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2 years ago

6

Use the drop-down menus to explain how noise cancellation works. Waves with similar are created to interfere with unwanted sound

waves. The of one wave overlap with the troughs of the other wave, causing the two waves to cancel out each other.

2 answers:

leonid [27]2 years ago

8 0

Amplitudes, and Crests

USPshnik [31]2 years ago

7 0

Answer:

1. Amplitudes and 2. Crests

Explanation:

I got it right I promise :)

You might be interested in

A.) Determine the work done by Zach on the bull.

tresset_1 [31]

Explanation:

The unit of work done is in Joules

Work done is a physical quantity that is defined as the force applied to move a body through a particular distance.

Work is only done when the force applied moves a body through a distance.

Work done = Force x distance

The maximum work is done when the force is parallel to the distance direction.

The minimum work is done when the force is at an angle of 90° to the distance direction.

So to solve this problem;

multiply the force applied by Zack and distance through which the bull was pulled.

8 0

2 years ago

A 50-cm-long spring is suspended from the ceiling. A 330 g mass is connected to the end and held at rest with the spring unstret

Nataly [62]

Answer:

a)32.34 N/m

b)10cm

c)1.6 Hz

Explanation:

Let 'k' represent spring constant

'm' mass of the object= 330g =>0.33kg

a) in order to find spring constant 'k', we apply Newton's second law to the equilibrium position 10cm below the release point.

ΣF=kx-mg=0

k=mg / x

k= (0.33 x 9.8)/ 0.1

k= 32.34 N/m

b) The amplitude, A, is the distance from the equilibrium (or center) point of motion to either its lowest or highest point (end points). The amplitude, therefore, is half of the total distance covered by the oscillating object.

Therefore, amplitude of the oscillation is 10cm

c)frequency of the oscillation can be determined by,

f= 1/2π $\sqrt{\frac{k}{m} }$

f= 1/2π $\sqrt{\frac{32.34}{0.33} }$

f= 1.57

f≈ 1.6 Hz

Therefore, the frequency of the oscillation is 1.6 Hz

5 0

3 years ago

You are 1.8 m tall and stand 2.8 m from a plane mirror that extends vertically upward from the floor. on the floor 1 m in front

velikii [3]

This problem must be solved using a sketch. I attached an illustration of the problem.
You must trace the ray that reflects from the top off the table to your eyes. This how eyesight works, light rays reflects off the objects into your eyes.
Law of reflection tells us that light ray reflects off the surface at the same angle in which it falls on it( i attached another illustration of this).
Now we can write tangens equations:
$tan(\theta)=\frac{h-0.8}{1}\\
tan(\theta)=\frac{1.8-h}{2.8}\\
\frac{h-0.8}{1}=\frac{1.8-h}{2.8}$
We solve for h:
$\frac{h-0.8}{1}=\frac{1.8-h}{2.8}\\
2.8h-0.8\cdot 2.8=1.8-h\\
3.8h=1.8+2.24\\
h=\frac{4.04}{3.8}\\
h=1.06m$

6 0

2 years ago

As a quality test of ball bearings, you drop bearings (small metal balls), with zero initial velocity, from a height of 1.94 m i

11Alexandr11 [23.1K]

Answer:

The average acceleration of the bearings is $0.77\times10^{3}\ m/s^2$

Explanation:

Given that,

Height = 1.94 m

Bounced height = 1.48 m

Time interval $t=14.86\times10^{-3}\ s$

Velocity of the ball bearing just before hitting the steel plate

We need to calculate the velocity

Using conservation of energy

$mgh=\dfrac{1}{2}mv^2$

Put the value into the formula

$9.8\times1.94=\dfrac{1}{2}\times v^2$

$v=\sqrt{2\times9.8\times1.94}$

$v=6.166\ m/s$

Negative as it is directed downwards

After bounce back,

We need to calculate the velocity

Using conservation of energy

$mgh=\dfrac{1}{2}mv^2'$

Put the value into the formula

$9.8\times1.48=\dfrac{1}{2}\times v^2'$

$v'=\sqrt{2\times9.8\times1.48}$

$v'=5.38\ m/s$

We need to calculate the average acceleration of the bearings while they are in contact with the plate

Using formula of acceleration

$a=\dfrac{v-v'}{t}$

Put the value into the formula

$a=\dfrac{5.38-(-6.166)}{14.86\times10^{-3}}$

$a=776.98\ m/s^2$

$a=0.77\times10^{3}\ m/s^2$

Hence,The average acceleration of the bearings is $0.77\times10^{3}\ m/s^2$

6 0

2 years ago

A small dog is trained to jump straight up a distance of 1.2 m. How much kinetic energy does the 7.2-kg dog need to jump this hi

LUCKY_DIMON [66]

Answer:84.672 joules.

Explanation:

1) Data:

m = 7.2 kg
h = 1.2 m
g = 9.8 m / s²

2) Physical principle

Using the law of mechanical energy conservation principle, you have that the kinetic energy of the dog, when it jumps, must be equal to the final gravitational potential energy.

3) Calculations:

The gravitational potential energy, PE, is equal to m × g × h

So, PE = m × g × h = 7.2 kg × 9.8 m/s² × 1.2 m = 84.672 joules.

And that is the kinetic energy that the dog needs.

8 0

2 years ago