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3 years ago

9

If a longitudinal wave passes a specific point seven times per second, and the distance between wave rarefaction points is 2 met

ers, what is the speed of the wave?

1 answer:

zavuch27 [327]3 years ago

4 0

Answer: 14 m/s

Explanation:

The speed $S$ of a sound wave is given by the following equation:

$S=\lambda f$

Where:

$\lambda=2 m$ is the wavelength of the sound wave

$f=7 times/s=7 Hz$

$S=(2 m)(7 Hz)$

Hence:

$S=14 m/s$

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A toy rocket launcher can project a toy rocket at a speed as high as 35.0 m/s.

Anestetic [448]

Answer:

(a) 62.5 m

(b) 7.14 s

Explanation:

initial speed, u = 35 m/s

g = 9.8 m/s^2

(a) Let the rocket raises upto height h and at maximum height the speed is zero.

Use third equation of motion

$v^{2}=u^{2}+2as$

$0^{2}=35^{2}- 2 \times 9.8 \times h$

h = 62.5 m

Thus, the rocket goes upto a height of 62.5 m.

(b) Let the rocket takes time t to reach to maximum height.

By use of first equation of motion

v = u + at

0 = 35 - 9.8 t

t = 3.57 s

The total time spent by the rocket in air = 2 t = 2 x 3.57 = 7.14 second.

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3 years ago

A 1/10th scale model of an airplane is tested in a wind tunnel. The reynolds number of the model is the same as that of the full

trasher [3.6K]

Answer:

of the velocity of a full size plane in the air

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3 years ago

9. If you placed the contents of a packet of powdered iced tea mix into a bottle of water and shook it, which of the following w

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The water would be solvent. Its your answer!

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3 years ago

Read 2 more answers

An electron is projected with an initial speed of 5 3.2 10 / ⋅ m s directly toward a proton that is fixed in place. If the elect

Dvinal [7]

Answer:

The distance is $d =1.66*10^{-9}m$

Explanation:

From the question we are told that

The initial speed of the electron is $v_i = 3.2 *10^5 m/s$

The mass of electron is $m = 9.11*10^{-31}kg$

Let $d$ be the distance between the electron and the proton when the speed of the electron instantaneously equal to twice the initial value

Let $KE_i$ be the initial kinetic energy of the electron \

Let $KE_d$ be the kinetic energy of the electron at the distance $d$ from the proton

Considering that energy is conserved,

The energy at the initial position of the electron = The energy at the final position of the electron

i.e

$KE_i +U_1 = KE_d + U_2$

$U_1 \ and \ U_2$ are the potential energy at the initial position of the electron and at distance d of the electron to the proton

Here $U_1 = 0$

So the equation becomes

$\frac{1}{2} mv_i^2 = \frac{1}{2} mv_d^2 + \frac{kq_1 q_2}{d}$

Here $q_1 \ and \ q_2$ are the charge on the electron and the proton and their are the same since a charge on an electron is equal to charge on a proton

$k$ is electrostatic constant with value $8.99*10^9 N \cdot m^2 /C^2$

i.e $q = 1.602 *10^{-19}C$

$v_d$ is the velocity at distance d from the proton = 2 $v_i$

So the equation becomes

$\frac{1}{2}mv_i^2 = \frac{1}{2} m (2v_i)^2 -\frac{k(q)^2}{d}$

$\frac{1}{2} mv_i^2 = 4 [\frac{1}{2}mv_i^2 ]- \frac{k(q)^2}{d}$

$3[\frac{1}{2}mv_i^2 ] = \frac{k(q)^2}{d}$

Making d the subject of the formula

$d = \frac{2k(q)^2}{3mv_i^2}$

$= \frac{2* 8.99*10^9 *(1.602*10^{-19}^2)}{3 * 9.11*10^{-31} *(3.2*10^5)^2}$

$=1.66*10^{-9}m$

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3 years ago

What wave occurs when you push a spring back and forth

kipiarov [429]

Moving a spring back and forth creates a longitudinal wave

4 0

3 years ago