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Savatey [412]
3 years ago
9

What is the number of the lowest energy level that contains an f sublevel?. . 3. . 4. . 5. . 6

Physics
1 answer:
DENIUS [597]3 years ago
7 0
The correct answer among all the other choices is 4. This is the number of the lowest energy level that contains an f sublevel. Thank you for posting your question. I hope that this answer helped you. Let me know if you need more help. 
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How can professional education improve life of people?<br>​
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7 0
3 years ago
If an object is dropped from a tall building and hits the ground 3.0 s later, how tall is the building?
allsm [11]
D = 1/2 g t^2. It works out to 44.1 meters.
6 0
3 years ago
A(n) 55.5 g ball is dropped from a height of 53.6 cm above a spring of negligible mass. The ball compresses the spring to a maxi
Serggg [28]

Answer:

The spring force constant is  k=243\ \frac{N}{m} .

Explanation:

We are told the mass of the ball is m=0.0555\ kg, the height above the spring where the ball is dropped is h=0.536\ m,  the length the ball compresses the spring is d=0.04897\ m and the acceleration of gravity is 9.8\ \frac{m}{s^{2}} .

We will consider the initial moment to be when the ball is dropped and the final moment to be when the ball stops, compressing the spring. We supose that there is no friction so the initial mechanical energy E_{mi} is equal to the final mechanical energy E_{mf} :

                                                    E_{mf}=E_{mi}

Initially there is only gravitational potential energy because the force of the spring isn't present and the speed is zero. In the final moment there is only elastic potential energy because the height is zero and the ball has stopped. So we have that:

                                                   \frac{1}{2}kd^{2}=mgh

If we manipulate the equation we have that:

                                                    k=\frac{2mgh}{d^{2} }

                                         k=\frac{2\ 0.0555\ kg\ 9.8\frac{m}{s^{2}}\ 0.536\ m}{(0.04897)^{2}m^{2}}

                                              k=\frac{0.58306\ \frac{kgm^{2}}{s^{2}}}{2.398x10^{-3}m^{2}}

                                                     k=243\ \frac{N}{m}

                                                   

                             

5 0
3 years ago
A 17926-lb truck enters an emergency exit ramp at a speed of 75.6 ft/s. It travels for 6.4 s before its speed is reduced to 30.3
I am Lyosha [343]

Answer:

F_{braking}=337299 pdl

Explanation:

Impulse-Momentum relation:

I=\Delta p\\ F_{total}*t=m(v_{f}-v{o})

F_{total}=-F_{braking}+mgsin{\theta}

We solve the equations in order to find the braking force:

F_{braking}=m(v_{o}-v{f})/t+mgsin{\theta}=17926(75.6-30.3)/6.4+17926*32.17*sin21.4=337299 pdl

6 0
3 years ago
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