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VLD [36.1K]
3 years ago
10

An object weighs 15N in air and 13N when completely

Physics
1 answer:
anastassius [24]3 years ago
8 0

Answer:

The volume of water displaced = 0.204 m³

Explanation:

<em>From Archimedes' principle,</em>

<em>Upthrust = lost in weight = weight of water displaced.</em>

<em>U = W₁ - W₂..................... Equation 1</em>

<em>Where U = upthrust, W₁ = weight in air, W₂ = weight when completely submerged in water.</em>

<em>Given: W₁ = 15 N, W₂ = 13 N</em>

<em>Substituting these values into equation 1</em>

<em>U = 15 - 13 </em>

<em>U = 2 N</em>

<em>But,</em>

<em>Weight of water displaced = mass of water displaced. × acceleration due to gravity.</em>

<em>Mass of water displaced = weight of water displaced/ acceleration due to gravity.................... Equation 2</em>

Where: acceleration due to gravity = 9.8 m/s², weight of water displaced = 2 N

Substituting these values into equation 2

Mass of water displaced = 2/9.8

Mass of water displaced = 0.204 kg.

Also,

Volume of water displaced = mass of water displaced/ Density of water.............. Equation 3

Where Density of water = 1.0 kg/m³, mass of water displaced = 0.204 kg

Substituting these values into equation 3 above,

Volume of water displaced = 0.204/1

volume of water displaced = 0.204 m³

Therefore the volume of water displaced = 0.204 m³

<em />

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Assume that the loop is initially positioned at θ=30∘θ=30∘ and the current flowing into the loop is 0.500 AA . If the magnitude
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Answer:\tau=1.03\times 10^{-4}\ N-m

Torque,

Explanation:

Given that,

The loop is positioned at an angle of 30 degrees.

Current in the loop, I = 0.5 A

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We need to find the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field. We know that the torque is given by :

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\tau=1\times 0.5\ A\times 0.0008\ m^2\times 0.3\ T\ \sin(60)\\\\\tau=1.03\times 10^{-4}\ N-m

So, the net torque about the vertical axis is 1.03\times 10^{-4}\ N-m. Hence, this is the required solution.

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