The answers to your questions are as written below:
- The objects that represents a negatively charged particle is : Image B
- The object that represents a positively charged molecule is : Image A
- The object that represents an uncharged molecule is : Image C
- The object the will not move when in an electric fied is : Image C
<h3>Different types of charges molecules</h3>
A negatively charged molecule move inwards when placed in an electric field while positively charged molecule placed in a electric field will move outwards the electric field.
A neutral/uncharged molecule will remains still when placd in an elctric field due to the absence of charges.
Hence we can concude that the answers to your questions are as listed above.
Learn more about electric charges :brainly.com/question/857179
#SPJ4
attached below is the missing image
The wave will decrease in its frequency due to a disturbing force acting upon it.
acceleration of car is 1.8 m/s^2
time = 14 s
initial speed = 14 m/s
so the final speed is calculated by
so the total distance moved in this interval of time is
now the average speed is given as
so the average speed will be 26.6 m/s
The boiling point of ethanol is at 78.37°C. So, the energy must include sensible heat to raise 19°C to the boiling point and latent heat to change liquid to gas. The equation would be
Energy = Sensible heat + Latent heat
Energy = mCpΔT + mΔH
For ethanol,
Cp = 46.068 + 102,460T - 139.63T² - 0.030341T³ + 0.0020386T⁴ J/kmol·K
ΔH = 38,560 J/mol
Integrate the Cp expression to determine CpΔT:
CpΔT = ∫₂₉₂³⁵²(46.068 + 102,460T - 139.63T² - 0.030341T³ + 0.0020386T⁴ )dT
The upper limit is (78.37+273) = 352 K, while the lower limit is (19 + 273) = 292.
CpΔT = 2384857192 J/kmol·K
2,000 J = m(2384857192 J/kmol)(1 kmol/1000 mol) + m(38,560 J/mol)
m = 8.253×10⁻⁴ moles of ethanol
Since the molar mass of ethanol is 46.07 g/mol,
Mass = (8.253×10⁻⁴ mol)(46.07 g/mol)
Mass = 0.038 g ethanol
Answer:
Making a vertical vector, we have a starting point at (-5,2) and an end point at (5,2) that will give us a vector of magnitude of 10 units.
Explanation:
In order to make vectors that have a magnitude of 10 units, the distance between the starting and ending points must be equal to 10.
The easiest way is to set points on either an horizontal or vertical line to make horizontal or vertical vectors.
We can have starting point at (-5,2) and then move up 10 units so we will be at the ending point (5,2), thus the distance between them is 10 units so the vector has a magnitude of 10 units.
We can verify that using the formula for the magnitude which requires first to find the vector.
So for the points we have
We can work with each component, for the x component we have 5-(-5) which give us 10 and for the y component we have 2-2 which give us 0, so the vector is
Thus its magnitude is
Thus we have verified our vector has a length 10.
