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4 years ago

The buoyant force on an object fully submerged in a liquid depends on (select all that apply)

1 answer:

8 0

The buoyant force on an object fully submerged in a liquid depends on
the density of the liquid, and the density of the object. But the density
of the object depends on the object's volume and the object's mass.

So the only item on this list that it DOESN't depend on is the mass of
the liquid.

I guess that means that the buoyant force on a fully submerged object is
the same whether it's submerged in a cup of water or the Pacific Ocean.

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Rank in order, from largest to smallest, the magnitudes of the electric field at the black dot. A. 2, 1, 3, 4 B. 1, 4, 2, 3 C. 3

sweet [91]

Given that,

Rank in order from largest to smallest the magnitude of the electric field at block dot.

Electric field :

Electric field is proportional to the charge divided by square of distance.

In mathematically,

$E\propto\dfrac{q}{r^2}$

Where, q = charge

r = distance

If the charge is greater then electric field will be greater.

If the distance is greater then electric field will be smaller.

We need to find the electric field at black dot

According to figure,

(I). The electric field at black dot due to positive charge point q to left. the distance is r.

The electric field will be

$E=\dfrac{kq}{r^2}$

The electric field will be largest.

(II). The electric field at black dot due to positive charge point 2q to left. The distance is 2r.

Then, the electric field will be

$E=\dfrac{k2q}{(2r)^2}$

$E=\dfrac{kq}{2r^2}$

The electric field will be smallest.

(III). The electric field at black dot due to positive charge point 2q to left. The distance is r.

Then, the electric field will be

$E=\dfrac{k2q}{(r)^2}$

The electric field will be very largest.

(IV). The electric field at black dot due to positive charge point q to left. The distance is 2r.

Then, the electric field will be

$E=\dfrac{kq}{(2r)^2}$

$E=\dfrac{kq}{4r^2}$

The electric field will be very smallest.

So, The electric field from largest to smallest will be

$E_{3}>E_{1}>E_{2}>E_{4}$

Hence, The ranking will be 3, 1, 2, 4.

(D) is correct option.

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3 years ago

An amplifier produces a sound wave with a power of 1.00 μW (micro-Watt). If it is a point source and radiates uniformly in all d

gtnhenbr [62]

Answer:

(a) $I=8.84*10^{-9}W/m^{2}$

(b) $\beta =39.46dB$

Explanation:

Given data

The source is 1.00 μW

The point is 3m away from the point source

For Part (a)

The intensity at a certain distance from a point source that emits sound wave is given as:

$I=\frac{P_{s} }{4\pi r^{2} }\\I=\frac{1*10^{-6}W }{4\pi (3m)^{2} } \\I=8.84*10^{-9}W/m^{2}$

For Part (b)

The Sound level is given by

β=(10dB)×log(I/I₀)

Where

I₀=1×10⁻¹²W/m²

Substitute the given values to find Sound level

$\beta =(10dB)log(\frac{8.84*10^{-9}W/m^{2} }{1*10^{-12}W/m^{2}} )\\\beta =39.46dB$

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3 years ago