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2 years ago

Which of the following statements apply to electric charges?

Physics

1 answer:

Gre4nikov [31]2 years ago

6 0

Answer:

The statement "If a positively charged rod is brought close to a positively charged object, the two objects will repel " applies to electric charges.

Explanation:

There are only two types of electric charges. Both having own magnitude but different charge.

1. Positive charge

2. Negative charge

Like charges repel each other and opposite charges always attract each other.

When a positively charged rod is brought close to a positively charged object, the rod and the object will repel.

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The repeating pattern of a minerals particles in a solid is called

Klio2033 [76]

Answer:

Crystal structure

Explanation:

The repeated pattern of similar particles in a material is called crystal. Crystal structure is the largest constituent unit of a solid matter.

The fundamental identity of a crystal structure is a unit cell that is formed by the arrangement of atoms or ions in a particular manner. A crystal is defined as a regular, long-ranged repeated arrangement of unit cells.

Crystal have a sharp melting and boiling point and they give a sharp edge on being cut with a knife.

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2 years ago

Two froghoppers sitting on the ground aim at the same leaf, located 35 cm above the ground. Froghopper A jumps straight up while

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2 years ago

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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

An experiment is set up to test the angular resolution of an optical device when red light (wavelength ????r ) shines on an aper

Neko [114]

Explanation:

As per Rayleigh criterion, the angular resolution is given as follows:

$\theta=\frac{1.22 \lambda}{D}$

From this expression larger the size of aperture, smaller will be the value of angular resolution and hence, better will be the device i.e. precision for distinguishing two points at very high angular difference is higher.

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3 years ago

How to charge a laptop battery without a charger?

Likurg_2 [28]

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2 years ago

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