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3 years ago

How does resonance occur when you play a horn?

2 answers:

8 0

Sound waves come from mechanical vibrations in solids, liquids, and gases. ... When a sound or light wave strikes an object, it is already vibrating at some particular frequency. If that frequency happens to match the resonant frequency of the object it's hitting, then you'll get what's called resonance.

Oliga [24]3 years ago

3 0

Answer:

When the frequency being produced by the horn matches the natural frequency of the horn then resonance occurs.

Explanation:

Every object has its own frequency called natural frequency. When the sound being produced by an instrument matches the natural frequency of the instrument then acoustic resonance is achieved. Wine glasses breaking by the application of sound is an example of this phenomenon.

So, when the frequency being produced by the horn matches the natural frequency of the horn then resonance occurs.

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If the force used to push a shopping cart increases, the cart's acceleration will

rewona [7]

Answer:

b) increase

Explanation:

4 0

3 years ago

Read 2 more answers

You are a passenger in a truck that gets stuck in mud. You and the driver use a tree branch as a lever to lift up the truck. You

Anika [276]

Answer:

i think it is 1,800

Explanation:

8 0

3 years ago

An object of mass 2kg is on an incline where there is an applied force of 15N. The

seraphim [82]

a) See free-body diagram in attachment

b) The acceleration is $2.46 m/s^2$

Explanation:

The free-body diagram of an object is a diagram representing all the forces acting on the object. Each force is represented by a vector of length proportional to the magnitude of the force, pointing in the same direction as the force.

The free-body diagram for this object is shown in the figure in attachment.

There are three forces acting on the object:

The weight of the object, labelled as $mg$ (where m is the mass of the object and g is the acceleration of gravity), acting downward
The applied force, $F_a$ , acting up along the plane
The force of friction, $F_f$ , acting down along the plane

In order to find the acceleration of the object, we need to write the equation of the forces acting along the direction parallel to the incline. We have:

$F_a - F_f - mg sin \theta = ma$

where:

$F_a = 15 N$ is the applied force, pushing forward

$F_f = 5 N$ is the frictional force, acting backward

$mg sin \theta$ is the component of the weight parallel to the incline, acting backward, where

m = 2 kg is the mass of the object

$g=9.8 m/s^2$ is the acceleration of gravity

$\theta=15^{\circ}$ is the angle between the horizontal and the incline (it is not given in the problem, so I assumed this value)

a is the acceleration

Solving for a, we find:

$a=\frac{F_a - F_f - mg sin \theta}{m}=\frac{15-5-(2)(9.8)(sin 15^{\circ})}{2}=2.46 m/s^2$

Learn more about inclined planes:

brainly.com/question/5884009

#LearnwithBrainly

8 0

3 years ago

An object of mass m = 4.0 kg, starting from rest, slides down an inclined plane of length l = 3.0 m. The plane is inclined by an

kirill [66]

Answer:

(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface

(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:

Wf= -20.4 J is negative

(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane:

Wg= 58.8 J is positive

Explanation:

Nomenclature

vf: final velocity

v₀ :initial velocity

a: acceleleration

d: distance

Ff: Friction force

W: weight

m:mass

g: acceleration due to gravity

Graphic attached

The attached graph describes the variables related to the kinetics of the object (forces and accelerations)

Calculation de of the components of W in the inclined plane

W=m*g

Wx₁ = m*g*sin30°

Wy₁= m*g*cos30°

Object kinematics on the inclined plane

vf₁²=v₀₁²+2*a₁*d₁

v₀₁=0

vf₁²=2*a₁*d₁

$v_{f1} = \sqrt{2*a_{1}*d_{1} }$ Equation (1)

Object kinetics on the inclined plane (μ= 0.2)

∑Fx₁=ma₁ :Newton's second law

-Ff₁+Wx₁ = ma₁ , Ff₁=μN₁

-μ₁N₁+Wx₁ = ma₁ Equation (2)

∑Fy₁=0 : Newton's first law

N₁-Wy₁= 0

N₁- m*g*cos30°=0

N₁ = m*g*cos30°

We replace N₁ = m*g*cos30 and Wx₁ = m*g*sin30° in the equation (2)

-μ₁m*g*cos30₁+m*g*sin30° = ma₁ : We divide by m

-μ₁*g*cos30°+g*sin30° = a₁

g*(-μ₁*cos30°+sin30°) = a₁

a₁ =9.8(-0.2*cos30°+sin30°)=3.2 m/s²

We replace a₁ =3.2 m/s² and d₁= 3m in the equation (1)

$v_{f1} = \sqrt{2*3.2*3} }$

$v_{f1} =\sqrt{2*3.2*3}$

$v_{f1} = 4.38 m/s$

Rough surface kinematics

vf₂²=v₀₂²+2*a₂*d₂ v₀₂=vf₁=4.38 m/s

0 =4.38²+2*a₂*d₂ Equation (3)

Rough surface kinetics (μ= 0.3)

∑Fx₂=ma₂ :Newton's second law

-Ff₂=ma₂

--μ₂*N₂ = ma₂ Equation (4)

∑Fy₂= 0 :Newton's first law

N₂-W=0

N₂=W=m*g

We replace N₂=m*g inthe equation (4)

--μ₂*m*g = ma₂ We divide by m

--μ₂*g = a₂

a₂ =-0.2*9.8= -1.96m/s²

We replace a₂ = -1.96m/s² in the equation (3)

0 =4.38²+2*-1.96*d₂

3.92*d₂ = 4.38²

d₂=4.38²/3.92

(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface

(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:

Wf = - Ff₁*d₁

Ff₁= μ₁N₁= μ₁*m*g*cos30°= -0.2*4*9.8*cos30° = 6,79 N

Wf= - 6.79*3 = 20.4 N*m

Wf= -20.4 J is negative

(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane

Wg=W₁x*d= m*g*sin30*3=4*9.8*0.5*3= 58.8 N*m

Wg= 58.8 J is positive

6 0

4 years ago

Determine the magnitude of the force for each direction. A. Add the forces that are applied in the same direction. B. Subtract f

dalvyx [7]

Answer:

A) R = F1 + F2

B) R = | F1 - F2 |

Explanation:

A) For forces acting in the same direction

we will add up the forces : lets assume the forces to be F1 , F2

Then the magnitude of the forces acting in the same direction will be

R = F1 + F2

B) For forces acting in the opposite direction

we will subtract the forces i.e. stronger force - weaker force . assuming the forces to be F1 , F2

Hence the magnitude of the resultant force = R = | F1 - F2 |

4 0

3 years ago