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2 years ago

A phoneme is the largest unit of sound in a word. TRUE or FALSE.

Physics

2 answers:

slamgirl [31]2 years ago

8 0

Answer:

Explanation:

just took the test

Degger [83]2 years ago

4 0

The answer to the question is False i.e A phoneme is not the largest unit of sound.

EXPLANATION:

Phoneme is related to linguistics or the sound system of language which is the language specific.

It is defined as the smallest contrastive unit of sound in speech which can give a distinct meaning. It distinguishes one word from the other word in a unique way .When we put all the phonemes of a particular word, the corresponding word is formed .

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A celebrated Mark Twain story has motivated contestants in the Calaveras County Jumping Frog Jubilee, where frog jumps as long a

IrinaK [193]

The frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.

To find the answer, we need to know about the time of flight and range of projectile motion.

<h3>What's the expression of range of a projectile motion?</h3>

Range = U²× sin(2θ)/g
U= initial velocity, θ= angle of projectile and g= acceleration due to gravity
U=√{Range×g/sin(2θ)}
Here, range= 2.20m, = 36.5°
U= √{2.20×9.8/sin(73)}

U= √{2.20×9.8/sin(73)} = 22.5m/s

<h3>What's the expression of time of flight in projectile motion?</h3>

Time of flight= (2×U×sinθ)/g
So, T= (2×22.5×sin36.5°)/9.8

= 2.73 s

Thus, we can conclude that the frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.

Learn more about the range and time period of projectile motion here:

brainly.com/question/24136952

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4 0

1 year ago

You illuminate the grating in a spectrometer at normal incidence θi=0° with a beam of light that has a wavelength of 6562.8 Å. T

monitta

Answer:

a) θ₁ = 23.14 ° , b) θ₂ = 51.81 °

Explanation:

An address network is described by the expression

d sin θ = m λ

Where is the distance between lines, λ is the wavelength and m is the order of the spectrum

The distance between one lines, we can find used a rule of proportions

d = 1/600

d = 1.67 10⁻³ mm

d = 1-67 10⁻³ m

Let's calculate the angle

sin θ = m λ / d

θ = sin⁻¹ (m λ / d)

First order

θ₁ = sin⁻¹ (1 6.5628 10⁻⁷ / 1.67 10⁻⁶)

θ₁ = sin⁻¹ (3.93 10⁻¹)

θ₁ = 23.14 °

Second order

θ₂ = sin⁻¹ (2 6.5628 10⁻⁷ / 1.67 10⁻⁶)

θ₂ = sin⁻¹ (0.786)

θ₂ = 51.81 °

3 0

2 years ago

When tightening a bolt, you push perpendicularly on a wrench with a force of 165 N at a distance of 0.140 m from the center of t

choli [55]

Answer:

Part a)

$\tau = 23.1 Nm$

Part b)

$\tau = 17.05 Foot pound force$

Explanation:

As we know that torque is defined as the product of force and its perpendicular distance from reference point

so here we have

$\tau = \vec r \times \vec F$

now we have

$\tau = (0.140)(165)$

$\tau = 23.1 Nm$

Part b)

Now we know the conversion as

1 meter = 3.28 foot

1 N = 0.225 Lb force

now we have

$\tau = 23.1 Nm$

$\tau = 23.1 (0.225 Lb)(3.28 foot)$

$\tau = 17.05 Foot pound force$

3 0

2 years ago

Light-rail passenger trains that provide transportation within and between cities speed up and slow down with a nearly constant

Elena L [17]

Answer:

19 m/s

Explanation:

The complete question requires the final speed to be calculated.

Velocity is the rate and direction at which an object moves. Acceleration is the rate of change of velocity per unit time and can be calculated by the difference in velocity over a given time.

For this question, first the unknown acceleration must be calculated and used to determine the final velocity

Step 1: Calculate the acceleration

$a=\frac{v_{2}-v{1}}{t_{1}}$

$a=\frac{11-7}{8}$

$a=\frac{4}{8}$

$a=0.5 m/s^{2}$

Step 2: Calculate the velocity using the acceleration calculated above

$a=\frac{v_{3}-v{2}}{t_{2}}$

$0.5=\frac{v_{3}-11}{16}$

$v_{3}=19 m/s$

3 0

2 years ago

If the energy bar was his only fuel, how far could a 68 kg person walk at 5 km/h?

goldfiish [28.3K]

An energy bar contains of 20grams of carbohydrates and 1gram is equal to 17 KJ. If the energy bar was his only fuel the total energy available is equal to (17,000 x 20) = 340,000j. Estimate KJ per hour is equal to 1539KJ, it has Time equal to (340,000/1539,000) is equal to 0.2209 hour and his Distance is equal to (5,000 x .2209) = 1.1045km.

3 0

3 years ago