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3 years ago

A 55 kg person falling with a velocity of 0.6

Physics

1 answer:

siniylev [52]3 years ago

3 0

Answer:

What are we supposed to find, if it is kinetic energy then this is the solution.

K.E=1/2mv^2

K.E= kinetic energy

M=mass

V=velocity

K.E =0.5*55*0.6^2

K.E=9.9J

Explanation:

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What common laboratory measuring device will we likely always use in experiments that measure enthalpy changes?

adelina 88 [10]

The most common measuring device to be used in measuring enthalpy changes is the thermometer.

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3 years ago

the density of ice is 917.what fraction of the volume of a piece of ice will be above the liquid when floating in fresh water

yulyashka [42]

Answer:

$8.3\,\%$ of that piece of ice would be above the freshwater. Assumptions:

the density of the ice is $\rho(\text{ice}) = 917\; \rm kg \cdot m^{-3}$ , and
the density of freshwater is $\rho(\text{water}) = 1.00 \times 10^3\; \rm kg \cdot m^{-3}$ .

Explanation:

The volume of that chunk of ice can be split into two halves: volume above water $V(\text{above})$ , and volume under water $V(\text{under})$ . The mass of the whole chunk of ice would be:

$m(\text{ice}) = \rho(\text{ice}) \cdot (V(\text{above}) + V(\text{under}))$ .

Let $g$ be the acceleration due to gravity. The gravity on the entire chunk of ice would be

$\begin{aligned}&W(\text{ice}) \\ &= m({\text{ice}}) \cdot g \\ &= \rho(\text{ice}) \cdot (V(\text{above}) + V(\text{under})) \cdot g\end{aligned}$ .

On the other hand, the size of buoyant force on an object is equal to the weight of the liquid that it displaces. That is: $F(\text{bouyancy}) = W(\text{water displaced})$ .

Recall that $V(\text{above})$ is the volume of the ice above the water, and $V(\text{under})$ is the volume of the ice under the water.

The mass of water displaced would be equal to:

$\begin{aligned}& m(\text{water displaced}) \\ &= \rho(\text{water}) \cdot V(\text{water displaced}) \\ &= \rho(\text{water}) \cdot V(\text{under})\end{aligned}$ .

The weight of that much water would be

$\begin{aligned} &W(\text{water displaced}) \\ &= m(\text{water displaced}) \cdot g \\ &= \rho(\text{water}) \cdot V(\text{under}) \cdot g \end{aligned}$ .

Apply the equation $F(\text{bouyancy}) = W(\text{water displaced})$ . The bouyant force on this chunk of ice would be equal to $\begin{aligned} &W(\text{water displaced}) = \rho(\text{water}) \cdot V(\text{under}) \cdot g \end{aligned}$ .

Since the ice is floating, the forces on it need to be balanced. In other words, $\begin{aligned}W(\text{ice}) &= F(\text{bouyancy}) \\ &= \rho(\text{water}) \cdot V(\text{under}) \cdot g\end{aligned}$ .

On the other hand, recall that

$\begin{aligned}&W(\text{ice}) = \rho(\text{ice}) \cdot (V(\text{above}) + V(\text{under})) \cdot g\end{aligned}$ .

Combine the two halves to obtain:

$\begin{aligned}& \rho(\text{ice}) \cdot (V(\text{above}) + V(\text{under})) \cdot g \\ &= W(\text{ice}) = \rho(\text{water}) \cdot V(\text{under}) \cdot g\end{aligned}$ .

$\begin{aligned}& \rho(\text{ice}) \cdot (V(\text{above}) + V(\text{under})) \cdot g = \rho(\text{water}) \cdot V(\text{under}) \cdot g\end{aligned}$ .

Divide both sides by $g$ (assume that $g \ne 0$ ) to obtain:

$\begin{aligned}& \rho(\text{ice}) \cdot (V(\text{above}) + V(\text{under})) = \rho(\text{water}) \cdot V(\text{under})\end{aligned}$ .

Rearrange to obtain:

$\begin{aligned}& \frac{V(\text{under})}{V(\text{above}) + V(\text{under})} = \frac{\rho(\text{water})}{\rho(\text{ice})}\end{aligned}$ .

However, the question is asking for $\displaystyle \frac{V(\text{above})}{V(\text{above}) + V(\text{under})}$ , the fraction of the volume above water. Note that

$\begin{aligned}& \frac{V(\text{under})}{V(\text{above}) + V(\text{under})} + \frac{V(\text{above})}{V(\text{above}) + V(\text{under})} = 1\end{aligned}$ .

Therefore,

$\begin{aligned} &\frac{V(\text{above})}{V(\text{above}) + V(\text{under})} \\ &= 1 - \frac{V(\text{under})}{V(\text{above}) + V(\text{under})} \\ &= 1 - \frac{\rho(\text{water})}{\rho(\text{ice})} = 1 - \frac{917}{10^3} = 0.083\end{aligned}$ .

That's equivalent to $8.3\,\%$ .

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3 years ago

If the mass of the sun is 2x, at least one planet will fall into the habitable zone if I place a planet in orbits___, ____, ____

nalin [4]

So, the complete sentence is If the mass of the sun is 2x, at least one planet will fall into the habitable zone if I place a planet in orbits 84, 1, and 5, and all planets will orbit the sun successfully for the best conditions.

When the mass of the sun is larger, Earth moves around the sun at a faster pace and When the mass of the sun is smaller, Earth moves around the sun at a slower pace.

When Earth is closer to the sun, its orbit becomes faster and When Earth is farther from the sun, its orbit becomes slower.

When Earth is closer to the sun, there will be a hotter climate. A little movement that takes one closer to the sun could lead to a huge impact, as the sun is very hot.

So, it can be concluded that If the mass of the sun is 2x, at least one planet will fall into the habitable zone if I place a planet in orbits 84, 1, and 5, and all planets will orbit the sun successfully for the best conditions.

Learn more about Sun here:

brainly.com/question/15837114

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2 years ago

A 55-liter tank is full and contains 40kg of fuel. Find using Sl units: • Density p. • Specific Weight y • Specific Gravity Answ

Pavlova-9 [17]

Answer:

p = 727.273 kg/m3, y = 7134.545 N/m^3, SG = 0.7273

Explanation:

Density is simply the amount of mass of a substance per unit of volume. It can be found by dividing the mass in kg by the volume im m^3:

$p = \frac{m}{v} = \frac{40kg}{55lt*\frac{1 m^3}{1000 lt}} = 727.273 kg/m^3$

Specific weight is the weight of the substance per unit of volume. The weight is the mass of the material times the gravity, and it represents the force that the earth exerts on an object. Another way of calculate this value, its multiplying the density of the fuel times the gravity. Then:

$y = p*g = 727.273 kg/m^3 * 9.81 m/s^2 = 7134.545 N/m^3$

Specific Gravity is the ratio of the density of the substance to the density of a reference substance. For liquids, the reference substance is water at 4°C, which has a density of about 1000 kg/m^3.

$SG =\frac{ p_{fuel}}{p_{water}} = \frac{727.273 kg/m^3}{1000 kg/m^3} = 0.7273$