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Licemer1 [7]
3 years ago
13

How many different values of ml are possible in the 5d sublevel?

Physics
1 answer:
m_a_m_a [10]3 years ago
4 0
<span>There are 5 different values of ml in the 5d sublevel (-2, -1, 0, 1, and 2).</span>
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which of the folloing statements about ionization energy is true? A elements toward the bottom of a group periodic table general
beks73 [17]
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accept the flow of electrons.resist the flow of electrons.accept the flow of protons.resist the flow of protons.It is one of these </span>
6 0
3 years ago
London lives 1200 meters south of Chick-fil-A and is thinking about going there for lunch. When she is ready to leave, she reali
saul85 [17]

Answer:

  1750 m

Explanation:

The distance traveled is the 750 meters to the Aunt's house plus the 1000 m from there to Chick-fil-A.

  750 +1000 = 1750 . . . meters traveled

6 0
3 years ago
A 10.2-kg mass is located at the origin, and a 4.6-kg mass is located at x = 8.1 cm. Assuming g is constant, what is the locatio
goldfiish [28.3K]

Answer:

center of mass of the two masses will lie at x = 2.52 cm

center of gravity of the two masses will lie at x = 2.52 cm

So center of mass is same as center of gravity because value of gravity is constant here

Explanation:

Position of centre of mass is given as

r_{cm} = \frac{m_1r_1 + m_2r_2}{m_1 + m_2}

here we have

m_1 = 10.2 kg

m_2 = 4.6 kg

r_1 = (0, 0)

r_2 = (8.1cm, 0)

now we have

r_{cm} = \frac{10.2 (0,0) + 4.6 (8.1 , 0)}{10.2 + 4.6}

r_{cm} = {(37.26, 0)}{14.8}

r_{cm} = (2.52 cm, 0)

so center of mass of the two masses will lie at x = 2.52 cm

now for center of gravity we can use

r_g_{cm} = \frac{m_1gr_1 + m_2gr_2}{m_1g + m_2g}

here we have

m_1 = 10.2 kg

m_2 = 4.6 kg

r_1 = (0, 0)

r_2 = (8.1cm, 0)

now we have

r_g_{cm} = \frac{10.2(9.8) (0,0) + 4.6(9.8) (8.1 , 0)}{10.2(9.8) + 4.6(9.8)}

r_g_{cm} = {(37.26, 0)}{14.8}

r_g_{cm} = (2.52 cm, 0)

So center of mass is same as center of gravity because value of gravity is constant here

3 0
3 years ago
A 0.49-kg cord is stretched between two supports, 7.8m apart. When one support is struck by a hammer, a transverse wave travels
katovenus [111]

To solve this problem we will apply the laws of Mersenne. Mersenne's laws are laws describing the frequency of oscillation of a stretched string or monochord, useful in musical tuning and musical instrument construction. This law tells us that the velocity in a string is directly proportional to the root of the applied tension, and inversely proportional to the root of the linear density, that is,

v = \sqrt{\frac{T}{\mu}}

Here,

v = Velocity

\mu= Linear density (Mass per  unit length)

T = Tension

Rearranging to find the Period we have that

T = v^2 \mu

T = v^2 (\frac{m}{L})

As we know that speed is equivalent to displacement in a unit of time, we will have to

T = (\frac{L}{t}) ^2(\frac{m}{L})

T = (\frac{7.8}{0.83})^2 (\frac{0.49}{7.8})

T = 5.54N

Therefore the tension is 5.54N

8 0
3 years ago
A ball rolls of buildings that is 100m high calculate the time that it takes for ball to hit the ground​
LUCKY_DIMON [66]

Answer:

2as=v2-u2

2000=v2

V=44

V=u+at

44/10=t

T=4.4seconds

5 0
3 years ago
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