What it looks to be that you found in A was the "initial"...b/c the question asks:
<span>"how much energy does the electron have 'initially' in the n=4 excited state?" </span>
<span>"final" would be where it 'finally' ends up at, ie. its last stop...as for this question...the 'ground state' as in its lowest energy level. </span>
The answer comes to: <span>−1.36×10^−19 J</span>
You use the same equation for the second part as for part a.
<span>just have to subract the 2 as in the only diff for part 2 is that you use 1squared rather than 4squared & subract "final -initial" & you should get -2.05*10^-18 as your answer. </span>
The final velocity of skater 1 is 3.7 m/s to the right. The right option is O A. 3.7 m/s to the right.
<h3>What is velocity?</h3>
Velocity can be defined as the ratio of the displacement and time of a body.
To calculate the final velocity of Skater 1 we use the formula below.
Formula:
- mu+MU = mv+MV............ Equation 1
Where:
- m = mass of the first skater
- M = mass of the second skater
- u = initial velocity of the first skater
- U = initial velocity of the second skater
- v = final velocity of the first skater
- V = final velocity of the second skater.
make v the subject of the equation.
- v = (mu+MU-MV)/m................ Equation 2
Note: Let left direction represent negative and right direction represent positive.
From the question,
Given:
- m = 105 kg
- u = -2 m/s
- M = 71 kg
- U = 5 m/s
- V = -3.4 m/s.
Substitute these values into equation 2
- v = [(105×(-2))+(71×5)-(71×(-3.4))]/105
- v = (-210+355+241.4)/105
- v = 386.4/105
- v = 3.68 m/s
- v ≈ 3.7 m/s
Hence, the final velocity of skater 1 is 3.7 m/s to the right. The right option is O A. 3.7 m/s to the right.
Learn more about velocity here: brainly.com/question/25749514
Answer
given,
range of the projectile = 4.3 m
time of flight = T = 0.829 s
v = 5.19 m/s
vertical component of velocity of projectile
v_y = gt'
a) Launch angle
θ = 38°
b) initial speed of projectile
v = 6.59 m/s
c) maximum height reached by the projectile
Answer:
Answer:
1.1 x 10^9 ohm metre
Explanation:
diameter = 1.5 mm
length, l = 5 cm
Potential difference, V = 9 V
current, i = 230 micro Ampere = 230 x 10^-6 A
radius, r = diameter / 2 = 1.5 / 2 = 0.75 x 10^-3 m
Let the resistivity is ρ.
Area of crossection
A = πr² = 3.14 x 0.75 x 0.75 x 10^-6 = 1.766 x 10^-6 m^2
Use Ohm's law to find the value of resistance
V = i x R
9 = 230 x 10^-6 x R
R = 39130.4 ohm
Use the formula for the resistance
ρ = 1.1 x 10^9 ohm metre
Explanation:
