Ethylene glycol is termed as the primary ingredients in antifreeze.
The ethylene glycol molecular formula is C₂H₆O₂.
Molar mass of C₂H₆O₂ is = (2×12) +(6×1) + (216) = 62g/mol
Now that antifreeze by mass is 50%, then there is 1kg of ethylene glycol which is present in 1kg of water.
ΔTf = Kf×m
ΔTf = depression in the freezing point.
= freezing point of water freezing point of the solution
= O°c - Tf
= -Tf
Kf = depression in freezing constant of water = 1.86°C/m
M is the molarity of the solution.
=(mass/molar mass) mass of solvent in kg
=1000g/62 (g/mol) /1kg
=16.13m
If we plug the value we get
-Tf = 1.86 × 16.13 = 30
Tf = -30°c
Answer:

Explanation:
According to Newton's first law:

The component of the force on the y-axis can be obtained through the Pythagorean Theorem. This is because the components are the cathetus of a right triangle and its hypotenuse is the magnitude of the force:

Replacing and solving for N:

Answer:
Magnetic field will be ZERO at the given position
Explanation:
As we know that the magnetic field due to moving charge is given as

so here we know that for the direction of magnetic field we will use

so we have

so magnetic field must be ZERO
So whenever charge is moving along the same direction where the position vector is given then magnetic field will be zero
Answer:
303 Ω
Explanation:
Given
Represent the resistors with R1, R2 and RT
R1 = 633
RT = 205
Required
Determine R2
Since it's a parallel connection, it can be solved using.
1/Rt = 1/R1 + 1/R2
Substitute values for R1 and RT
1/205 = 1/633 + 1/R2
Collect Like Terms
1/R2 = 1/205 - 1/633
Take LCM
1/R2 = (633 - 205)/(205 * 633)
1/R2 = 428/129765
Take reciprocal of both sides
R2 = 129765/428
R2 = 303 --- approximated
The answer is 1.01 x 10^(-11) N. I arrived to this answer through calculating the GPEs of both balls. Bjorn's ball has a GPE of 1.402 x 10^(-11) N. Billie Jean's ball has a GPE of <span>2.503 x 10^(-11) N. I subtracted the two and I found that Billie Jean's tennis ball has a GPE of 1.01 x 10^(-11) more than Bjorn's tennis ball.</span>