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3 years ago

Most metallic elements, such as copper

Physics

1 answer:

nata0808 [166]3 years ago

5 0

Answer:

G. it enables people to use pots and pans made of metal to cook food

Explanation:

Metals are known to be substances with a relative high melting point, which means that it takes a very high temperature to melt them. As used in this question, copper (Cu) and iron (Fe) at metallic elements that have high melting points.

One advantage of this property based on the options provided is that "it enables people to use pots and pans made of metal to cook food". This is because cooking food requires subjecting them to a constant temperature from a heat source e.g cooker or stove.

The material used to hold the food being cooked must possess a very high melting point to withstand the heat, hence, the importance of these metallic substances. Most pots and pans are made of steel, an alloy of iron, hence, they have high melting point.

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A point charge of 9.00 × 10−9 C is located at the origin of a coordinate system. A positive charge of 3.00 × 10−9 C is brought i

dlinn [17]

A point charge is located at the origin of a coordinate system. A positive charge is brought in from infinity to a point. The charges are at distance for given electrical potential energy is 3.34 x 10⁷ m.

<h3>What is electric potential energy?</h3>

The electric potential energy is the work done by a test charge to bring it from infinity to a particular location.

The electric potential energy is given by the relation,

V = kQ/r

where k = 9 x 10⁹ J.m/C ,Q = 3 x 10⁻⁹ C, V =8.09 × 10⁻⁷ J.

Substitute the values into the expression to get the distance between the charges.

8.09 × 10⁻⁷ = 9 x 10⁹ x 3 x 10⁻⁹ / r

r =3.34 x 10⁷ m

Thus, the distance between the charges will be 3.34 x 10⁷ m.

Learn more about electric potential energy.

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2 years ago

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Tema [17]

Answer:

Explanation:

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3 years ago

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A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the

Maurinko [17]

Answer:

(a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity $R_{0}=10\ mCi$

Time $t_{1}=4\ hours$

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

$R=R_{0}e^{-\lambda t}$

$\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})$

Put the value into the formula

$\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})$

$\lambda=0.0000154\ s^{-1}$

$\lambda=1.55\times10^{-5}\ s^{-1}$

We need to calculate the half life

Using formula of half life

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}$

Put the value into the formula

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}$

$T_{\dfrac{1}{2}}=44.719\times10^{3}\ s$

$T_{\dfrac{1}{2}}=11.3\ hr$

(b). We need to calculate the value of N₀

Using formula of $N_{0}$

$N_{0}=\dfrac{3.70\times10^{6}}{\lambda}$

Put the value into the formula

$N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}$

$N_{0}=2.38\times10^{11}\ nuclei$

(c). We need to calculate the sample's activity

Using formula of activity

$R=R_{0}e^{-\lambda\times t}$

Put the value intyo the formula

$R=10e^{-(1.55\times10^{-5}\times30\times3600)}$

$R=1.87\ mCi$

Hence, (a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

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3 years ago

Which statement most completely describes a Lewis electron dot diagram? A.

andrezito [222]

Answer:

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2 years ago

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What is an example of an electromagnetic wave? Select all that apply. *

Afina-wow [57]

Answer:

A gamma ray, light from a lamp, and a microwave are examples of electromagnetic radiation.

Explanation:

Gamma radiation, visible light (light from a lamp), and microwaves are all apart of the electromagnetic spectrum.

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