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sleet_krkn [62]
3 years ago
13

A solid 200-g block of lead and a solid 200-g block of copper are completely submerged in an aquarium filled with water. Each bl

ock is suspended just above the bottom of the aquarium by a thread. Which of the following is true?
A. More information is needed to choose the correct answer.
B. The buoyant force on the copper block is greater than the buoyant force on the lead block.
C. The buoyant force on the lead block is greater than the buoyant force on the copper block.
D. The buoyant force is the same on both blocks.
Physics
1 answer:
finlep [7]3 years ago
3 0

Answer:

B. The buoyant force on the copper block is greater than the buoyant force on the lead block.

Explanation:

Given;

mass of lead block, m₁ = 200 g = 0.2 kg

mass of copper block, m₂ = 200 g = 0.2 kg

density of water, ρ = 1 g/cm³

density of lead block, ρ₁ = 11.34 g/cm³

density of copper block, ρ₂ = 8.96 g/cm³

The buoyant force on each block is calculated as;

F_B = mg(\frac{density \ of \ fluid}{density \ of \ object} )

The buoyant force of lead block;

F_{lead} = 0.2*9.8(\frac{1}{11.34} )\\\\F_{lead} = 0.173  \ N

The buoyant force of copper block

F_{copper} = 0.2*9.8(\frac{1}{8.96})\\\\F_{copper} = 0.219  \ N

Therefore, the buoyant force on the copper block is greater than the buoyant force on the lead block

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A 3200-lb car is moving at 64 ft/s down a 30-degree grade when it runs out of fuel. Find its velocity after that if friction exe
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The velocity is 40 ft/sec.

Explanation:

Given that,

Force = 3200 lb

Angle = 30°

Speed = 64 ft/s

The resistive force with magnitude proportional to the square of the speed,

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Where, k = 1 lb s²/ft²

We need to calculate the velocity

Using balance equation

F\sin\theta-F_{r}=m\dfrac{d^2v}{dt^2}

Put the value into the formula

3200\sin 30-kv^2=m\dfrac{d^2v}{dt^2}

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3200\times\dfrac{1}{2}-v^2=m\dfrac{d^2v}{dt^2}

1600-v^2=m\dfrac{d^2v}{dt^2}

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. A child drops a ball from a window. The ball strikes the ground in 3.0 seconds. What is the. A child drops a ball from a windo
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Capacitors C1 = 6.45 µF and C2 = 2.50 µF are charged as a parallel combination across a 250 V battery. The capacitors are discon
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Answer:

For C1, Q =  1.6125×10⁻³ C

For C2, Q =  6.25×10⁻⁴ C

Explanation:

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From the question,

Q = CV........................ Equation 1

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For the first capacitor,

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Where C1 = 6.45 μF= 6.45×10⁻⁶ F, V = 250 V

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Q = 1.6125×10⁻³ C.

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Q = (2.5×10⁻⁶ )(250)

Q = 6.25×10⁻⁴ C

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