a lead bullet is placed in a pool of mercury. what fractional part of the volume of the bullet is submerged?
well, the lead displaces mercury such that the weight of the mercury is equal to the weight of the lead.
weight of mercury displaced= densitymercury*volumeleadsubmerged
weight of lead= densitylead*volumelead
so set them equal
Volumesubmerged/totalvolume= densitylead/densitymercury
Perfect!
Since the density of lead is 11.3 x 10^3 kg/m^3 and the density of mercury is 13.6 x 10^3 kg/m^3, the answer is 0.831.
Thanks much!
Well, let's take a plunge into some science humor, shall we?
If you put a lead bullet in a pool of mercury, the bullet will feel right at "elemental" home! But as for the fractional part of the volume submerged, it's a bit like a fish out of water, or in this case, a bullet out of mercury. It's all about density!
Since mercury is significantly denser than lead, the bullet will float on top, like a boss! So, technically speaking, no fraction of the bullet's volume will be submerged in the mercury pool. It'll just be chilling on the liquid's surface, enjoying its lofty position like it's on a bullet-worthy vacation! 🌊💥
To determine the fractional part of the volume of a lead bullet that is submerged in a pool of mercury, we need to understand the concept of buoyancy and density.
When an object is submerged in a fluid, the buoyant force exerted on the object is equal to the weight of the fluid displaced by the object. If the buoyant force is greater than the weight of the object, it will float; if it is less, the object will sink.
In this case, since lead is denser than mercury, the lead bullet will sink in the pool of mercury.
Let's assume the volume of the lead bullet is V_bullet and the density of lead is d_lead, and the density of mercury is d_mercury.
The buoyant force acting on the lead bullet is given by the equation:
Buoyant Force = Weight of Displaced Fluid
The weight of the displaced fluid is equal to the volume of the displaced fluid multiplied by the density of that fluid.
Weight of Displaced Fluid = V_submerged * d_mercury
Since the entire bullet is submerged, the weight of the bullet is equal to its volume multiplied by its density.
Weight of Bullet = V_bullet * d_lead
Setting these two equations equal to each other, we have:
V_submerged * d_mercury = V_bullet * d_lead
We want to find the fractional part of the volume of the bullet submerged, so we can rewrite this equation as:
V_submerged / V_bullet = d_lead / d_mercury
Therefore, the fractional part of the volume of the bullet submerged is equal to the ratio of the density of lead to the density of mercury.
Keep in mind that the density of lead is approximately 11,340 kg/m³, and the density of mercury is approximately 13,534 kg/m³.
To determine the fraction of the volume of the lead bullet that is submerged in the pool of mercury, you need to understand the concept of buoyancy and the relationship between the density of the objects involved.
Here's how you can calculate it:
1. Identify the density of lead and mercury:
- The density of lead is typically around 11.34 grams per cubic centimeter (g/cm³).
- The density of mercury is approximately 13.6 g/cm³.
2. Compare the densities:
- Because the density of mercury (13.6 g/cm³) is higher than the density of lead (11.34 g/cm³), the bullet will sink in the mercury pool.
3. Use the density ratio to find the fraction submerged:
- The fraction of the volume submerged can be calculated by comparing the densities of the materials involved.
- In this case, the fraction submerged can be found by using the formula:
Fraction submerged = (Density of object) / (Density of fluid)
- Plugging in the variables, we get:
Fraction submerged = (Density of lead) / (Density of mercury)
Fraction submerged = 11.34 g/cm³ / 13.6 g/cm³
Fraction submerged ≈ 0.834
Therefore, approximately 0.834, or about 83.4%, of the volume of the lead bullet will be submerged in the pool of mercury.