = The new radius = d₂/2

∴ V₂ = 1.0018 cm³ = (4/3)×π×(d₂/2)³

∴ d₂ = ∛(2³ × 1.0018 cm³/((4/3) × π)) ≈ 1.241445 cm

d₁ = d₂/(1 + α·ΔT)

∴ d₁ ≈ 1.241445 cm/(1 + 2.0 × 10⁻⁵·K × 30 K) ≈ 1.24070058 cm

The original volume of the metal ball, V₁ = (4/3)×π×(d₁/2)³

∴ V₁ = (4/3)×π×(1.24070058/2)³ ≈ 0.99999902845 cm³ ≈ 1 cm³

The original volume of the metal sphere, V₁ ≈ 1 cm³.

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3 years ago

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3 years ago

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