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hoa [83]
3 years ago
12

Two sound waves are traveling through the same medium. They have the same amplitude, wavelength, and direction of travel. If the

phase difference between them is 6π, what type of interference do they exhibit? Why is that the case?
Physics
1 answer:
Sloan [31]3 years ago
7 0

Answer:

Constructive interference

Explanation:

Here, the medium is same, same wavelength, same frequency, same amplitude and same direction of propagation.

Let the intensity of waves be I which is same for both

The formula for the net intensity is

I^{2}=I_{1}^{2}+I_{2}^{2}+2I_{1}I_{2}Cos\phi

where, Ф be the phase difference

So, I_{R}^{2}=I^{2}+I^{2}+2IICos6\pi

Here, IR is maximum so the interference is constructive in nature.

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Assuming Adam is on earth g= 9.8 m/s and m= weight/ gravity = 667/9.8 = 68 kg
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Which change will always result in an increase in the gravitational force between 2 objects
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Suppose that on earth you can throw a ball vertically upward a distance of 1.20 m. Given that the acceleration of gravity on the
tatuchka [14]

Answer:

7.04 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity = 0

s = Displacement on Earth = 1.2 m

a = Acceleration due to gravity on Moon = 1.67 m/s²

a = Acceleration due to gravity Earth= 9.81 m/s²

Accelration going up is considered as negetive

Initial Velocity of the ball

v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow -u^2=2\times -9.81\times 1.2-0^2\\\Rightarrow u=\sqrt{2\times 9.81\times 1.2}\\\Rightarrow u=4.85\ m/s

Assuming that the ball is thrown with the same velocity on the Moon, displacement of the ball is

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-4.85^2}{2\times -1.67}\\\Rightarrow s=7.04\ m

The displacement of the ball on the moon is 7.04 m

6 0
3 years ago
In 2005 astronomers announced the discovery of a large black hole in the galaxy Markarian 766 having clumps of matter orbiting a
IRISSAK [1]

A. 4.64\cdot 10^{11}m

The orbital speed of the clumps of matter around the black hole is equal to the ratio between the circumference of the orbit and the period of revolution:

v=\frac{2\pi r}{T}

where we have:

v=30,000 km/s = 3\cdot 10^7 m/s is the orbital speed

r is the orbital radius

T=27 h \cdot 3600 =97,200 s is the orbital period

Solving for r, we find the distance of the clumps of matter from the centre of the black hole:

r=\frac{vT}{2\pi}=\frac{(3\cdot 10^7 m/s)(97200 s)}{2\pi}=4.64\cdot 10^{11}m

B. 6.26\cdot 10^{36}kg, 3.13\cdot 10^6 M_s

The gravitational force between the black hole and the clumps of matter provides the centripetal force that keeps the matter in circular motion:

m\frac{v^2}{r}=\frac{GMm}{r^2}

where

m is the mass of the clumps of matter

G is the gravitational constant

M is the mass of the black hole

Solving the formula for M, we find the mass of the black hole:

M=\frac{v^2 r}{G}=\frac{(3\cdot 10^7 m/s)^2(4.64\cdot 10^{11} m)}{6.67\cdot 10^{-11}}=6.26\cdot 10^{36}kg

and considering the value of the solar mass

M_s = 2\cdot 10^{30}kg

the mass of the black hole as a multiple of our sun's mass is

M=\frac{6.26\cdot 10^{36} kg}{2\cdot 10^{30} kg}=3.13\cdot 10^6 M_s

C. 9.28\cdot 10^9 m

The radius of the event horizon is equal to the Schwarzschild radius of the black hole, which is given by

R=\frac{2MG}{c^2}

where M is the mass of the black hole and c is the speed of light.

Substituting numbers into the formula, we find

R=\frac{6.26\cdot 10^{36} kg)(6.67\cdot 10^{-11})}{(3\cdot 10^8 m/s)^2}=9.28\cdot 10^9 m

8 0
2 years ago
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