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marta [7]
3 years ago
14

How many electrons can a nucleas hold

Physics
1 answer:
Hoochie [10]3 years ago
5 0
A nuclear can hold 2 electrons
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which criteria are responsible for deciding whether a heterogeneous mixture is a colloid or a suspension
ad-work [718]
I think you forgot to give the options along with the question. I am answering the question based on my knowledge and research. The criteria responsible for deciding whether a heterogeneous mixture is a colloid or a suspension is whether the <span>particles remain suspended for an extended period of time. I hope it helps you.</span>
6 0
3 years ago
If a source of sound waves is rapidly approaching a person, the sound heard by the person appears to have
velikii [3]

Correct answer choice is:

D. A frequency higher than the original frequency.

Explanation:

This is a true case of Doppler's effect. The Doppler effect can be defined as the effect originated by a traveling source of waves in which there is a visible higher variation in pulse for observers towards what the source is progressing and a visible descending shift in rate for observers from what the source is dropping.

8 0
3 years ago
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What type of organization is used in a paragraph that lists similarities between two objects?
alekssr [168]
<span>A. Comparison 

</span>What type of organization is used in a paragraph that lists similarities between two objects? Comparison


NOT:
<span>B. Contrast 
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3 0
3 years ago
A pendulum oscillates 12 times in 4 seconds. what is the length of the pendulum?
seraphim [82]

Answer:

L = 2.8 cm

Explanation:

Period T = 4 / 12 = 1/3 s

T = 2π√(L/g)

L = (T/2π)²g

L = ((1/3)/2π)²9.8 = 0.02758... ≈ 2.8 cm

6 0
2 years ago
A 5-g lead bullet traveling in 20°C air at 300 m/s strikes a flat steel plate and stops.
densk [106]

To solve this problem it is necessary to apply the concepts related to the Kinetic Energy and the Energy Produced by the heat loss. In mathematical terms kinetic energy can be described as:

KE = \frac{1}{2} mv^2

Where,

m = Mass

v = Velocity

Replacing we have that the Total Kinetic Energy is

KE = \frac{1}{2} mv^2

KE = \frac{1}{2} (5*10^{-3})(300)^2

KE =  225J

On the other hand the required Energy to heat up t melting point is

Q_1 = mC_p \Delta T

Q_2 = L_f m

Where,

m = Mass

C_p =Specific Heat

\Delta T =Change at temperature

L_f = Latent heat of fussion

Heat required to heat up to melting point,

Q = Q_1+Q_2

Q = mC_p \Delta T+L_f m

Q = 5*0.128*(327-20) + 5*24.7

Q = 310J

The energy required to melt is larger than the kinetic energy. Therefore the heat of fusion of lead would be 327 ° C: The melting point of lead.

4 0
3 years ago
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