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3 years ago

Does air play a role in the propagation of the human voice from one end of a lecture hall to the other

Physics

1 answer:

love history [14]3 years ago

4 0

Answer: Yes

Air serves as a propagation medium.

Explanation:

Propagation of sound wave require a medium for transmission. Sound is produced by vibrating objects, The matter or substance through which sound is transmitted is called a medium. It can be solid, liquid or gas. Therefore to propagate sounds (human voice) from one end of a lecture hall to the other we need a propagation medium which is air. Air serves as a propagation medium.

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Which color has the most energetic photons?

Ksju [112]

Red or violet is the color with the most energetic photons!! Hope this helped!!

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3 years ago

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a toboggan loaded with vacationing students (total weight 1300 N) slides down a slope at 30 degrees and there is no friction. wh

sasho [114]

Answer:

The acceleration of the sliding toboggan is, a = 4.9 m/s²

Explanation:

Given data,

The total weight of the toboggan, W = 1300 N

The slope is, Ф = 30°

The acceleration of a body under the influence of the gravitational field does not depend on its mass, size and shape in the absence of the air resistance.

Therefore,

The acceleration of the toboggan is given by the formula,

a = g Sin Ф

Substituting the given values in the above equation,

a = 9.8 x Sin 30°

= 4.9 m/s²

Hence, the acceleration of the sliding toboggan is, a = 4.9 m/s²

4 0

3 years ago

A cylinder of radius R, length L, and mass M is released from rest on a slope inclined at angle θ. It is oriented to roll straig

inna [77]

Answer:

$\mu_s=\frac{1}{3}\tan \theta$

Explanation:

Let the minimum coefficient of static friction be $\mu_s$ .

Given:

Mass of the cylinder = $M$

Radius of the cylinder = $R$

Length of the cylinder = $L$

Angle of inclination = $\theta$

Initial velocity of the cylinder (Released from rest) = 0

Since, the cylinder is translating and rolling down the incline, it has both translational and rotational motion. So, we need to consider the effect of moment of Inertia also.

We know that, for a rolling object, torque acting on it is given as the product of moment of inertia and its angular acceleration. So,

$\tau =I\alpha$

Now, angular acceleration is given as:

$\alpha = \frac{a}{R}\\Where, a\rightarrow \textrm{linear acceleration of the cylinder}$

Also, moment of inertia for a cylinder is given as:

$I=\frac{MR^2}{2}$

Therefore, the torque acting on the cylinder can be rewritten as:

$\tau = \frac{MR^2}{2}\times \frac{a}{R}=\frac{MRa}{2}------ 1$

Consider the free body diagram of the cylinder on the incline. The forces acting along the incline are $mg\sin \theta\ and\ f$ . The net force acting along the incline is given as:

$F_{net}=Mg\sin \theta-f\\But,\ f=\mu_s N\\So, F_{net}=Mg\sin \theta -\mu_s N-------- 2$

Now, consider the forces acting perpendicular to the incline. As there is no motion in the perpendicular direction, net force is zero.

So, $N=Mg\cos \theta$

Plugging in $N=Mg\cos \theta$ in equation (2), we get

$F_{net}=Mg\sin \theta -\mu_s Mg\cos \theta\\F_{net}=Mg(\sin \theta-\mu_s \cos \theta)--------------3$

Now, as per Newton's second law,

$F_{net}=Ma\\Mg(\sin \theta-\mu_s \cos \theta)=Ma\\\therefore a=g(\sin \theta-\mu_s \cos \theta)------4$

Now, torque acting on the cylinder is provided by the frictional force and is given as the product of frictional force and radius of the cylinder.

$\tau=fR\\\frac{MRa}{2}=\mu_sMg\cos \theta\times R\\\\a=2\times \mu_sg\cos \theta\\\\But, a=g(\sin \theta-\mu_s \cos \theta)\\\\\therefore g(\sin \theta-\mu_s \cos \theta)=2\times \mu_sg\cos \theta\\\\\sin \theta-\mu_s \cos \theta=2\mu_s\cos \theta\\\\\sin \theta=2\mu_s\cos \theta+\mu_s\cos \theta\\\\\sin \theta=3\mu_s \cos \theta\\\\\mu_s=\frac{\sin \theta}{3\cos \theta}\\\\\mu_s=\frac{1}{3}\tan \theta............(\because \frac{\sin \theta}{\cos \theta}=\tan \theta)$

Therefore, the minimum coefficient of static friction needed for the cylinder to roll down without slipping is given as:

$\mu_s=\frac{1}{3}\tan \theta$

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3 years ago

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Which type of experiment involves changing only one variable at a time

BlackZzzverrR [31]

Controlled Experiment

6 0

3 years ago

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Given a 45 45 90 prism with index of 1.5, immersed in air. The hypotenuse acts as the reflecting face by TIR. A ray of light ent

Genrish500 [490]

Answer:

83.6°

Explanation:

For the ray to be totally internally reflected, at the boundary, the angle of refraction is 90. Using the law of refraction where

n₁sinθ₁ = n₂sinθ₂ where n₁ = refractive index of prism = 1.5, θ₁ = critical angle in prism, n₂ = refractive index of air = 1 and θ₂ = refractive angle = 90°.

So, substituting these values into the equation,

n₁sinθ₁ = n₂sinθ₂

1.5 × sinθ₁ = 1 × sin90

1.5 × sinθ₁ = 1

sinθ₁ = 1/1.5

sinθ₁ = 0.6667

θ₁ = sin*(0.6667)

θ₁ = 41.8°

So, for total internal reflection, an incidence angle of 41.8° is required. So, a full convergence angle of 2 × 41.8° = 83.6° is required for the whole bundle of rays.

5 0

3 years ago