Answer:
Since the area of the perfect square is 11650, and all of a squares sides ar equal, we just need to find the square root.
The square root of 11650 is 107.935166.
One side of the square is 107.935166
107.935166 x 107.935166 = 11650
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Here light ray strikes to interface at an angle of 45 degree and then refracts into other medium such that it will bend towards boundary.
So here the angle of incidence will be less than the angle of refraction as light moves towards the boundary after refraction which mean it will bend away from the normal
here it can be said that medium 2 will be rarer then medium 1
So here the possible options are
1. Water
Air
2. Diamond
Air
So in above two options medium 1 is denser and medium 2 is rarer
Answer:
The ratio of the resistances of second coil to the first coil is the ratio of square of radius of the first coil to the square of radius of second coil.
And
The ratio of the resistances of fourth coil to the third coil is the ratio of square of radius of the third coil to the square of radius of fourth coil.
Explanation:
The resistance of the coil is directly proportional to the length of the coil and inversely proportional to the area of coil and hence inversely proportional to the square of radius of the coil.
So, the ratio of the resistances of second coil to the first coil is the ratio of square of radius of the first coil to the square of radius of second coil.
And
The ratio of the resistances of fourth coil to the third coil is the ratio of square of radius of the third coil to the square of radius of fourth coil.
Answer:=14,160 kJ
Explanation: Let m1 and m2 be the initial and final amounts of mass within the tank, respectively. The steam properties are listed in the table below
Specific Internal SpecificTemp Pressure Volume Energy Enthalpy Quality Phase
C MPa m^3/kg kJ/kg kJ/kg
1 260 4.689 0.02993 2158 2298 0.7 Liquid Vapor Mixture
2 260 4.689 0.0422 2599 2797 1 Saturated Vapor
The mass initially contained in the tank is m1 = V/v1
m1 =0.85 m^3 /0.02993 m^3 /kg
= 28.4 kg
The mass finally contained in the tank is
m2 =V2/v
= 0.85 m^3 /0.0422 m^3 /kg
= 20.14 kg
The heat transfer is then
Qcv = m2u2 − m1u1 − he(m2 − m1)
Qcv = (20.14)(2599) − (28.4)(2158) − (2797)(20.14 − 28.4) = 14,160 kJ
Answer: 5.68g/ML.
Explanation; Divide the mass of the unknown substance and the volume of the unknown substance and you will have your answer.
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