A team of engineering students is testing their newly designed raft in the pool where the diving team practices.

Physics

1 answer:

Yuliya22 [3.3K]3 months ago

6 0

Answer:

Lower than

Explanation:

When the cube is placed on the raft, the displaced water equals the combined weight of both the cube and raft. In contrast, when submerged in water, the displaced water corresponds to the raft's weight plus the cube's volume. Since the cube's volume is smaller than what is needed to displace its weight with water, the water level is lower.

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Three point charges are positioned on the x axis. If the charges and corresponding positions are +32 µC at x = 0, +20 µC at x =

Sav [3153]

Response:

Magnitude of the electrostatic force acting on the +32 µC charge, $F_{net} = 12 N$

Clarification:

Let q₁ = +32 µC, located at x₁ = 0

q₂ = +20 µC, positioned at x₂ = 40 cm = 0.4 m

q₃ = -60 µC, placed at x₃ = 60 cm = 0.6 m

Define the force magnitude on the +32 µC charge from the +20 µC charge as F₁ (the force on q₁ due to q₂).

$F_{2} = \frac{kq_{1}q_{2} }{x_{2}^2 }$

$F_{2} = \frac{9 * 10^{9} * 32 * 10^{-6} * 20 * 10^{-6} }{0.4^2 }\\F_{2} = 36 N$

Define the force magnitude on the +32 µC charge from the -60 µC charge as F₂ (the force on q₁ due to q₃).

$F_{3} = \frac{kq_{1}q_{3} }{x_{3}^2 }$

$F_{3} = -\frac{9 * 10^{9} * 32 * 10^{-6} * 60 * 10^{-6} }{0.6^2 }\\F_{3} =-48 N$

The resultant electrostatic force on the 32 µC charge is $F_{net} = |F_{2} + F_{3}|$

$F_{net} =| 36 + (-48)| \\F_{net} =|- 12 N| \\ F_{net} = 12 N$

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3 months ago

If the potential difference across the bulb in a certain flashlight is 3.0 V, What is the potential difference across the combin

Yuliya22 [3333]

Answer:

The voltage across the bulb measures 3.0 V,

Explanation:

The bulb's voltage aligns with the voltage of the batteries, as they are the only power source for the bulb. Therefore, the voltage across the batteries is 3.0 V.

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A graph of the net force F exerted on an object as a function of x position is shown for the object of mass M as it travels a ho

Softa [3030]

The alteration in kinetic energy is $\Delta K = 3Fd$

Clarification:

According to the work-energy principle, the task performed on an object corresponds to the alteration in its kinetic energy. In mathematical terms:

$W=K_f -K_i= \Delta K$