Isabela creates a demonstration showing a sound echo. How could she demonstrate this?

Light travels at 3.0 × 108 m/s in a vacuum and slows to 2.0 × 108 m/s in glass. What is the index of refraction of glass?

Slav-nsk [51]

We are asked to solve for the index of refraction and the formula is n = c/v where "n" represents the index of refraction, "c" represents the speed of light in the vacuum while "v" represents the speed of another medium.
In the problem, we have the given values below:
c = 3 x 10^8 m/s
v = 2 x 10^8 m/s
n =?

Solving for n, we have the solution below:
n = 3x10^8 / 2x10^8
n = 1.5

The answer is 1.5 for the index of refraction.

8 0

2 years ago

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A heavy stone of mass m is hung from the ceiling by a thin 8.25-g wire that is 65.0 cm long. When you gently pluck the upper end

Triss [41]

Answer: m= 35.6 kg

Explanation:

For finding the mass of the stone we have the formula

v= $\sqrt{\frac{Tension}{Linear. Mass. density} }$

Here, Tension= m*g = m*9.81

and linear mass density= $\frac{8.25 g}{65 cm}$

Linear mass density= $\frac{8.25*10^-3}{65*10^-2}$

Linear mass density= 0.0127 kg/m

Velocity= $2*\frac{l}{t}$

Velocity= 2 * $\frac{65*10^-2}{7.84}$

Velocity= 165.8 m/s

So putting all these values in equation we get

v= $\sqrt{\frac{Tension}{Linear. Mass. density} }$

165.8= $\sqrt{\frac{m*9.81}{0.0127} }$

Solving we get

m= 35.58 kg

or m= 35.6 kg

3 0

3 years ago

The speed of sound waves

neonofarm [45]

B. Is faster in solids than liquids would be the correct answer because the molecules in solids are much closer and can pass along energy faster and more effectively.

4 0

2 years ago

State the factors that affect tha gravitational force

sweet [91]

Explanation:

1. Mass of an object

2. Distance between the objects

7 0

2 years ago

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At the instant the traffic light turns green, an automobile starts with a constant acceleration a of 2.5 m/s2. At the same insta

Licemer1 [7]

Answer:

6.96 s

Explanation:

<u>Given:</u>

u = initial speed of the automobile = 0 m/s

a = constant acceleration of the automobile = $2.5\ m/s^2$
v = constant speed of the truck = 8.7 m/s

<u>Assume:</u>

t = time instant at which the automobile overtakes the truck.

At the moment the automobile and the truck both meat each other the distance travel by both vehicles must be the same.

$\therefore \textrm{Distance traveled by the automobile }=\textrm{Distance traveled by the truck}\\\Rightarrow ut+\dfrac{1}{2}at^2=vt\\\Rightarrow (0)t+\dfrac{1}{2}\times 2.5\times t^2=8.7t\\\Rightarrow 1.25t^2=8.7t\\\Rightarrow 1.25t^2-8.7t=0\\\Rightarrow t(1.25t-8.7)=0\\\Rightarrow t = 0\,\,\,or\,\,\, t = \dfrac{8.7}{1.25}\\\Rightarrow t = 0\,\,\,or\,\,\, t = 6.96\\$

Since t = 0 s is the initial condition. So, they both meet again at t = 6.96 s such that the automobile overtakes the truck.

6 0

2 years ago