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kupik [55]
3 years ago
9

True or False an overtone occurs when two or more sound waves are produced at the same time.

Physics
2 answers:
Setler79 [48]3 years ago
5 0

Answer:

True, overtone occurs when two or more sounds are produced at the same time.

Ket [755]3 years ago
3 0

Answer:

This statement is true.

Explanation:

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Consider An equilateral prism made of silica glass with n=1.5. Light is incident on one face at an angle of π/4 with respect to
erastovalidia [21]

Answer:

The angle at which the light emerges from the opposite face of the prism with respect to the original incident beam is 53.3°.

Explanation:

Given that,

Angle \theta=\dfrac{\pi}{4}

\theta=45

Refractive index n=1.5

For the refraction at the first surface

We need to calculate the angle

Using formula of refraction

n_{a}\sin\theta_{1}=n_{g}\sin\theta_{2}

Put the value into the formula

1\sin\dfrac{\pi}{4}=1.5\sin\theta_{2}

\sin\theta_{2}=\dfrac{1\times\sin45}{1.5}

\sin\theta_{2}=0.46520

\theta_{2}=\sin^{-1}(0.46520)

\theta_{2}=27.7^{\circ}

We need to calculate the angle on incidence at the second surface

(90-\theta_{2})+(90-\theta_{3})+A=180

\theta_{3}=A-\theta_{2}

Put the value into the formula

\theta_{3}=60-27.7

\theta_{3}=32.3

For the refraction at the second surface,

We need to calculate the angle at which the light emerges from the opposite face of the prism with respect to the original incident beam direction

Using formula of refraction

n_{g}\sin\theta_{3}=n_{a}\sin\theta_{4}

Put the value into the formula

1.5\sin32.3=1\times\sin\theta_{4}

\theta=\sin^{-1}(\dfrac{1.5\sin32.3}{1})

\theta=53.3^{\circ}

The angle is 53.3° from the original

Hence, The angle at which the light emerges from the opposite face of the prism with respect to the original incident beam is 53.3°.

8 0
4 years ago
Read 2 more answers
In 1932 Albert Dremel of Racine, Wisconsin, created his rotary tool that has come to be known as a dremel.
kramer

Answer:

3054.32618 rad/s²

-431.1989 rad/s²

29080

Explanation:

Converting angular speed to rad/s

\omega=35000\times \frac{2\pi}{60}=3665.19142\ rad/s

\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{3665.19142-0}{1.2}\\\Rightarrow a=3054.32618\ rad/s^2

The average acceleration while speeding up is 3054.32618 rad/s²

The number of turns in the 1.2 seconds

\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow \theta=0\times t+\frac{1}{2}\times 3054.32618\times 1.2^2\\\Rightarrow \theta=2199.11484\ rad=\frac{2199.11484}{2\pi}=349.99\ rotations

The number of rotations in the 1.2 seconds is 349.99

Number of rotations in the 45 seconds

\frac{35000}{60}\times 45=26250\ rotations

\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{0-3665.19142}{8.5}\\\Rightarrow a=-431.1989\ rad/s^2

Average angular acceleration while slowing down -431.1989 rad/s²

\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\frac{\omega_f^2-\omega_i^2^2}{2\alpha}\\\Rightarrow \theta=\frac{0^2-3665.19142^2}{2\times -431.1989}\\\Rightarrow \theta=15577.0668\ rad=\frac{15577.0668}{2\pi}\\ =2479.16718\ rotations

Number of rotations while slowing down is 2479.16718

Total number of rotations is 349.99+26250+2479.16718 = 29079.15718 = 29080

3 0
3 years ago
The greater force is applied the the acceleration
sweet [91]
<span>Newton's 1st Law of Motion is also known as the Law of. ... Newton's 2nd Law says that when an is applied to a  it causes it to  Newton's 2nd Law says that when an UNBALANCED FORCE is applied to a MASS, it causes it to ACCELERATE. The greater the force that is applied, the the acceleration.</span>
6 0
3 years ago
Amber applies a net force of 35 N to pull her younger brother (32 kg) in a wagon (3 kg). at what rate does the wagon accerlerate
Ket [755]

Answer:

C. 1.00 m/s^2

Explanation:

We can solve the problem by using Newton's second law:

F=ma

where

F is the net force acting on an object

m is the mass of the object

a is its acceleration

In this case, the net force applied by Amber is F=35 N, while the total mass of the brother+the wagon is

m=32 kg+3 kg=35 kg

Therefore, we can re-arrange the previous equation to find the acceleration:

a=\frac{F}{m}=\frac{35 N}{35 kg}=1.00 m/s^2

6 0
3 years ago
With an initial velocity of 20km per hour a car accelerated at 8m/s2 for 10s,what is the position of the car at the end of 10s
Dima020 [189]
  • initial velocity=u=20km/h=5.5m/s
  • Acceleration=a=8m/s^2
  • Time=t=10s

\\ \sf\longmapsto s=ut+\dfrac{1}{2}at^2

\\ \sf\longmapsto s=5.5(10)+\dfrac{1}{2}(8)(5.5)^2

\\ \sf\longmapsto s=55+4(5.5)^2

\\ \sf\longmapsto s=55+4(30.25)

\\ \sf\longmapsto s=55+121

\\ \sf\longmapsto s=176m

4 0
3 years ago
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