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3 years ago

A box of mass 3.6 kg is lifted 5.4 m above the

Physics

1 answer:

horsena [70]3 years ago

7 0

So, the energy change that occurs is 190.512 J.

<h3>Introduction</h3>

Hello ! I am Deva from Brainly Indonesia will help you regarding energy and its transformation. In this case, it's the use of energy from the lifter to be equivalent to the change in the object's potential energy. Why potential energy? Because the box undergoes a change in height and the potential energy specializes at a certain height. Work (W) due to change in potential energy ( $\sf{\Delta PE}$ ) can be realized in the equation :

$\sf{W = \Delta PE}$

$\sf{W = m \cdot g \cdot h_2 - m \cdot g \cdot h_2}$

$\boxed{\sf{\bold{W = m \cdot g \cdot (h_2 - h_1)}}}$

With the following condition :

W = work of subject (J)
$\sf{\Delta PE}$ = change of potential energy (J)
m = mass (kg)
g = acceleration of the gravity (m/s²)
$\sf{h_2}$ = final height (m)
$\sf{h_1}$ = initial height (m)

<h3>Problem Solving</h3>

We know that :

m = mass = 3.6 kg
g = acceleration of the gravity = 9.8 m/s²
$\sf{h_2}$ = final height = 5.4 m
$\sf{h_1}$ = initial height = 0 m

What was asked :

W = work of subject = ... J

Step by step :

$\sf{W = \Delta PE}$

$\sf{W = m \cdot g \cdot (h_2 - h_1)}$

$\sf{W = 3.6 \cdot 9.8 \cdot (5.4 - 0)}$

$\sf{W = 3.6 \cdot 9.8 \cdot 5,4}$

$\boxed{\sf{W = 190.512 \: J}}$

So, the energy change that occurs is 190.512 J.

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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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App182.studyisland.com/cfw/test/practice-session/a56do?CFID=14068

Alex

The amount of matter contained by an object is called mass.

A. mass

<u>Explanation:</u>

Mass is fundamentally a property of any physical amount and it is additionally the estimation of the resistance from the acceleration when force is applied on an object. The mass equals the quality of the gravitational force on a body.

Mass, in material science, the quantitative proportion of idleness, a crucial property of all matter. The greater the mass of a body, the littler the change created by an applied power. The mass of an object can be portrayed by its capacity to oppose a given power (we once in a while call this a body's inertial mass and subsequently mass is personally connected with the idea of inertia).

This is a straightforward result of Newton's second law where the power F, on a body is equivalent to the mass m, times the speeding up an, it encounters, ie:

F=ma or m=F/a

Mass of an object can not be zero but weight can be zero. The mass and weight of an object are different things.

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Answer:

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