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kap26 [50]
3 years ago
9

Make a simple pendulum and find its period for a given length of the pendulum.

Physics
1 answer:
Kruka [31]3 years ago
7 0

Answer:

The period of a simple pendulum is T=2π√Lg T = 2 π L g , where L is the length of the string and g is the acceleration due to gravity.

Explanation:

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Answer:

The speed of light measured in any frame is c = 3.00E8 m/s.

This is one  of Einstein's  postulates of special relativity.

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2 years ago
A box of books has _______inertia than a box of cotton balls.
White raven [17]

Talking about a quantity of inertia is exactly the same as talking about a quantity of mass.  So, if the boxes are anywhere near the same size, then the box of books has <u><em>more</em></u> inertia than the box of cotton balls, because books have more mass than an identical volume of cotton.

4 0
3 years ago
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What information is given by the formula of an ionic compound?
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4 0
3 years ago
Elements that typically give up electrons CHECK ALL THAT APPLY
lys-0071 [83]

Answer:

B. have a lower ionization energy

D. are metals

Explanation:

An atom can be defined as the smallest unit comprising of matter that forms all chemical elements. Thus, atoms are basically the building blocks of matters and as such determines or defines the structure of a chemical element.

Generally, atoms are typically made up of three distinct particles and these are protons, neutrons and electrons.

In Chemistry, electrons can be defined as subatomic particles that are negatively charged and as such has a magnitude of -1.

Valence electrons can be defined as the number of electrons present in the outermost shell of an atom. Valence electrons are used to determine whether an atom or group of elements found in a periodic table can bond with others. Thus, this property is typically used to determine the chemical properties of elements.

Valency can be defined as a measure of the combining power of a chemical element with other atoms to form a molecule or chemical compound.

Typically, valency is measured by the amount of hydrogen atoms that a chemical element can combine with or displace to form a molecule or chemical compound.

Ionization energy can be defined as the minimum energy required to remove or detach an electron from a neutral atom in a gaseous state.

Generally, the ionization energy of chemical elements tend to increase from left to right across a period on the periodic table. This increase is due to the fact that the atomic radius of chemical elements generally decreases across the periodic table, typically from alkali metals (group one elements such as hydrogen, lithium and sodium) to noble gases (group eight elements such as argon, helium and neon) i.e from left to the right of the periodic table. Also, the atomic radius of a chemical element increases down each group of the periodic table, typically from top to bottom (column).

This ultimately implies that, atoms with relatively large atomic radii tend to have a low electron affinity and a low ionization energy.

In conclusion, chemical elements that typically give up electrons are metals because their outermost shell contains excess electrons and have a lower ionization energy.

4 0
3 years ago
Moist air initially at 1258C, 4 bar, and 50% relative humidity is contained in a 2.5-m3 closed, rigid tank. The tank contents ar
brilliants [131]

Here is the missing part of the question

To Determine the heat transfer, in kJ  if the final temperature in the tank is 110 deg C

Answer:

Explanation:

The image attached below shows the process on T - v diagram

<u>At State 1:</u>

The first step is to find the vapor pressure

P_{v1} = \rho_1 P_g_1

= \phi_1 P_{x  \ at \ 125^0C}

= 0.5 × 232 kPa

= 116 kPa

The initial specific volume of the vapor is:

P_{v_1} v_{v_1} = \dfrac{\overline R}{M_v}T_1

116 \times 10^3 \times v_{v_1} = \dfrac{8314}{18} \times (125 + 273)

116 \times 10^3 \times v_{v_1} = 183831.7778

v_{v_1} = 1.584 \ m^3/kg

<u>At State 1:</u>

The next step is to determine the mass of water vapor pressure.

m_{v1} = \dfrac{V}{v_{v1}}

= \dfrac{2.5}{1.584}

= 1.578 kg

Using the ideal gas equation to estimate the mass of the dry air m_aP_{a1} V = m_a \dfrac{\overline R}{M_a}T_1

(P_1-P_{v1})  V = m_a \dfrac{\overline R}{M_a}T_1

(4-1.16) \times 10^5 \times 2.5 = m_a \dfrac{8314}{28.97}\times ( 125 + 273)

710000= m_a \times 114220.642

m_a = \dfrac{710000}{114220.642}

m_a = 6.216 \ kg

For the specific volume v_{v_1} = 1.584 \ m^3/kg , we get the identical value of saturation temperature

T_{sat} = 100 + (110 -100) \bigg(\dfrac{1.584-1.673}{1.210 - 1.673}\bigg)

T_{sat} =101.92 ^0\ C

Thus, at T_{sat} =101.92 ^0\ C, condensation needs to begin.

However, since the exit temperature tends to be higher than the saturation temperature, then there will be an absence of condensation during the process.

Heat can now be determined by using the formula

Q = ΔU + W

Recall that: For a rigid tank, W = 0

Q = ΔU + 0

Q = ΔU

Q = U₂ - U₁

Also, the mass will remain constant given that there will not be any condensation during the process from state 1 and state 2.

<u>At State 1;</u>

The internal energy is calculated as:

U_1 = (m_a u_a \ _{ at \ 125^0 C})+ ( m_{v1} u_v \ _{ at \ 125^0 C} )

At T_1 = 125° C, we obtain the specific internal energy of air

SO;

U_{a \ at \ 125 ^0C } = 278.93 + ( 286.16 -278.93) (\dfrac{398-390}{400-390}   )

=278.93 + ( 7.23) (\dfrac{8}{10}   )

= 284.714 \ kJ/kg\\

At T_1 = 125° C, we obtain the specific internal energy of  water vapor

U_{v1 \ at \ 125^0C} = u_g = 2534.5 \ kJ/kg

U_1 = (m_a u_a \ at \ _{  125 ^0C }) + ( m_{v1} u_v  \ at \ _{125^0C} )

= 6.216 × 284.714 + 1.578 × 2534.5

= 5768.716 kJ

<u>At State 2:</u>

The internal energy is calculated as:

U_2 = (m_a u_a \ _{ at \ 110^0 C})+ ( m_{v1} u_v \ _{ at \ 110^0 C} )

At temperature 110° C, we obtain the specific internal energy of air

SO;

U_{a \ at \ 110^0C } = 271.69+ ( 278.93-271.69) (\dfrac{383-380}{390-380}   )

271.69+ (7.24) (0.3)

= 273.862 \ kJ/kg\\

At temperature 110° C, we obtain the specific internal energy of  water vapor

U_{v1 \ at \ 110^0C}= 2517.9 \ kJ/kg

U_2 = (m_a u_a \ at \ _{  110 ^0C }) + ( m_{v1} u_v  \ at \ _{110^0C} )

= 6.216 × 273.862 + 1.578 × 2517.9

= 5675.57 kJ

Finally, the heat transfer during the process is

Q = U₂ - U₁

Q = (5675.57 - 5768.716 ) kJ

Q = -93.146 kJ

with the negative sign, this indicates that heat is lost from the system.

6 0
3 years ago
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