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4 years ago

6O POINTS!!!!!!!

Physics

2 answers:

galben [10]4 years ago

7 0

16.34 km/2.34 h = 6.98 km/h

STALIN [3.7K]4 years ago

5 0

Average speed = (total distance covered) / (time to cover the distance)

You only need the first and last points from the table.

First point: Position=0, Time=0

Last point: Position= 16.34, Time = 2.34 hours

Total distance covered = 16.34 (miles ? km ?)

Time to cover the distance = 2.34 hours

Average speed = (16.34 miles) / (2.34 hours)

<em>Average speed = 6.983 mile/hour </em>

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3 years ago

A sound wave with a waveenght of 2.5 meters traves 660 meters in 2 seconds calculate the frequemcy of the wave to cacuation the

finlep [7]

The frequency of the wave is 132 Hz

Explanation:

To calculate the speed of the wave, we can use the following formula:

$v = \frac{d}{t}$

where

d is the distance travelled by the wave

t is the time elapsed

For the sound wave in this problem, we have:

d = 660 m is the distance travelled

t = 2 s is the time interval considered

Substituting and solving for v, we find the speed of the sound wave:

$v=\frac{660}{2}=330 m/s$

Now we can calculate the frequency of the wave by using the wave equation:

$v=f\lambda$

where

v = 330 m/s is the speed of the wave

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f is the frequency

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$f=\frac{v}{\lambda}=\frac{330}{2.5}=132 Hz$

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7 0

3 years ago

A rocket engine provides 28,913 Newtons of thrust. The rocket has a mass of 2,350 kilograms. Calculate its acceleration if it mo

Dafna11 [192]

1. The unknown is acceleration
2. The givens are the force(28,913N) and the mass(2350kg)
3. The equation is a=f/m where a is acceleration,f is force and m is mass
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3 years ago

Help!!!<br>middle school physics also please explain :)

Dmitriy789 [7]

Answer:

hope its helps

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8 0

3 years ago

Two strings on a musical instrument are tuned to play at 196 hz (g) and 523 hz (c). (a) what are the first two overtones for eac

Tems11 [23]

(a) first two overtones for each string:
The first string has a fundamental frequency of 196 Hz. The n-th overtone corresponds to the (n+1)-th harmonic, which can be found by using
$f_n = n f_1$
where f1 is the fundamental frequency.

So, the first overtone (2nd harmonic) of the string is
$f_2 = 2 f_1 = 2 \cdot 196 Hz = 392 Hz$
while the second overtone (3rd harmonic) is
$f_3 = 3 f_1 = 3 \cdot 196 Hz = 588 Hz$

Similarly, for the second string with fundamental frequency $f_1 = 523 Hz$ , the first overtone is
$f_2 = 2 f_1 = 2 \cdot 523 Hz = 1046 Hz$
and the second overtone is
$f_3 = 3 f_1 = 3 \cdot 523 Hz = 1569 Hz$

(b) The fundamental frequency of a string is given by
$f= \frac{1}{2L} \sqrt{ \frac{T}{\mu} }$
where L is the string length, T the tension, and $\mu = m/L$ is the mass per unit of length. This part of the problem says that the tension T and the length L of the string are the same, while the masses are different (let's calle them $m_{196}$ , the mass of the string of frequency 196 Hz, and $m_{523}$ , the mass of the string of frequency 523 Hz.
The ratio between the fundamental frequencies of the two strings is therefore:
$\frac{523 Hz}{196 Hz} = \frac{ \frac{1}{2L} \sqrt{ \frac{T}{m_{523}/L} } }{\frac{1}{2L} \sqrt{ \frac{T}{m_{196}/L} }}$
and since L and T simplify in the equation, we can find the ratio between the two masses:
$\frac{m_{196}}{m_{523}}=( \frac{523 Hz}{196 Hz} )^2 = 7.1$

(c) Now the tension T and the mass per unit of length $\mu$ is the same for the strings, while the lengths are different (let's call them $L_{196}$ and $L_{523}$ ). Let's write again the ratio between the two fundamental frequencies
$\frac{523 Hz}{196 Hz}= \frac{ \frac{1}{2L_{523}} \sqrt{ \frac{T}{\mu} } }{\frac{1}{2L_{196}} \sqrt{ \frac{T}{\mu} }}$
And since T and $\mu$ simplify, we get the ratio between the two lengths:
$\frac{L_{196}}{L_{523}}= \frac{523 Hz}{196 Hz}=2.67$

(d) Now the masses m and the lenghts L are the same, while the tensions are different (let's call them $T_{196}$ and $T_{523}$ . Let's write again the ratio of the frequencies:
$\frac{523 Hz}{196 Hz}= \frac{ \frac{1}{2L} \sqrt{ \frac{T_{523}}{m/L} } }{\frac{1}{2L} \sqrt{ \frac{T_{196}}{m/L} }}$
Now m and L simplify, and we get the ratio between the two tensions:
$\frac{T_{196}}{T_{523}}=( \frac{196 Hz}{523 Hz} )^2=0.14$

7 0

3 years ago