Answer:
The area that remains is 4.201 m²
Explanation:
Provided that
AB=D= 4 m (R=2 m)
The area of the half-circle AB is
A=2π
The area of the smaller half-circle is
a=5π/6 + 2√3/4 m²
a=5π/6 + √3/2 m²
<pThus, the remaining area = A - a= 2π - (5π/6 + √3/2) m²
The remaining area is expressed as 2π - (5π/6 + √3/2) m²
Answer:
Explanation:
Within a duration of 60 seconds, six waves are observed.
With a total of 6 waves,
this equates to 3 wavelengths.
As a result,
the period for each wavelength is calculated as 60 divided by 3.
Thus, period = 20 seconds.
According to the frequency-period relationship,
f = 1 / T
f = 1 / 20
f = 0.05 Hz
Answer:
155.38424 K
2.2721 kg/m³
Explanation:
= Reservoir pressure = 10 atm
= Reservoir temperature = 300 K
= Exit pressure = 1 atm
= Exit temperature
= Specific gas constant = 287 J/kgK
= Specific heat ratio = 1.4 for air
Assuming isentropic flow
Flow temperature at exit is 155.38424 K
Density at exit can be derived using the ideal gas equation
Flow density at exit measures 2.2721 kg/m³
Answer:
Electric flux is calculated as
Explanation:
We start with the given parameters:
The electric field impacting the circular surface is
Our objective is to ascertain the electric flux passing through a circular region with a radius of 1.83 m situated in the xy-plane. The area vector is oriented in the z direction. The formula for electric flux is expressed as:
Applying properties of the dot product, we calculate the electric flux as:
Consequently, the electric flux for the circular area is . Thus, this represents the required answer.