A trombone can produce pitches ranging from 85 Hz to 660 Hz approximately. When the trombone is producing a 357 Hz tone, what is the wavelength of that tone in air at standard conditions?

Answers

Answer 1

Answer:

Answer:

The wavelength of that tone in air at standard condition is 0.96 m.

Explanation:

Given that, a trombone can produce pitches ranging from 85 Hz to 660 Hz approximately. We need to find the wavelength of that tone in air when the trombone is producing a 357 Hz tone.

We know that the speed of sound in air is approximately 343 m/s. Speed of a wave is given by :

$v=f\lambda\n\n\lambda=(v)/(f)\n\n\lambda=(343\ m/s)/(357\ Hz)\n\n\lambda=0.96\ m$

So, the wavelength of that tone in air at standard condition is 0.96 m. Hence, this is the required solution.

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The electric field at a point 2.8 cm from a small object points toward the object with a strength of 180,000 N/C. What is the object's charge q? ( k = 1/4πε 0 = 8.99 × 10 9 N ∙ m 2/C 2)

Answers

Answer:

Charge, $Q=1.56* 10^(-8)\ C$

Explanation:

It is given that,

Electric field strength, E = 180000 N/C

Distance from a small object, r = 2.8 cm = 0.028 m

Electric field at a point is given by :

$E=(kQ)/(r^2)$

Q is the charge on an object

$Q=(Er^2)/(k)$

$Q=(180000\ N/C* (0.028\ m)^2)/(9* 10^9\ Nm^2/C^2)$

$Q=1.56* 10^(-8)\ C$

So, the charge on the object is $1.56* 10^(-8)\ C$ . Hence, this is the required solution.

A convex mirror with a focal length of 0.25 m forms a 0.080 m tall image of an automobile at a distance of 0.24 m behind the mirror. What is the magnification of the image? Where is the car located, and what is its height? Is the image real or virtual? Is the image upright or inverted? Draw a ray diagram to show where the image forms and how large it is with respect to the object

Answers

Answer:

The distance and height of the object is 6 m and 2 m.

The image is virtual and upright.

Explanation:

Given that,

Focal length = 0.25 m

Length of image = 0.080 m

Image distance = 0.24 m

We need to calculate the distance of the object

Using formula of lens

$(1)/(v)=(1)/(f)+(1)/(u)$

Put the value into the formula

$(1)/(0.24)=(1)/(0.25)+(1)/(u)$

$(1)/(u)=(1)/(0.24)-(1)/(0.25)$

$(1)/(u)=(1)/(6)$

$u=6\ m$

We need to calculate the magnification

Using formula of magnification

$m=-(v)/(u)$

Put the value into the formula

$m=-(0.24)/(-6)$

$m=0.04$

We need to calculate the height of the object

Using formula of magnification

$m=(h')/(h)$

$h=(0.080)/(0.04)$

$h=2\ m$

A convex mirror produce a virtual and upright image behind the mirror.

Hence, The distance and height of the object is 6 m and 2 m.

The image is virtual and upright.

Answer:

Distance of the object = 6 m

Height of the object = 2 m

Explanation:

Thinking process:

Given that,

Focal length = 0.25 m

Length of image = 0.080 m

Image distance = 0.24 m

We need to calculate the distance of the object

Therefore, using formula of lens:

$(1)/(u) = (1)/(f) + (1)/(u)$

$(1)/(u) = (1)/(6)$

solving, gives u = 6

The magnification is calculated as follows:

m = -0.24/-6

= 0.04

The height = 2 m

The diagram yields an image behind the mirror which is upright.

what is the necessary condition on a force the result the conservation of angular momentum for a particle affected by that force?

Answers

Answer:

The force must be applied on the axis of rotation

Explanation:

A rotating system conserves its angular momentum only if there are no external torques on the system. In other words, the external torque must be equal to zero.

T=0

T=Fxd

Torque is equal to the vector product of a force by the distance between the axis of rotation and where the force is applied.

For this product to be zero, the force must be applied on the axis of rotation (d=0).

The position of a particle changes linearly with time, i.e. as one power of t, as given by the following: h(t) = ( 4.1 t + 5.5 ) meters. Find the speed of the particle, in meters per second.

Answers

Answer:

v = 4.1 m / s

Explanation:

Velocity is defined by the relation

v = $(dx)/(dt)$

we perform the derivative

v = 4.1 m / s

Another way to find this magnitude is to see that the velocity on the slope of a graph of h vs t

v = $(\Delta x)/(\Delta t)$

Δx = v Δdt + x₀

h= 4.1 t + 5.5

v = 4.1 m / s

x₀ = 5.5 m

The Speed of a Particle is 4.1 meters per second.

The position of a particle can be represented by a linear equation of the form h(t) = (at + b) where a and b are constants.

In this case, the equation is h(t) = (4.1t + 5.5).

To find the speed of the particle, we can take the derivative of the position equation with respect to time.

The derivative of h(t) is the rate of change of position with respect to time, which represents the velocity of the particle.

In this case, the derivative is 4.1 meters per second.

Therefore, the speed of the particle is 4.1 meters per second.

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A triply charged ion with velocity 7.00 × 10^6 m/s moves in a path of radius 36.0 cm in a magnetic field of 0.55 T in a mass spectrometer. What is the mass of this ion?

Answers

Answer:

Mass of ion will be $22* 10^(-13)kg$

Explanation:

We have given ion is triply charged that is $q=3* 1.6* 10^(-19)=4.8* 10^(-19)C$

Radius r = 36 cm = 0.36 m

Velocity of the electron $v=7* 10^6m/sec$

Magnetic field B = 0.55 T

We know that radius of the path is given by $r=(mv)/(qB)$

$m=(rqB)/(v)=(0.36* 4.8* 10^(-19)* 7* 10^6)/(0.55)=22* 10^(-13)kg$

Dogs can hear higher-pitched whistles that humans do. How do you
think the sound frequencies that dogs can
hear compare to the frequencies that humans
can hear?

Answers

Dogs can hear sounds at higher frequencies than humans. The range of sound frequencies that dogs can hear is approximately 40 Hz to 60,000 Hz, while the range for humans is 20 Hz to 20,000 Hz. This means that dogs can hear ultrasonic sounds that are beyond the range of human hearing.

What is sound about?

In terms of physics, sound is a vibration that travels through a transmission medium like a gas, liquid, or solid as an acoustic wave.

Sound is the reception of these waves and the brain's perception of them in terms of human physiology and psychology. Dogs have the ability to hear ultrasonic sounds that are audible only to them.

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