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balu736 [363]
3 years ago
10

What do elements in the same group have?

Physics
2 answers:
Leviafan [203]3 years ago
7 0

Answer:

the same number of electron shells

Explanation:

Natalija [7]3 years ago
5 0
The same number of electron shells, the atomic number is given to each individual element, same as the atomic mass, and properties have nothing to do with groups
You might be interested in
How do I find the cosine of theta.
NNADVOKAT [17]

The value of cos θ in the given figure is 0.98.

<h3>What is cosine of an angle?</h3>

The cosine of an angle is defined as the sine of the complementary angle.

The complementary angle equals the given angle subtracted from a right angle, 90.

cos θ = sin(90 - θ)

For example, if the angle is 30°, then its complement is 60°

cos 30 = sin(90 - 30)

cos 30 = sin 60

0.866 = 0.866

<h3>Cosine of an angle with respect to sides of a right triangle</h3>

cos θ = adjacent side / hypotenuse side

adjacent side of the given right triangle is calculated as follows;

adj² = 10² - 2²

adj² = 100 - 4

adj² = 96

adj = √96

adj = 9.8

cos θ = 9.8/10

cos θ = 0.98

Thus, the value of cos θ in the given figure is 0.98.

Learn more about cosine of angles here: brainly.com/question/23720007

#SPJ1

5 0
1 year ago
Say that you are in a large room at temperature TC = 300 K. Someone gives you a pot of hot soup at a temperature of TH = 340 K.
DiKsa [7]

Answer:0.061

Explanation:

Given

T_C=300 k

Temperature of soup T_H=340 K

heat capacity of soup c_v=33 J/K

Here Temperature of soup is constantly decreasing

suppose T is the temperature of soup at any  instant

efficiency is given by

\eta =\frac{dW}{Q}=1-\frac{T_C}{T}

dW=Q(1-\frac{T_C}{T})

dW=c_v(1-\frac{T_C}{T})dT

integrating From T_H to T_C

\int dW=\int_{T_C}^{T_H}c_v(1-\frac{T_C}{T})dT

W=\int_{T_C}^{T_H}33\cdot (1-\frac{300}{T})dT

W=c_v\left [ T-T_C\ln T\right ]_{T_H}^{T_C}

W=c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]

Now heat lost by soup is given by

Q=c_v(T_C-T_H)

Fraction of the total heat that is lost by the soup can be turned is given by

=\frac{W}{Q}

=\frac{c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]}{c_v(T_C-T_H)}

=\frac{T_C-T_H-T_C\ln (\frac{T_C}{T_H})}{T_C-T_H}

=\frac{300-340-300\ln (\frac{300}{340})}{300-340}

=\frac{-40+37.548}{-40}

=0.061

4 0
3 years ago
An object is dropped from a vertical distance of 25.5 m above the ground, and it takes 2.28 sec to fall that distance. A second
DENIUS [597]

Answer:

The second object takes 2.28 s to fall the 25.5 m.

Explanation:

In this case, both objects take the same time to fall, since <em>no vertical velocity is added </em>to any of them.

You can also confirm this by sepparating the second's object movement into its two directions: in the horizontal one, we have <em>linear uniform motion, </em>and in the vertical one, we have <em>free fall, </em>with exactly the same characteristics as for the first object.

4 0
3 years ago
______ is most resistant to thermal energy transfers. a. copper wire b. aluminum foil c. water d. foam insulation
Reika [66]

Answer:

Water

Explanation:

Because it does not conduct much energy.

8 0
3 years ago
Three moles of an ideal gas with a molar heat capacity at constant volume of 4.9 cal/(mol∙K) and a molar heat capacity at consta
Sergeu [11.5K]

Answer:

The heat flows into the gas during this two-step process is 120 cal.

Explanation:

Given that,

Number of moles = 3

Heat capacity at constant volume = 4.9 cal/mol.K

Heat capacity at constant pressure = 6.9 cal/mol.K

Initial temperature = 300 K

Final temperature = 320 K

We need to calculate the heat flow in to gas at constant pressure

Using formula of heat

\Delta H_{1}=nC_{p}\times\Delta T

Put the value into the formula

\Delta H_{1}=3\times6.9\times(320-300)

\Delta H_{1}=414\ cal

We need to calculate the heat flow in to gas at constant volume

Using formula of heat

\Delta H_{1}=nC_{v}\times\Delta T

Put the value into the formula

\Delta H_{1}=3\times4.9\times(300-320)

\Delta H_{1}=-294\ cal

We need to calculate the heat flows into the gas during two steps

Using formula of total heat

\Delta H_{T}=\Delta H_{1}+\Delta H_{2}

\Delta H_{T}=414-294

\Delta H_{T}=120\ cal

Hence, The heat flows into the gas during this two-step process is 120 cal.

7 0
3 years ago
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