Answer:D
Explanation: i found the answer in an answer key!
Explanation:
In a vacuum (no air resistance), it doesn't. All falling objects, regardless of mass, accelerate at the same rate.
However, when air resistance is taken into account, heavier objects indeed fall faster than lighter objects, provided they have the same shape and size. For example, a lead ball falls faster than a styrofoam ball.
To understand why, first look at what factors affect air resistance:
D = ½ρv²CA
where ρ is air density,
v is velocity,
C is drag coefficient,
and A is cross sectional area.
As falling objects accelerate, they eventually reach a maximum velocity where air resistance equals weight. This is called terminal velocity.
D = W
½ρv²CA = mg
v = √(2mg/(ρCA))
If we increase m while holding everything else constant, v increases. So two objects with the same size and shape but different masses will have different terminal velocities, with the heavier object falling faster.
Answer:
The second distance of the sound from the source is 431.78 m..
Explanation:
Given;
first distance of the sound from the source, r₁ = 1.48 m
first sound intensity level, I₁ = 120 dB
second sound intensity level, I₂ = 70.7 dB
second distance of the sound from the source, r₂ = ?
The intensity of sound in W/m² is given as;
![dB = 10 Log[\frac{I}{I_o} ]\\\\For \ 120 dB\\\\120 = 10Log[\frac{I}{1*10^{-12}}]\\\\12 = Log[\frac{I}{1*10^{-12}}]\\\\10^{12} = \frac{I}{1*10^{-12}}\\\\I = 10^{12} \ \times \ 10^{-12}\\\\I = 1 \ W/m^2](https://tex.z-dn.net/?f=dB%20%3D%2010%20Log%5B%5Cfrac%7BI%7D%7BI_o%7D%20%5D%5C%5C%5C%5CFor%20%5C%20120%20dB%5C%5C%5C%5C120%20%3D%2010Log%5B%5Cfrac%7BI%7D%7B1%2A10%5E%7B-12%7D%7D%5D%5C%5C%5C%5C12%20%3D%20%20Log%5B%5Cfrac%7BI%7D%7B1%2A10%5E%7B-12%7D%7D%5D%5C%5C%5C%5C10%5E%7B12%7D%20%3D%20%5Cfrac%7BI%7D%7B1%2A10%5E%7B-12%7D%7D%5C%5C%5C%5CI%20%3D%2010%5E%7B12%7D%20%5C%20%5Ctimes%20%5C%2010%5E%7B-12%7D%5C%5C%5C%5CI%20%3D%201%20%5C%20W%2Fm%5E2)
![For \ 70.7 dB\\\\70.7 = 10Log[\frac{I}{1*10^{-12}}]\\\\7.07 = Log[\frac{I}{1*10^{-12}}]\\\\10^{7.07} = \frac{I}{1*10^{-12}}\\\\I = 10^{7.07} \ \times \ 10^{-12}\\\\I = 1 \times \ 10^{-4.93} \ W/m^2](https://tex.z-dn.net/?f=For%20%5C%2070.7%20dB%5C%5C%5C%5C70.7%20%3D%2010Log%5B%5Cfrac%7BI%7D%7B1%2A10%5E%7B-12%7D%7D%5D%5C%5C%5C%5C7.07%20%3D%20%20Log%5B%5Cfrac%7BI%7D%7B1%2A10%5E%7B-12%7D%7D%5D%5C%5C%5C%5C10%5E%7B7.07%7D%20%3D%20%5Cfrac%7BI%7D%7B1%2A10%5E%7B-12%7D%7D%5C%5C%5C%5CI%20%3D%2010%5E%7B7.07%7D%20%5C%20%5Ctimes%20%5C%2010%5E%7B-12%7D%5C%5C%5C%5CI%20%3D%201%20%5Ctimes%20%5C%2010%5E%7B-4.93%7D%20%5C%20W%2Fm%5E2)
The second distance, r₂, can be determined from sound intensity formula given as;
Therefore, the second distance of the sound from the source is 431.78 m.
1.Iron atoms tend to form regions, clusters. These are what become magnetized and then affected by magnetic fields
Answer 2
.When two magnets are brought near each other, like poles repel; opposite poles attract.
2) When a magnet is brought near a piece of iron, the iron also gets attracted to the magnet, and it acquires the same ability to attract other pieces of iron.
