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BARSIC [14]
3 years ago
13

Which change will increase the speed of a sound wave traveling in a solid aluminum rod? changing the rod from a solid to a liqui

d changing the rod from a solid to a gas decreasing the temperature of the rod increasing the temperature of the rod
Physics
1 answer:
Pepsi [2]3 years ago
4 0

Answer:

increasing the temperature of the rod

Explanation:

Sound wave is a longitudinal wave and its speed in a solid rod is given by the formula

v = \sqrt{\frac{Y}{\rho}}

here we know that

Y = young's modulus

\rho = density of the medium

so as we increase the temperature of rod the density of the rod will decrease while the elasticity will remain same

So on increasing the temperature we can say that speed will increase due to decrease in the density

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3 years ago
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3. A worker pushes a box across the floor to the right at a constant speed with a force of 25N. What
WITCHER [35]

Answer:

Friction between the box and the floor is 25N to the left.

Explanation:

According to Newton's second law of motion, the net force acting on an object is equal to the produce between the object's mass and its acceleration:

F_{net}=ma

where

m is the mass of the object

a is its acceleration

In this problem, we have two forces acting on the object:

- The applied force, F = 25 N, to the right

- The force of friction F_f, opposing the motion of the box, so to the left

So we can write the net force as

F_{net}=F-F_f

Also, we know that the box is moving at constant speed: this means its acceleration is zero, so

a=0

Therefore

F_{net}=0

WHich means:

F-F_f=0

And therefore,

F_f=F=25 N

which means that the force of friction is also 25 N.

6 0
3 years ago
In which scenario will the two objects have the least gravitational force between them?​
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Answer:

A

Explanation:

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3 years ago
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A green light is submerged 2.70 m beneath the surface of a liquid with an index of refraction 1.31. What is the radius of the ci
Anettt [7]

Answer:

The radius is  r = 3.1905 \ m

Explanation:

From the question we are told that

        The  distance  beneath the liquid  is  d =  2.70 \ m

        The refractive index of the liquid is  n_i  =  1.31

Now the critical value is mathematically represented as

         \theta =  sin ^{-1} [\frac{1}{n_i} ]

substituting values

         \theta =  sin ^{-1} [\frac{1}{131} ]

         \theta  =  49.76^o

Using SOHCAHTOA rule we have that

         tan \theta =  \frac{ r}{d}

=>     r =  d * tan \theta

substituting values  

        r =  2.7 * tan (49.76)

        r = 3.1905 \ m

         

5 0
3 years ago
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