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ANEK [815]
3 years ago
15

An atom contains 15 neutrons and has a mass number of 30. What is its atomic number? How did you determine the atomic number?

Physics
1 answer:
irinina [24]3 years ago
8 0

Answer:

15

Explanation:

The mass number is defined as the total number of protons and neutrons. The atomic number is total number of protons. And since we have the mass number and number of neutrons, we can use simple subtraction (30 - 15) to find that the atomic number is 15.

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Dr. Matthews has submitted a proposal to the institutional review board (IRB) of a university. At this university, she intends t
OlgaM077 [116]
The IRB at the university will decide whether her study meets ethical guidelines before it is initiated. The importance of these codes of conduct is to safeguard research participants, the status of psychology and the researchers or psychologists themselves. Moral issues hardly yield a simple, unequivocal, right or wrong answer. It is consequently often a matter of judgment whether the research is justified or not. For instance, it might be that a study roots psychological or physical uneasiness to participants, maybe they agonize pain or maybe even come to solemn harm.
6 0
3 years ago
Suppose that the sound level of a conversation is initially at an angry 70 db and then drops to a soothing 50 db. assuming that
stepan [7]
Angry sound level = 70 db
Soothing sound level = 50 db
Frequency, f = 500 Hz
Assuming speed of sound = 345 m/s
Density (assumed) = 1.21 kg/m^3
Reference sound intensity, Io = 1*10^-12 w/m^2

Part (a): Initial sound intensity (angry sound)
10log (I/Io) = Sound level
Therefore,
For Ia = 70 db
Ia/(1*10^-12) = 10^(70/10)
Ia = 10^(70/10)*10^-12 = 1*10^-5 W/m^2

Part (b): Final sound intensity (soothing sound)
Is = 50 db
Therefore,
Is = 10^(50/10)*10^-12 = 18*10^-7 W/m^2

Part (c): Initial sound wave amplitude
Now,
I (W/m^2) = 0.5*A^2*density*velocity*4*π^2*frequency^2

Making A the subject;
A = Sqrt [I/(0.5*density*velocity*4π^2*frequency^2)]

Substituting;
A_initial = Sqrt [(1*10^-5)/(0.5*1.21*345*4π^2*500^2)] = 6.97*10^-8 m = 69.7 nm

Part (d): Final sound wave amplitude
A_final = Sqrt [(1*10^-7)/(0.5*1.21*345*4π^2*500^2)] = 6.97*10^-9 m = 6.97 nm
4 0
3 years ago
How fast would 40 Newtons of force accelerate a 2 kg object?
Digiron [165]

Answer:

20 m/s^2

Explanation:

We can solve this problem by using Newton's second law of motion, which states that the net force acting on an object is equal to the product between its mass and its acceleration:

F=ma

where

F is the net force on the object

m is its mass

a is its acceleration

In this problem:

F = 40 N is the force on the object

m = 2 kg is its mass

Therefore, the acceleration of the object is

a=\frac{F}{m}=\frac{40}{2}=20 m/s^2

8 0
3 years ago
) A steel guitar string with a diameter of 1.00 mm is stretched between supports 80.0 cm apart. The temperature is 0.0°C. (a) Fi
ladessa [460]

Answer: a. Mass per unit length =0.0245kg/m

b. Tension =2.45x10^-8N

C. Tension = 2.45 x10^-8N

Fundamental frequency =200Hz

Explanation:

7 0
3 years ago
Something that can not be used up or depleted​
Jet001 [13]

Answer:

oxygen

Explanation:

as it is in the air it can't be depleted or used up

5 0
3 years ago
Read 2 more answers
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