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3 years ago

13

A sound wave with a frequency of 400 Hz is moving through a solid object. If the wavelength of the sound wave is 8 m, what is th

e speed of sound traveling through the solid object?

1 answer:

Paraphin [41]3 years ago

4 0

Answer:

v = 3200 m/s

Explanation:

As we know that the frequency of the sound wave is given as

$f = 400 Hz$

wavelength of the sound wave is given as

$\lambda = 8 m$

so now we have

$speed = wavelength \times frequency$

so we will have

$v = (8m) \times (400 Hz)$

$v = 3200 m/s$

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3 years ago

Overtime how did matter and energy in a system naturally change in physics

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4 years ago

A rock is thrown straight up into the air with an initial speed of 55 m/s at time t = 0. Ignore air resistance in this problem.

Alekssandra [29.7K]

Answer:

After 1.938 sec velocity of rock will be 36 m/sec

Explanation:

We have given initial velocity at which rock is thrown u = 55 m/sec

Final velocity v = 36 m/sec

Acceleration due to gravity $g=9.8m/sec^2$

According to first equation of motion we know that $v=u+gt$ , here v is final velocity, u is initial velocity, g is acceleration due to gravity and t is time

So $36=55-9.8t$ ( Negative sign is due to rock is thrown upward )

So $9.8t=19$

t = 1.938 sec

So after 1.938 sec velocity of rock will be 36 m/sec

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4 years ago

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3 years ago

Read 2 more answers

The earth rotates once per day about an axis passing through the north and south poles, an axis that is perpendicular to the pla

mina [271]

Answer:

(a) v = 463.97 m/s, $a_{c} = 0.0337\ m/s^{2}$

(b) v' = 196.02 m/s, $a'_{c} = 0.0143\ m/s^{2}$

Solution:

As per the question:

Time period of the rotation of earth, T = 24 h = $24\times 3600 = 86400\ s$

Radius of the earth, R = $6.38\times 10^{6}\ m$

Angle, $\theta = 65.0^{\circ}$

Now,

Angular velocity is given by:

$\omega = \frac{2\pi}{T} = \frac{2\pi}{86400} = 7.27\times 10^{- 5} \rad$

(a) To calculate the speed and the centripetal acceleration at the equator:

Linear velocity or the speed, v = $R\omega$

$v = 6.38\times 10^{6}\times 7.27\times 10^{- 5} = 463.967\ m/s$

Centripetal Acceleration, $a_{c} = \frac{v^{2}}{R}$

$a_{c} = \frac{463.967^{2}}{6.38\times 10^{6}} = 0.0337\ m/s^{2}$

(b) To calculate the speed and acceleration at an altitude of $65.0^{\circ}$ N:

The horizontal component of the radius, R' = $Rcos\theta$

$R' = 6.38\times 10^{6}cos65.0^{\circ} = 2.69\times 10^{6}\ m$

Now,

For the speed, v' = $R'\omega = 2.69\times 10^{6}\times 7.27\times 10^{- 5} = 196.02\ m/s$

For the centripetal acceleration,

$a'_{c} = \frac{v'^{2}}{R'}$

$a'_{c} = \frac{196.02^{2}}{2.69\times 10^{6}} = 0.01428\ m/s^{2}$

8 0

4 years ago