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3 years ago

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

Physics

2 answers:

lukranit [14]3 years ago

7 0

This question is incomplete; here is the complete question:

A phoneme is the largest unit of sound in a word. Please select the best answer from the choices provided

A. T

B. F

The correct answer to this question is F (False)

Explanation:

The word "phoneme" is used to refer to the minimal unit of sound in words, and therefore in language. For example, the first phoneme in the word "man" is "m". These units of sound are essential in language because they make each word unique in meaning and sound. For example, "fan" and "man" are different due to the phonemes "m" and "f". According to this, the phone is not the largest unit of sound but the smallest unit.

kvasek [131]3 years ago

5 0

Answer:

False!

Explanation:

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As part of his job, Warren attends seminars where he must listen to presentations given by experts in his field. When he is list

jeka57 [31]

Answer:

Option (B)

Explanation:

In terms of communication, a receiver is usually referred to as a person who listens, reads as well as observes. In simple words, a receiver is an individual or it can be a group of individuals, to whom any type of message is being diverted. The other name for the receiver is 'audience'.

In the given condition, Warren is attending a seminar in which he is listening to the speaker, as a part of the audience. So, it can be concluded that Warren is a receiver who is receiving information or hearing the speaker.

Thus, the correct answer is option (B).

7 0

3 years ago

7. A mother pushes her 9.5 kg baby in her 5kg baby carriage over the grass with a force of 110N @ an angle

jasenka [17]

Weight of the carriage $=(m+M)g =142.1\ N$

Normal force $=Fsin(\theta) + W = 197.1\ N$

Frictional force $=\mu N=27.59\ N$

Acceleration $=4.66\ m\ s^{-2}$

Explanation:

We have to look into the FBD of the carriage.

Horizontal forces and Vertical forces separately.

To calculate Weight we know that both the mass of the baby and the carriage will be added.

So Weight(W) $=(m+M)\times g =(9.5+5)\ kg \times 9.8 =142.1\ Newton\ (N)$

To calculate normal force we have to look upon the vertical component of forces, as Normal force is acting vertically.We have weight which is a downward force along with $F_x$ , force of $110\ N$ acting vertically downward.Both are downward and Normal is upward so Normal force $=Summation\ of\ both\ forces$

Normal force (N) $= Fsin(\theta)+W=110sin(30) + 142.1 =197.1\ N$

Frictional force (f) $=\mu N=0.14\times 197.1 =27.59\ N$

To calculate acceleration we will use Newtons second law.

That is Force is product of mass and acceleration.

We can see in the diagram that $F_y=Horizontal$ and $F_x=Vertical$ component of forces.

So Fnet = Fy(Horizontal) - f(friction) $= m\times a$

Acceleration (a) = $\frac{Fcos(\theta)-\mu N}{mass(m)} =\frac{(95.26-27.59)}{14.5}= 4.66\ m\ s{^2 }$

So we have the weight of the carriage, normal force,frictional force and acceleration.

3 0

3 years ago

A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The

EleoNora [17]

Answer:

The tank is losing $4.976*10^{-4} m^3/s$

$v_g = 19.81 \ m/s$

Explanation:

According to the Bernoulli’s equation:

$P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2$

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume $v_1$ ≅ 0 ;

then $v_2$ can be determined as: $\sqrt{[2g (h_1- h_2)]$

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

$v_2 = \sqrt{[2*9.81*(20 - 15)]$

$v_2 = \sqrt{[2*9.81*(5)]$

$v_2= 9.9 \ m/s$ as it leaves the hole at the base.

radius r = d/2 = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = $A_1v_2$

J = πr² $v_2$

J = $\pi *(2*10^{-3})^{2}*9.9$

J = $1.244*10^{-4} m^3/s$

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : $v_g = \sqrt{392.31}$

$v_g = 19.81 \ m/s$

4 0

3 years ago

A rocket is fired at 100 m/s at an angle of 37, what was its speed at the top of its path?

Nadusha1986 [10]

Answer:

0m/s

Explanation:

Since its fired at an angle, at the top there will be a split second where the velocity will be 0, as it has a parabolic shape, so the speed at the top of its path is 0

4 0

3 years ago

Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P P and volume V V satisfy th

Verdich [7]

Answer:

the volume decreases at the rate of 500cm³ in 1 min

Explanation:

given

v = 1000cm³, p = 80kPa, Δp/t= 40kPa/min

PV=C

vΔp + pΔv = 0

differentiate with respect to time

v(Δp/t) + p(Δv/t) = 0

(1000cm³)(40kPa/min) + 80kPa(Δv/t) = 0

40000 + 80kPa(Δv/t) = 0

Δv/t = -40000/80

= -500cm³/min

the volume decreases at the rate of 500cm³ in 1 min

3 0

3 years ago