7
Explanation:
its obv 7 cuz each shell contains a certain mass
When drawing a Bohr diagram for a hydrogen atom in the 4th excited state, five energy shells must be drawn. This includes the atom's base state plus the four excited states.
Explanation:In the context of a Bohr diagram for a hydrogen atom, when the atom is in the 4th excited state, you would draw five energy shells. The reason is that the base state of the atom is counted as the first energy level, or shell. So, for each excited state, you're essentially adding one more energy shell. Hence, for the 4th excited state, we count the base state plus the four excited states, which gives us a total of five energy shells.
Learn more about Bohr diagram here:https://brainly.com/question/39266877
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A ball of mass 0.16 kg is moving forwards at a speed of 0.50 m/s. A second ball of mass 0.10kg is stationary. The first ball strikes the second ball. The second ball moves forwards at a speed of
0.50 m/s.
What is the speed of the first ball after the collision?
Answer:
0.19 m/s
Explanation:
Momentum before collision = momentum after collision
m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂
(0.16 kg) (0.50 m/s) + (0.10 kg) (0 m/s) = (0.16 kg) v₁ + (0.10 kg) (0.50 m/s)
0.08 kg m/s = (0.16 kg) v₁ + 0.05 kg m/s
0.03 kg m/s = (0.16 kg) v₁
v₁ ≈ 0.19 m/s
Final answer:
After applying the conservation of momentum principle to the collision, the final velocity of the first ball after the collision with the second stationary ball is calculated to be 0.1875 m/s.
Explanation:
The subject of this question is Physics, specifically the conservation of momentum during collisions. When analyzing collisions, we use the principle of conservation of momentum, which states that the total momentum of a closed system before the collision is equal to the total momentum after the collision, provided no external forces act on the system. In the scenario given, a 0.16 kg ball moving at 0.50 m/s hits a stationary 0.10 kg ball which afterwards moves forward at 0.50 m/s. To find the speed of the first ball after the collision, we can set the total initial momentum equal to the total final momentum:
Initial momentum = (0.16 kg * 0.50 m/s) + (0.10 kg * 0 m/s)
Final momentum = (0.16 kg * v) + (0.10 kg * 0.50 m/s)
By solving the equation, we can find the final speed v of the first ball. The solution involves isolating v on one side of the equation.
Calculation:
0.08 kg·m/s = (0.16 kg * v) + 0.05 kg·m/s
0.08 kg·m/s - 0.05 kg·m/s = 0.16 kg * v
0.03 kg·m/s = 0.16 kg * v
v = 0.03 kg·m/s / 0.16 kg
v = 0.1875 m/s
The final velocity of the first ball after the collision is therefore 0.1875 m/s.
please help please please please please
Answer:
Your answer would be B.
Explanation:
Ionic bonding involves transfer of electrons to complete the octet valence shell.
Please mark Brainliest if this helps!
Answer:
I think it is B
Explanation:
Ionic bonding is the complete transfer of valence electron(s) between atoms. It is a type of chemical bond that generates two oppositely charged ions. In ionic bonds, the metal loses electrons to become a positively charged cation, whereas the nonmetal accepts those electrons to become a negatively charged anion.
List 3 different things a bus driver can do while driving to make the bus accelerate.
This is for a physics class for 9th grade.
Answer:
Step on the gas pedal (increase in speed).
Step on the brakes (decrease in speed).
Turn the steering wheel (change in direction).
A bus driver can make the bus accelerate by pressing the gas pedal, releasing the brakes, or changing to a higher gear.
To make the bus accelerate, a bus driver can perform several actions. Three ways a bus driver can cause the bus to accelerate are:
1. Pressing down on the gas pedal, which increases the engine's output and propels the bus forward, increasing its speed.
2. Releasing the brakes, if they were partially engaged, reducing the force opposing the bus's forward motion and allowing it to speed up.
3. Changing the gear ratio, if applicable, by shifting to a higher gear in vehicles with a manual transmission, which can increase acceleration at different speeds.
An aircraft travelling at 600 km/h accelerates steadily
at 10km/h per second. Taking the speed of sound as
1100 km/h at the aircraft's altitude, how long will it take to
reach the 'sound barrier'?
Answer: 50.14 s
Explanation:
We can solve this problem by the following equation:
[tex]V=V_{o}+at[/tex] (1)
Where:
[tex]V=1100\frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=305.55 m/s[/tex] is the final velocity of the aircraft.
[tex]V_{o}=600\frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=166.66 m/s[/tex] is the initial velocity of the aircraft
[tex]a=2.77 m/s^{2}[/tex] is the acceleration of the aircraft (taking into account 10 km/h=2.77 m/s and acceleration is \frac{2.77 m/s}{1 s})
[tex]t[/tex] is the time it takes to the aircraft to reach the sound barrier
Isolating [tex]t[/tex] from (1):
[tex]t=\frac{V-V_{o}}{a}[/tex] (2)
[tex]t=\frac{305.55 m/s-166.66 m/s}{2.77 m/s^{2}}[/tex] (3)
Finally:
[tex]t=50.14 s[/tex] (4)