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2 years ago

A wooden piece is made in different shapes take length (l) = radius (r) = 2m Calculate its volume as a:

Physics

1 answer:

In-s [12.5K]2 years ago

7 0

This question deals with the volume of different shapes.

a) volume of the sphere is "33.51 m³".

b) volume of the cylinder is "25.13 m³".

The volume of a sphere is given by the following formula:

$Volume = \frac{4}{3}\pi r^3\\\\Volume = \frac{4}{3}\pi (2\ m)^3$

Volume = 33.51 m³

The volume of a cylinder is given by the following formula:

$Volume = \pi r^2l\\\\Volume =\pi (2\ m)^2(2\ m)$

Volume = 25.13 m³

Learn more about volume here:

brainly.com/question/16686115?referrer=searchResults

The attached picture shows the formulae of the volume of different shapes.

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What do echolocation and ultrasounds have in common?

Natali5045456 [20]

They both are mechanical waves.

Explanation:

Echolocation and ultrasounds are both mechanical waves that require a medium to trace through. They both process and transfer information through waves; the difference is that we humans cannot hear ultrasound waves.

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1 year ago

What happens to the particles of a liquid when energy is removed from them?

KonstantinChe [14]

Answer:

D: The distance between the particles decreases

Explanation:

Taking away energy slows down molecules, like how you slow down when you are cold (I think)

3 0

3 years ago

Find the quantity of heat needed

krok68 [10]

Answer:

Approximately $3.99\times 10^{4}\; \rm J$ (assuming that the melting point of ice is $0\; \rm ^\circ C$ .)

Explanation:

Convert the unit of mass to kilograms, so as to match the unit of the specific heat capacity of ice and of water.

$\begin{aligned}m&= 100\; \rm g \times \frac{1\; \rm kg}{1000\; \rm g} \\ &= 0.100\; \rm kg\end{aligned}$

The energy required comes in three parts:

Energy required to raise the temperature of that $0.100\; \rm kg$ of ice from $(-10\; \rm ^\circ C)$ to $0\; \rm ^\circ C$ (the melting point of ice.)
Energy required to turn $0.100\; \rm kg$ of ice into water while temperature stayed constant.
Energy required to raise the temperature of that newly-formed $0.100\; \rm kg$ of water from $0\; \rm ^\circ C$ to $10\;\ rm ^\circ C$ .

The following equation gives the amount of energy $Q$ required to raise the temperature of a sample of mass $m$ and specific heat capacity $c$ by $\Delta T$ :

$Q = c \cdot m \cdot \Delta T$ ,

where

$c$ is the specific heat capacity of the material,
$m$ is the mass of the sample, and
$\Delta T$ is the change in the temperature of this sample.

For the first part of energy input, $c(\text{ice}) = 2100\; \rm J \cdot kg \cdot K^{-1}$ whereas $m = 0.100\; \rm kg$ . Calculate the change in the temperature:

$\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (0\; \rm ^\circ C) - (-10\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}$ .

Calculate the energy required to achieve that temperature change:

$\begin{aligned}Q_1 &= c(\text{ice}) \cdot m(\text{ice}) \cdot \Delta T\\ &= 2100\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 2.10\times 10^{3}\; \rm J\end{aligned}$ .

Similarly, for the third part of energy input, $c(\text{water}) = 4200\; \rm J \cdot kg \cdot K^{-1}$ whereas $m = 0.100\; \rm kg$ . Calculate the change in the temperature:

$\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (10\; \rm ^\circ C) - (0\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}$ .

Calculate the energy required to achieve that temperature change:

$\begin{aligned}Q_3&= c(\text{water}) \cdot m(\text{water}) \cdot \Delta T\\ &= 4200\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 4.20\times 10^{3}\; \rm J\end{aligned}$ .

The second part of energy input requires a different equation. The energy $Q$ required to melt a sample of mass $m$ and latent heat of fusion $L_\text{f}$ is:

$Q = m \cdot L_\text{f}$ .

Apply this equation to find the size of the second part of energy input:

$\begin{aligned}Q_2&= m \cdot L_\text{f}\\&= 0.100\; \rm kg \times 3.36\times 10^{5}\; \rm J\cdot kg^{-1} \\ &= 3.36\times 10^{4}\; \rm J\end{aligned}$ .

Find the sum of these three parts of energy:

$\begin{aligned}Q &= Q_1 + Q_2 + Q_3 = 3.99\times 10^{4}\; \rm J\end{aligned}$ .

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2 years ago

Use the drop-down menus to identify the parts of DNA. Label A: Label B:

love history [14]

Label A:

Nitrogen bases

Label B:

Sugar-phosphate backbone

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2 years ago

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When Trinity pulls on the rope with her weight, Newton's Third Law of Motion tells us that the rope will _____

Bingel [31]

When Trinity pulls on the rope with her weight, Newton's Third Law of Motion tells us that the rope will "pull back".

Newton's third law of motion expresses that, at whatever point a first question applies a power on a second object, the first object encounters a power meet in extent however inverse in heading to the power that it applies.

Newton's third law of movement reveals to us that powers dependably happen in sets, and one question can't apply a power on another without encountering a similar quality power consequently. We once in a while allude to these power matches as "action-reaction" sets, where the power applied is the activity, and the power experienced in kind is the response (despite the fact that which will be which relies upon your perspective).

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2 years ago

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