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Nataly_w [17]
3 years ago
11

The number of energy levels to which an electron can jump depends on the

Physics
1 answer:
Anettt [7]3 years ago
8 0
C. The amount of energy it absorbs.

Electrons can only transition between energy levels when they absorb or release certain quanta of energy.
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Though some self-report error may exist, surveys/interviews are useful in that they allow researchers to collect large amounts o
Leokris [45]
Definitely true, surveys and interviews aren’t flawless but you can collect lots of data from them
4 0
2 years ago
A particle of mass 4.00 kg is attached to a spring with a force constant of 100 N/m. It is oscillating on a frictionless, horizo
zloy xaker [14]

Solution :

Given :

Mass attached to the spring = 4 kg

Mass dropped = 6 kg

Force constant = 100 N/m

Initial amplitude = 2 m

Therefore,

a). $v_{initial} = A w$

          $= 2 \times \sqrt{\frac{100}{4}}$

          = 10 m/s

Final velocity, v at equilibrium position, v = 5 m/s

Now, $\frac{1}{2}(4+4)5^2 = \frac{1}{2} kA'$

A' = amplitude = 1.4142 m

b). $T=2 \pi \sqrt{\frac{m}{k}}$

    m' = 2m

    Hence, $T'=\sqrt2 T$

c). $\frac{\frac{1}{2}(4+4)5^2 + \frac{1}{2}\times 4 \times 10^2}{\frac{1}{2} \times 4 \times 10^2}$

  $=\frac{1}{2}$

Therefore, factor $=\frac{1}{2}$

Thus, the energy will change half times as the result of the collision.

7 0
3 years ago
Durante uma corrida de carros, o piloto da escuderia ganhadora percorreu a primeira volta da pista com velocidade media de 310 k
lesya692 [45]
A velocidade mínima é 0. A velocidade máxima é inferior ou igual a 620 km/h.
<span>buena suerte mi amigo</span>
3 0
3 years ago
I need the solution to this
posledela

Answer:

He could jump 2.6 meters high.

Explanation:

Jumping a height of 1.3m requires a certain initial velocity v_0. It turns out that this scenario can be turned into an equivalent: if a person is dropped from a height of 1.3m in free fall, his velocity right before landing on the ground will be v_0. To answer this equivalent question, we use the kinematic equation:

v_0 = \sqrt{2gh}=\sqrt{2\cdot 9.8\frac{m}{s^2}\cdot 1.3m}=5.0\frac{m}{s}

With this result, we turn back to the original question on Earth: the person needs an initial velocity of 5 m/s to jump 1.3m high, on the Earth.

Now let's go to the other planet. It's smaller, half the radius, and its meadows are distinctly greener. Since its density is the same as one of the Earth, only its radius is half, we can argue that the gravitational acceleration g will be <em>half</em> of that of the Earth (you can verify this is true by writing down the Newton's formula for gravity, use volume of the sphere times density instead of the mass of the Earth, then see what happens to g when halving the radius). So, the question now becomes: from which height should the person be dropped in free fall so that his landing speed is 5 m/s ? Again, the kinematic equation comes in handy:

v_0^2 = 2g_{1/2}h\implies \\h = \frac{v_0^2}{2g_{1/2}}=\frac{25\frac{m^2}{s^2}}{2\cdot 4.9\frac{m}{s^2}}=2.6m

This results tells you, that on the planet X, which just half the radius of the Earth, a person will jump up to the height of 2.6 meters with same effort as on the Earth. This is exactly twice the height he jumps on Earth. It now all makes sense.

6 0
2 years ago
You are on a sled at the top of a hemispherical, snowy hill of radius 13 m. You begin to slide down the hill. How fast are you m
8_murik_8 [283]

Answer:

Explanation:

There will be loss of potential energy due to loss of height and gain of kinetic energy .

loss of height = R - R cos 14 ,    R is radius of hemisphere .

R ( 1 - cos 12 )

= 13 ( 1 - .978 )

h = .286 m

loss of potential energy

= mgh

= m x 9.8 x .286

= 2.8 m

gain of kinetic energy

1/2 m v ² = mgh

v² = 2 g h

v²  = 2 x 9.8 x 2.8

v = 7.40 m /s

4 0
3 years ago
Read 2 more answers
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