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3 months ago

A 100 cm3 block of lead weighs 11N is carefully submerged in water. One cm3 of water weighs 0.0098 N.

Physics

1 answer:

Keith_Richards [3.2K]3 months ago

3 0

The volume of lead measures 100 cm^3

with a density of lead at 11.34 g/cm^3

. Thus, the mass of the lead block equals density multiplied by volume

$m = 100 * 11.34 = 1134 g$

$m = 1.134 kg$

Therefore, its weight in air is noted as

$W = mg = 1.134* 9.8 = 11.11 N$

Next, the buoyant force acting on the lead is defined as

$F_B = W - F_{net}$

$F_B = 11.11 - 11 = 0.11 N$

We know that

$F_B = \rho V g$

$0.11 = 1000* V * 9.8$

After solving, we find

V = 11.22 cm^3

(ii) This corresponding volume of water exerts the same weight as the buoyant force, resulting in 0.11 N

(iii) The buoyant force measures 0.11 N

(iv) The lead block sinks in water due to its density being greater than that of water.

The buoyant force acting on the lead block counterbalances its weight

$F_B = W$

$\rho V g = W$

$13* 10^3 * V * 9.8 = 11.11$

$V = 87.2 cm^3$

(ii) This volume of mercury corresponds to the buoyant force weight, confirming that the block floats within mercury, resulting in 11.11 N as its weight.

(iii) The buoyant force is recorded as 11.11 N

(iv) Given that lead's density is less than mercury's, the lead will float in the mercury medium.

Indeed, an object that has lesser density than a liquid will float; otherwise, it will sink in the liquid.

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3 months ago

If an electronic circuit experiences a loss of 3 decibels with an input power of 6 watts, what would its output power be, to the

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Answer:

The output power of the circuit is 3 Watts.

Given:

a loss in decibels = 3 dB

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To find:

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3 months ago

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