1) The buoyant force acting on an object submerged in a fluid can be described as:
where
indicates the fluid's density,
represents the volume of the fluid displaced, and
signifies the gravitational acceleration.
2) To determine the volume of the displaced fluid, we note that the titanium object is entirely submerged in the fluid (air), thus this volume matches the volume of 1 Kg of titanium, which has a density of
. Using the correlation between density, volume, and mass, we derive
3) We can now revisit the equation in step 1) to compute the buoyant force. Given that the air density is
, this provides us with
4) The weight of 1 Kg of titanium is:
Therefore, the buoyant force is negligible when compared to the weight.
Response:D
Clarification:
Provided
mass of object 
Distance traveled 
resulting velocity 
energy conservation occurs starting when the object begins its descent and reaches a speed of 12 m/s
Initial Energy
Final Energy
where 
is the work done by friction, if any
As friction is present, this indicates an open system with a net external force of zero.
An open system allows for the exchange of energy and mass, and the presence of friction indicates that it is indeed an open system.
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Answer with Explanation:
Concepts and reasoning
The principle for addressing this question is that a capacitor in an RC circuit allows current to flow until fully charged. Once charged, it prevents any further current from moving through. Conversely, the situation is different with an inductor in an RL circuit. In accordance with Faraday's law, an inductor generates an electromagnetic force to counteract the applied voltage, but when no change in flux occurs, it behaves akin to a regular wire as if the inductor is absent.
In the accompanying diagram, a resistor is connected in series with a capacitor.
As we observe
the voltage across both the capacitor and the source.
Voltage across a resistor in an RC circuit.
Voltage across a resistor in an RL circuit.
Answer:
(1.6 × 10¹⁸) /s
Explanation:
The current in a wire is calculated using the formula:
I = (Q/t)
where Q represents the total charge of the electrons in the wire
and t denotes time.
Moreover, Q can be expressed as nq
where n is the number of electrons present in the wire
and q is the charge of a single electron, 1.602 × 10⁻¹⁹ C.
This leads us to the equation I = nq/t.
Thus, (n/t) = (I/q) describes the number of electrons flowing per second through a cross-section.
Inserting our current, I = 260 mA or 0.26 A, along with q = 1.6 × 10⁻¹⁹ C, yields:
(n/t) = (0.26/(1.602×10⁻¹⁹)) = (1.62 × 10¹⁸) /s = (1.60 × 10¹⁸) /s
