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Ivenika [448]
3 years ago
14

Please send me the solution to these questions.

Physics
1 answer:
Oxana [17]3 years ago
4 0

Answer:

sorry I am not getting dear friend

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It is Limiting Reactant

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An echo is a reflection of sound against another surface. Suppose you are standing on one side of a canyon and you
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10-1 Temperature and Expansion. 135 ... text, the laboratory work that you do, or your physics teacher. ... Assume that the speed of sound in air is 340 m/s. How.

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3 years ago
Which sentence most likely explains why the rock has large open pores
hoa [83]

ur answer is C) Bubbles formed in the magma when it was still in liquid form.
6 0
3 years ago
In the United States, household electric power is provided at a frequency of 60 HzHz, so electromagnetic radiation at that frequ
grigory [225]

Answer:

the maximum intensity of an electromagnetic wave at the given frequency is 45 kW/m²

Explanation:

Given the data in the question;

To determine the maximum intensity of an electromagnetic wave, we use the formula;

I = \frac{1}{2}ε₀cE_{max²

where ε₀ is permittivity of free space ( 8.85 × 10⁻¹² C²/N.m² )

c is the speed of light ( 3 × 10⁸ m/s )

E_{max is the maximum magnitude of the electric field

first we calculate the maximum magnitude of the electric field ( E_{max  )

E_{max = 350/f kV/m

given that frequency of 60 Hz, we substitute

E_{max = 350/60 kV/m

E_{max = 5.83333 kV/m

E_{max = 5.83333 kV/m × ( \frac{1000 V/m}{1 kV/m} )

E_{max = 5833.33 N/C

so we substitute all our values into the formula for  intensity of an electromagnetic wave;

I = \frac{1}{2}ε₀cE_{max²

I = \frac{1}{2} × ( 8.85 × 10⁻¹² C²/N.m² ) × ( 3 × 10⁸ m/s ) × ( 5833.33 N/C )²

I = 45 × 10³ W/m²

I = 45 × 10³ W/m² × ( \frac{1 kW/m^2}{10^3W/m^2} )

I = 45 kW/m²

Therefore, the maximum intensity of an electromagnetic wave at the given frequency is 45 kW/m²

7 0
3 years ago
An ideal Diesel cycle has a compression ratio of 18 and a cutoff ratio of 1.5. Determine the (1) maximum air temperature and (2)
weqwewe [10]

Answer:

(1) The maximum air temperature is 1383.002 K

(2) The rate of heat addition is 215.5 kW

Explanation:

T₁ = 17 + 273.15 = 290.15

\frac{T_2}{T_1} =r_v^{k - 1} =18^{0.4} =3.17767

T₂ = 290.15 × 3.17767 = 922.00139

\frac{T_3}{T_2} =\frac{v_3}{v_2} = r_c = 1.5

Therefore,

T₃ = T₂×1.5 = 922.00139 × 1.5 = 1383.002 K

The maximum air temperature = T₃ = 1383.002 K

(2)

\frac{v_4}{v_3} =\frac{v_4}{v_2} \times \frac{v_2}{v_3}  = \frac{v_1}{v_2} \times \frac{v_2}{v_3} = 18 \times \frac{1}{1.5} = 12

\frac{T_3}{T_4} =(\frac{v_4}{v_3} )^{k-1} = 12^{0.4} = 2.702

Therefore;

T_4 = \frac{1383.002}{2.702} =511.859 \ k

Q_1 = c_p(T_3-T_2)

Q₁ = 1.005(1383.002 - 922.00139) = 463.306 kJ/jg

Heat rejected per kilogram is given by the following relation;

c_v(T_4-T_1)  = 0.718×(511.859 - 290.15) = 159.187 kJ/kg

The efficiency is given by the following relation;

\eta = 1-\frac{\beta ^{k}-1}{\left (\beta -1  \right )r_{v}^{k-1}}

Where:

β = Cut off ratio

Plugging in the values, we get;

\eta = 1-\frac{1.5 ^{1.4}-1}{\left (1.5 -1  \right )18^{1.4-1}}= 0.5191

Therefore;

\eta = \frac{\sum Q}{Q_1}

\therefore 0.5191 = \frac{150}{Q_1}

Heat supplied = \frac{150}{0.5191}  = 288.978 \ hp

Therefore, heat supplied = 215491.064 W

Heat supplied ≈ 215.5 kW

The rate of heat addition = 215.5 kW.

7 0
4 years ago
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