Individuals inherit one factor per trait from each parent, and a dominant factor can conceal the expression of a recessive one when both are present.
Explanation:
The formula illustrating the relationship between resistance and temperature is as follows:
R =
![R_{o} + \alpha [T_{2} - T_{1}]](https://tex.z-dn.net/?f=R_%7Bo%7D%20%2B%20%5Calpha%20%5BT_%7B2%7D%20-%20T_%7B1%7D%5D)
where, R = final resistance
= initial resistance
= temperature coefficient of resistivity
= final temperature 
= initial temperature
Given data as follows.
R = 36 ohm,
= 3 ohm
= 0.0045
Substituting the provided values into the above formula gives us the following.
R = 
36 =
=
![3 + 0.0045 \times [T_{2} - 293]](https://tex.z-dn.net/?f=3%20%2B%200.0045%20%5Ctimes%20%5BT_%7B2%7D%20-%20293%5D)
= 7626.33 K
Thus, it can be concluded that
the temperature of the light bulb at 12.0 V is 7626.33 K.
Ethylene glycol is known as the main component found in antifreeze.
The molecular formula for ethylene glycol is C₂H₆O₂.
Its molar mass is calculated as C₂H₆O₂ = (2×12) +(6×1) + (216) = 62g/mol
Given that antifreeze comprises 50% by weight, there exists 1 kg of ethylene glycol mixed with 1 kg of water.
ΔTf = Kf×m
ΔTf refers to the change in the freezing point.
= starting temperature of water - freezing temperature of the solution
= 0°C - Tf
= -Tf
Kf stands for the freezing point depression constant of water, which is 1.86°C/m
m is the molarity of the solution.
=(mass/molar mass) where mass of solvent is in kg
=1000g/62 (g/mol) /1kg
=16.13m
Substituting the value into the equation gives us
-Tf = 1.86 × 16.13 = 30
thus Tf = -30°C
Answer:
v = 66.4 m/s
Explanation:
We know that the aircraft starts off moving at a speed of

now we have




in the Y direction, we can apply kinematic equations



as there is no acceleration along the x-axis, the velocity in this direction remains unchanged
thus yielding



1 hour = 3,600 seconds
1 km = 1,000 meters
75 km/hour = (75,000/3,600) m/s = 20-5/6 m/s
The mean speed during the deceleration is
(1/2)(20-5/6 + 0) = 10-5/12 m/s.
Traveling at this average speed for 21 seconds,
the bus covers
(10-5/12) × (21) = 218.75 meters.