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sammy [17]
3 years ago
12

PLS HELP DUE IN 30 MINS!!! True or false: A large-amplitude pulse travels at a faster speed than a small-amplitude pulse. Explai

n your answer.
Physics
2 answers:
OLga [1]3 years ago
8 0

Answer:

False

Explanation:

Larger pulse with more energy means that the wave has large amplitude. As in the given question, both the waves are travelling in the same medium, hence their speeds will remain same and therefore the larger pulse will not overtake the smaller pulse. Remember the amplitude of a wave does not affect the speed at which the wave travels

Inga [223]3 years ago
8 0

Answer:

False

Explanation:

The amplitude of a wave does not affect the speed at which the wave travels.  The speed of a wave is only altered by alterations in the properties of the medium through which it travels.

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The range of audible frequencies is from about 20.0 hz to 2.00×104 hz . what is range of the wavelengths of audible sound in air
TiliK225 [7]
The wavelength is related to the frequency by the relationship:
\lambda= \frac{v}{f}
where v is the wave speed and f is its frequency.

The speed of sound in air is v=344 m/s. The lowest frequency is f=20.0 Hz, so the corresponding wavelength is
\lambda_1 =  \frac{v}{f_1}= \frac{344 m/s}{20.0 Hz}=17.2 m
The highest frequency is f_2 = 2.00 \cdot 10^4 Hz, so the corresponding wavelength is
\lambda_2 =  \frac{v}{f_2}= \frac{344 m/s}{2.00 \cdot 10^4 Hz}=0.017 m

Therefore, the range of wavelengths of audible sound in air is
[0.017 m - 17.2 m]
4 0
3 years ago
The diagram shows the parabolic path of a projectile that leaves the foot of a kicker with a horizontal velocity of 15 m/s and a
rusak2 [61]

Answer:

I would increase the horizontal velocity or the vertical velocity or both to make the ball go the extra distance to cross the goal line.

Explanation:

In order to increase the horizontal distance covered by the ball, we need to examine the variables involved in the formula of range of projectile. The formula for the range of projectile is given as follows:

R = V₀² Sin 2θ/g

where, g is a constant on earth (acceleration due to gravity) and θ is the angle of ball with ground at the time of launching. The value of θ should be 45° for maximum range. In this case we do not know the angle so, we can not tell if we should change it or not.

The only parameter here which we can increase to increase the range is launch velocity (V₀). The formula for V₀ in terms of horizontal and vertical components is as follows:

V₀ = √(V₀ₓ² + V₀y²)

where,

V₀ₓ = Horizontal Velocity

V₀y = Vertical Velocity

Hence, it is clear from the formula that we can increase both the horizontal and vertical velocity to increase the initial speed which in turn increases the horizontal distance covered by the ball.

<u>Therefore, I would increase the horizontal velocity or the vertical velocity or both to make the ball go the extra distance to cross the goal line.</u>

4 0
3 years ago
How many hours of daylight occur on the first day of spring
Elza [17]

twelve hours and ten minuets

5 0
3 years ago
Figure P2.23 is a somewhat simplified velocity graph for Olympic sprinter Carl Lewis starting a 100 m dash. Estimate his acceler
ehidna [41]

A) Acceleration in part A: 6.1 m/s^2

B) Acceleration in part B: 2.7 m/s^2

C) Acceleration in part C: 1.5 m/s^2

Explanation:

A)

The picture of the problem is missing: find it in attachment.

The acceleration of a body is equal to the rate of change of its velocity:

a=\frac{v-u}{\Delta t}

where

v is the final velocity

u is the initial velocity

\Delta t is the time it takes for the velocity to change from u to v

In part A of the race, we have:

u = 0

v = 5.5 m/s (estimate)

\Delta t = 0.9 - 0 = 0.9 s

So the acceleration is

a=\frac{5.5-0}{0.9}=6.1 m/s^2

B)

In part B of the race, we have:

u = 5.5 m/s is the initial velocity (estimate)

v = 9.5 m/s is the final velocity (estimate)

\Delta t = 2.4 - 0.9 = 1.5 s is the time interval between the two points considered

Therefore, using the equation for the acceleration, we can find the acceleration in part B:

a=\frac{9.5-5.5}{1.5}=2.7 m/s^2

C)

In part C of the race, we have:

u = 9.5 m/s is the initial velocity (estimate)

v = 11 m/s is the final velocity (estimate)

\Delta t = 3.4 - 2.4 = 1 s is the time interval between the two points considered

And therefore, the acceleration in part C of the race is:

a=\frac{11-9.5}{1}=1.5 m/s^2

Learn more about acceleration:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

3 0
3 years ago
A student bangs a brick at the head of a table. Three students are positioned equal distance from the head with their hands on t
iris [78.8K]

Explanation:

that the people closer too the head of the table will feel more vibrations than the people at the end of the table. since the vibrations will slow down as they travel farther down the table

Hope this helps!!

5 0
3 years ago
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