Answer:
As it moves farther and farther away from Q, it's speed will keep increasing.
Explanation:
This is so because the point charges are both positive charges and positive charges repel.
As the charges repel, they do so at an increasing speed.
The Balmer light series comes under the visible light.
<u>Explanation:</u>
The transition of electrons from higher to energy level with 2 as principal quantum number results in the spectral emission lines of hydrogen atom and this series of lines are known as Balmer series.
Mostly, these lines has the wavelength of more than 400 nm but lesser than 700 nm. Generally of the four categories namely, 410, 434, 486, 656 nm which comes under the type of visible light. So, it can be concluded that the Balmer series light falls under visible light.
In astronomy, Balmer lines occur in various stellar (celestial or astronomical) objects due to the higher content of hydrogen in the universe. Therefore, they are commonly seen and relatively strong when compared to other element lines.
Note: nm is nanometer (one billionth of a meter in length)
The pressure of nitrogen which is needed to maintain a N2 concentration of 0. 53 m is 3.2 × 10^(4).
<h3>What is pressure? </h3>
It is defined as the continuous physical force applied on or against an object by something which is in contact with it.
It is also defined as the force per unit area.
<h3>What is henry's law? </h3>
The henry law constant is thr ratio of the partial pressure of compound in air to the concentration of compound in water at given temperature.
C= kp
where,
C is the concentration of compound = 0.53m
k is the henry constant = 6. 2×10−4matm
p is the pressure of compound
By substituting all the value we get,
C = 6. 2×10−4 × p
0.53 = 6. 2×10−4 × p
p = 0.53/6. 2×10−4
p = 3.2 × 10^(4)
Thus we find that the pressure needed to maintain a N2 concentration of 0. 53 m is 3.2 × 10^(4).
learn more about Henry's law:
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Answer: a) 10.4km
b) 0.62h
C) 17km/h
d) 9.1km/h
Explanation:
A) suppose you move in the positive direction of an x axis, from a first position of x1 = 0 to a second position of x2 at the station. That second position must be at x2 = 8.4 km + 2.0 km = 10.4 km.
B) time interval Dtwlk (= 0.50 h), but we don't know the driving time interval Dt/dr. However, we know that for the drive the displacement Dx/dr is 8.4 km and the average velocity vavg,dr is 70 km/h.
This average velocity is the ratio of the displacement for the drive to the time interval for the drive:
dv = dx/ dt
dt = dx/dv = 8.4/70 = 0.12h
t = ti + two
t = 0.12 + 0.50 = 0.62h
C) avg speed for the entire trip is the ratio of the displacement of 10.4 km for the entire trip to the time interval of 0.62 h for the entire trip.
Avg v = distance/time
10.4/0.62 = 16.8km/h
D) average speed is the ratio of the total distance you move to the total time interval you take to make that move. The total distance is 8.4 km + 2.0 km + 2.0 km = 12.4 km. The total time interval is 0.12 h + 0.50 h + 0.75 h = 1.37 h.
Avg speed = 12.4/1.37 = 9.1km/h
Answer:
0.055 N
Explanation:
From coulomb's law,
F α 1/r²
F = k/r²
F₁r₁² = F₂r₂²......................... Equation 1
Where F₁ = Initial force, r₁ = initial distance, F₂ = Final force, r₂ = Final distance.
making F₂ the subject of the equation
F₂ = F₁r₁²/r₂²..................... Equation 2
Given: F₁ = 5.50 N.
Let: r₁ = x m, then r₂= 10x m
Substitute into equation 2
F₂ = 5.5(x²)/(10x)²
F₂ = 5.5x²/100x²
F₂ = 5.5/100
F₂ = 0.055 N.
Hence the force becomes 0.055 N
