PHYSICS COLLEGE

A trumpet player hears 5 beats per second when she plays a note and simultaneously sounds a 440 Hz tuning fork. After pulling her tuning valve out to slightly increase the length of her trumpet, she hears 3 beats per second against the tuning fork. Was her initial frequency 435 Hz or 445 Hz? Explain.

Answers

Answer 1

Answer:

Answer:

her initial frequency is 445 Hz

Explanation:

Given;

initial beat frequency, $F_B$ = 5

observed frequency, F = 440 Hz

let the initial frequency = F₁

F₁ = F ± 5 Hz

F₁ = 440 Hz ± 5 Hz

F₁ = 435 or 445 Hz

This result obtained shows that her initial frequency can either be 435 Hz or 445 Hz

The last beat frequency will be used to determine the actual initial frequency.

F = v/λ

Frequency (F) is inversely proportional to wavelength. That is an increase in length will cause a proportional decrease in frequency.

This shows that the final frequency is smaller than the initial frequency because of the increase in length.

Initial frequency - frequency of tuning fork = 5 beat frequency

Reduced initial frequency - frequency of tuning fork = 3 beat frequency

Initial frequency = 5Hz + 440 Hz = 445 Hz

Final frequency (Reduced initial frequency) = 440 + 3 = 443 Hz

Check: 445 Hz - 440 Hz = 5 Hz

443 Hz - 440 Hz = 3 Hz

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How do you define 'heat' and 'temperature'

Answers

Temperature -a measure of the average kinetic energy of the individual particles in an object

Heat is the movement of thermal energy from a substance of higher temperature to a substance of lower temperature

Answer:

Heat and temperature are related but very different.

Explanation:

Heat: The total energy of molecular motion in a substance

Temperature: A measure of the average energy of molecular motion in a substance

For further help:

Examples

Heat Temperature

-Heat is a form of energy that can -The degree of hotness and

transfer from hot body to cold body coldness of the body

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-Heat flows from hot body to cold -It rises when heated and falls down

body when an object is cooled down

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The surface tension of a liquid is to be measured using a liquid film suspended on a U-shaped wire frame with an 12-cm-long movable side. If the force needed to move the wire is 0.096 N, determine the surface tension of this liquid in air.

Answers

Answer:

σ = 0.8 N/m

Explanation:

Given that

L = 12 cm

We know that 1 m = 100 cm

L = 0.12 m

The force ,F= 0.096 N

Lets take surface tension = σ

We know that surface tension is given as

$\sigma =(F)/(L)\n\sigma =(F)/(L)\nNow\ by\ putting\ the\ values\n\sigma =(0.096)/(0.12)\ N/m\n\sigma=0.8\ N/m$

Therefore the surface tension σ will be 0.8 N/m .

σ = 0.8 N/m

Final answer:

The surface tension of the liquid in air is 0.8 N/m.

Explanation:

To determine the surface tension of the liquid, we need to use the formula F = yL, where F is the force needed to move the wire, y is the surface tension, and L is the length of the wire. In this case, F = 0.096 N and L = 12 cm. We can rearrange the formula to solve for y: y = F / L. Plugging in the values, we get y = 0.096 N / 0.12 m = 0.8 N/m. So, the surface tension of the liquid in air is 0.8 N/m.

Learn more about Surface tension here:

brainly.com/question/11348644

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What would the position of arrows on a target need to be to illustrate measurements that are neither accurate nor precise

Answers

Answer:

The position of the arrows will not be on the target i.e. outside the bull's eye, neither will they be close to one another (widely scattered).

Explanation:

Accuracy refers to the closeness of a measurement to an actual or accepted value while precision refers to the closeness of measurements to one another.

Using archery as an illustration of precision and accuracy, measurements (arrows) that are neither accurate not precise are those arrows that will be far away or outside the bull's eye region (target) of the board and also far apart from one another.

In a nutshell, the arrows will be distant from the bull's eye or target (not accurate) and also distant from one another (not precise).

Two non-conducting slabs of infinite area are given a charge-per-unit area of σ = -16 C/m^2 and σb =+6.0 C/m^2 respectively. A third slab, made of metal, is placed between the first two plates. The charge density σm on the metal slab is 0 (i.e., the slab is uncharged).Required:
Find the magnitude and direction of the electric field.

Answers

Answer:

$E_(C) = -5.65 * 10^(11) \hat{x}$ N/C

$E_(A) = -1.24 * 10^(12) \hat{x}$ N/C

Explanation:

The charge per unit area of the two non-conducting slabs are given by:

$\sigma_(a) = -16 C/m^2$

$\sigma_(b) = 6 C/m^2$

The charge density on the metal $\sigma_(m) = 0$

ε0 = 8.854 x 10-12 C2/N m2

Note that the electric field inside the conductor is zero because it is an equipotential surface.

The diagram attached to this solution typifies the description given in the question:

The electric field in the region C can be calculated by:

$E_(C) = ( |\sigma_(b) |- |\sigma_(a)| )/(2 \epsilon_(0) ) \nE_(C) = (6 - 16 )/(2 * 8.854 * 10^(-12) ) \nE_(C) = -5.65 * 10^(11) \hat{x}$

The electric field in the region A can be calculated by:

$E_(A) = (- |\sigma_(a) |- |\sigma_(b)| )/(2 \epsilon_(0) ) \nE_(A) = (-16 - 6 )/(2 * 8.854 * 10^(-12) ) \nE_(A) = -1.24 * 10^(12) \hat{x}$

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O A.
B.
C.
OD.
OE
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