Answer:
m<2 = 86
Explanation:
Vertical angles means the angles are equal to each other so:
<1 = <2
17x + 1 = 20x - 14
17x - 20x = -14 - 1
-3x = -15
x = 5
m<2:
20x - 14
20(5) - 14
100 - 14
86
-- Depending on the time of the year, the sun's rays strike Earth
most directly somewhere between the Tropic of Cancer and the
Tropic of Capricorn.
That's a belt around the Earth's "middle" called the "Tropic Zone".
The equator is in the middle of it, the Tropic of Cancer is 23.5° North
of the equator, and the Tropic of Capricorn is 23.5° degrees South of it.
The sun's rays can never be totally direct, straight down onto the Earth's
surface, anywhere outside this belt.
-- The sun's rays strike Earth least directly wherever, and whenever,
the sun is setting.
Answer:
To identify the electron transition that has the highest frequency in the hydrogen atom.Answer: The frequency of light produced by electron transition is highest among the given electron transition, therefore, the option (a) is correct.Explanation:The energy of emitted light by the electron transition is given as: is the energy of the initial level. is the energy of the final level.The energy difference for the electron transition is highest among given electronic transition.The energy of the light is given as:Here, is the plank’s constant. is the frequency of the light.The energy of the light is directly proportional to the frequency of the light.Therefore the frequency of light by the electron transition is highest among given electronic transition.
Explanation:
Answer:
The measured value of the constant is known with some certainty to four significant digits. In SI units, its value is approximately 6.674×10−11 m3⋅kg−1⋅s−2. The modern notation of Newton's law involving G was introduced in the 1890s by C. V. Boys.
Answer:
298,220 N
Explanation:
Let the force on car three is T_23-T_34
Since net force= ma
from newton's second law we have
T_23-T_34 = ma
therefore,
T_23-T_34 = 37000×0.62
T_23= 22940+T_34
now, we need to calculate
T_34
Notice that T_34 is accelerating all 12 cars behind 3rd car by at a rate of 0.62 m/s^2
F= ma
So, F= 12×37000×0.62= 22940×12= 275280 N
T_23 =22940+T_34= 22940+ 275280= 298,220 N
therefore, the tension in the coupling between the second and third cars
= 298,220 N
