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4 years ago

Question 1

Physics

1 answer:

solmaris [256]4 years ago

8 0

Answer:

4.0 m/s²

Explanation:

Draw two free body diagrams, one for the cart and one for the hanging mass.

The cart has one force acting on it in the x direction: tension to the right.

The hanging mass has two forces acting on it in the y direction: tension up and weight down.

The hanging mass accelerates downward, so if we take up to be positive, then the acceleration is negative.

Apply Newton's second law to the hanging mass:

∑F = ma

T − m₂g = m₂(-a)

T − m₂g = -m₂a

Apply Newton's second law to the cart:

∑F = ma

T = m₁a

Substitute and solve for a:

m₁a − m₂g = -m₂a

m₁a + m₂a = m₂g

(m₁ + m₂) a = m₂g

a = m₂g / (m₁ + m₂)

Given m₁ = 2.3 kg and m₂ = 1.6 kg:

a = (1.6 kg × 9.8 m/s²) / (2.3 kg + 1.6 kg)

a = 4.0 m/s²

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Answer:

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Explanation:

We are given that

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$v^2_f=2(\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2})$

$v_f=\sqrt{2(\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2})}$

Using the formula

$v_f=\sqrt{v^2_i+2GM(\frac{1}{d_2}-\frac{1}{d_1})}$

$v_f=\sqrt{(9.5\times 10^4)^2+2\times 6.7\times 10^{-11}\times 1.98\times 10^{30}(\frac{1}{6\times 10^{12}}-\frac{1}{4.7\times 10^{10})}$

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3 years ago

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3 years ago

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swat32

Answer:

the terminal velocity v_t= 202.96 m/s≅203 m/s

Explanation:

The expression for the terminal velocity

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4 years ago

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Answer:

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Explanation:

From the question we are told that

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Generally the frequency of the note played by the first player is mathematically represented as

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=> $f_1 = 490 + 2.0$

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3 years ago