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Alika [10]
4 years ago
14

An object with a density of 0.85 g/cc is dropped into each of the two beakers shown below. Beaker 1 has a density of 0.5 g/cc. B

eaker 2 has a density of 1 g/cc. What will happen to the object in each case?

Physics
1 answer:
Oliga [24]4 years ago
8 0

Answer:

The object will sink in the liquid in beaker 1.

The object will float in the liquid in beaker 2

Explanation:

The density of an object relative to the density of a fluid determines if the object floats or sink in a fluid. The density of a material is the measure of the amount of mass of that material packed into a unit volume of that material.

For the beaker 1, the liquid in this beaker has a density of 0.5 g/cc, which is lesser than the density of the object (0.85 g/cc). This means that the object will add more mass than there should be to the volume of the space it displaces within the field. This results in the object sinking in the fluid.

For beaker 2, the liquid in this beaker has a density of 1 g/cc, which is more than the density of the object (0.85 g/cc). This means that the object will add less mass than there should be to the volume of the space it displaces within the field. This results in the object floating in the fluid.

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Explanation:

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More precisely, the strength of the field at a distance r from the centre of the sphere is

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This can also be seen from the density of field lines: in fact, the closer the field lines, the stronger the field. Point A is the point where the lines have highest density, therefore it is also the point where the field is strongest.

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A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.50 Ω is in a 1.0 mT magnetic field, with the coil orien
n200080 [17]

Answer:

The voltage across the capacitor is 1.57 V.

Explanation:

Given that,

Number of turns = 10

Diameter = 1.0 cm

Resistance = 0.50 Ω

Capacitor = 1.0μ F

Magnetic field = 1.0 mT

We need to calculate the flux

Using formula of flux

\phi=NBA

Put the value into the formula

\phi=10\times1.0\times10^{-3}\times\pi\times(0.5\times10^{-2})^2

\phi=7.85\times10^{-7}\ Tm^2

We need to calculate the induced emf

Using formula of induced emf

\epsilon=\dfrac{d\phi}{dt}

Put the value into the formula

\epsilon=\dfrac{7.85\times10^{-7}}{dt}

Put the value of emf from ohm's law

\epsilon =IR

IR=\dfrac{7.85\times10^{-7}}{dt}

Idt=\dfrac{7.85\times10^{-7}}{R}

Idt=\dfrac{7.85\times10^{-7}}{0.50}

Idt=0.00000157=1.57\times10^{-6}\ C

We know that,

Idt=dq

dq=1.57\times10^{-6}\ C

We need to calculate the voltage across the capacitor

Using formula of charge

dq=C dV

dV=\dfrac{dq}{C}

Put the value into the formula

dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}

dV=1.57\ V

Hence, The voltage across the capacitor is 1.57 V.

5 0
3 years ago
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