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3 years ago

When the amplitude of a wave triples, the intensity changes by a factor of [blank](1/3,1/9,3,9)

Physics

2 answers:

trasher [3.6K]3 years ago

6 0

Answer:

1. It becomes 9 times.

2. it becomes one fourth times.

3. It becomes one hundredth times.

4. It becomes one ninth times.

Explanation:

Intensity of a wave is directly proportional to the square of amplitude of the wave and inversely proportional to the square of the distance between the source and observation point.

1. As the amplitude of the wave is tripled, the intensity becomes 9 times.

2. As the amplitude becomes half, the intensity becomes one fourth.

3. As the distance between the source and the ear is one tenth, the intensity becomes one hundredth.

4. As the distance between source and ear is one third, the intensity becomes one ninth.

sladkih [1.3K]3 years ago

5 0

1. 9

2. 1/4

3. 1/100

4. 9

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From the gravitational law calculate the weight W (gravitational force with respect to the earth) of a 89-kg man in a spacecraft

zhannawk [14.2K]

Answer:

$W=\frac{773}{4.45}=173.76 l b f$

Explanation:

$W=\frac{G \cdot m_{e} \cdot m}{(R+h)^{2}}$

The law of gravitation

$G=6.673\left(10^{-11}\right) m^{3} /\left(k g \cdot s^{2}\right)$

Universal gravitational constant [S.I. units]

$m_{e}=5.976\left(10^{24}\right) k g$

Mass of Earth [S.I. units]

$m=89 kg$

Mass of a man in a spacecraft [S.I. units]

$R=6371 \mathrm{~km}$

Earth radius [km]

Distance between man and the earth's surface

$h=261 \mathrm{~km} \quad[\mathrm{~km}]$

ESULT $W=\frac{6.673\left(10^{-11}\right) \cdot 5.976\left(10^{24}\right) \cdot 89}{\left(6371 \cdot 10^{3}+261 \cdot 10^{3}\right)^{2}}=773.22 \mathrm{~N}$

$W=\frac{773}{4.45}=173.76 l b f$

4 0

2 years ago

If Star A is magnitude 1.0 and Star B is magnitude 9.6 , which is brighter and by what factor?

ludmilkaskok [199]

Answer:

Star A is brighter than Star B by a factor of 2754.22

Explanation:

Lets assume,

the magnitude of star A = m₁ = 1

the magnitude of star B = m₂ = 9.6

the apparent brightness of star A and star B are b₁ and b₂ respectively

Then, relation between the difference of magnitudes and apparent brightness of two stars are related as give below: $(m_{2} - m_{1}) = 2.5\log_{10}(b_{1}/b_{2})$

The current magnitude scale followed was formalized by Sir Norman Pogson in 1856. On this scale a magnitude 1 star is 2.512 times brighter than magnitude 2 star. A magnitude 2 star is 2.512 time brighter than a magnitude 3 star. That means a magnitude 1 star is (2.512x2.512) brighter than magnitude 3 bright star.

We need to find the factor by which star A is brighter than star B. Using the equation given above,

$(9.6 - 1) = 2.5\log_{10}(b_{1}/b_{2})$