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Karo-lina-s [1.5K]

2 years ago

5

To navigate, a porpoise emits a sound wave that has a wavelength of 2.2 cm. The speed at which the wave travels in seawater is 1

522 m/s. Find the period of the wave.

1 answer:

Alenkasestr [34]2 years ago

3 0

To solve this problem we will apply the concept related to frequency, by which it is defined as the relationship between the speed of the wave and the wavelength. Since the frequency is the inverse of the period, this relationship will also be reversed. That is to say,

$f = \frac{v}{\lambda}$

Here,

v = Velocity

$\lambda$ = Wavelength

Now the period is defined as,

$T = \frac{1}{f}$

Therefore

$T = \frac{\lambda}{v}$

Replacing we have that,

$T = \frac{2.2*10^{-2}}{1522}$

$T = 1.445*10^{-5}s$

Therefore the period of the wave is $1.445*10^{-5}s$

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A ball weighing 1 lb is attached to a string 2 feet long and is whirled in a vertical circle at a constant speed of 10 ft/sec.

fredd [130]

Explanation:

It is given that,

Mass of the ball, m = 1 lb

Length of the string, l = r = 2 ft

Speed of motion, v = 10 ft/s

(a) The net tension in the string when the ball is at the top of the circle is given by :

$F=\dfrac{mv^2}{r}-mg$

$F=m(\dfrac{v^2}{r}-g)$

$F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}-1\ lb\times 32\ ft/s^2)$

F = 18 N

(b) The net tension in the string when the ball is at the bottom of the circle is given by :

$F=\dfrac{mv^2}{r}+mg$

$F=m(\dfrac{v^2}{r}+g)$

$F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}+1\ lb\times 32\ ft/s^2)$

F = 82 N

(c) Let h is the height where the ball at certain time from the top. So,

$T=mg(\dfrac{r-h}{r})+\dfrac{mv^2}{r}$

$T=\dfrac{m}{r}(g(r-h)+v^2)$

Since, $v^2=u^2-2gh$

$T=\dfrac{m}{r}(u^2-3gh+gr)$

Hence, this is the required solution.

6 0

2 years ago

Ferdinand the frog is hopping from lily pad to lily pad in search of a good fly

loris [4]

Answer: $36.86\°$

Explanation:

According to the described situation we have the following data:

Horizontal distance between lily pads: $d=2.4 m$

Ferdinand's initial velocity: $V_{o}=5 m/s$

Time it takes a jump: $t=0.6 s$

We need to find the angle $\theta$ at which Ferdinand jumps.

In order to do this, we first have to find the <u>horizontal component (or x-component)</u> of this initial velocity. Since we are dealing with parabolic movement, where velocity has x-component and y-component, and in this case we will choose the x-component to find the angle:

$V_{x}=\frac{d}{t}$ (1)

$V_{x}=\frac{2.4 m}{0.6 s}$ (2)

$V_{x}=4 m/s$ (3)

On the other hand, the x-component of the velocity is expressed as:

$V_{x}=V_{o}cos\theta$ (4)

Substituting (3) in (4):

$4 m/s=5 m/s cos\theta$ (5)

Clearing $\theta$ :

$\theta=cos^{-1} (\frac{4 m/s}{5 m/s})$

$\theta=36.86\°$ This is the angle at which Ferdinand the frog jumps between lily pads

4 0

2 years ago

What is the explanation for how a modern transmission electron microscope (TEM) can achieve a resolution of about 0.2 nanometers

IgorC [24]

Answer:

Explanation:

A simple light microscope uses light for imaging of objects where as a transmission electron microscope uses a monochromatic beam of electrons.

This beam is passed by a magnetic field which is very strong and thus act as a lens.

Its resolution of very high which is about 0.2 nanometers because of the separation between two atoms.

Because of this reason its resolution is about 1000 times greater than light microscope.

3 0

2 years ago

If the rocket has an initial mass of 6300 kg and ejects gas at a relative velocity of magnitude 2000 m/s , how much gas must it

Rzqust [24]

Answer:

The amount of gas that is to be released in the first second in other to attain an acceleration of 27.0 m/s2 is

$\frac{\Delta m}{\Delta t} = 83.92 \ Kg/s$

Explanation:

From the question we are told that

The mass of the rocket is m = 6300 kg

The velocity at gas is being ejected is u = 2000 m/s

The initial acceleration desired is $a = 27.0 \ m/s$

The time taken for the gas to be ejected is t = 1 s

Generally this desired acceleration is mathematically represented as

$a = \frac{u * \frac{\Delta m}{\Delta t} }{M -\frac{\Delta m}{\Delta t}* t}$

Here $\frac{\Delta m}{\Delta t }$ is the rate at which gas is being ejected with respect to time

Substituting values

$27 = \frac{2000 * \frac{\Delta m}{\Delta t} }{6300 -\frac{\Delta m}{\Delta t}* 1}$

=> $170100 -27* \frac{\Delta m}{\Delta t} = 2000 * \frac{\Delta m}{\Delta t}$

=> $170100 = 2027 * \frac{\Delta m}{\Delta t}$

=> $\frac{\Delta m}{\Delta t} = \frac{170100}{2027}$

=> $\frac{\Delta m}{\Delta t} = 83.92 \ Kg/s$

3 0

2 years ago

A square conducting loop 8.4 cm on a side is placed in a uniform B-field so that the plane of the loop is perpendicular to the d

arsen [322]

Answer:

Explanation:

area of square loop A = side²

= 8.4² x 10⁻⁴

A = 70.56 x 10⁻⁴ m²

when it is converted into rectangle , length = 14.7 , width = 2.1

area = length x width

= 14.7 x 2.1 x 10⁻⁴

= 30.87 x 10⁻⁴ m²

Let magnetic field be B

Change in flux = magnetic field x change in area

= B x ( 70.56 x 10⁻⁴ - 30.87 x 10⁻⁴ )

= 39.69 x 10⁻⁴ B

rate of change of flux = change in flux / time taken

= 39.69 x 10⁻⁴ B / 6.5 x 10⁻³

= 6.1 x 10⁻¹ B

emf induced = 6.1 x 10⁻¹ B

6.1 x 10⁻¹ B = 14.7 ( given )

B = 2.41 x 10

= 24.1 T

B ) magnetic flux is decreasing , so it needs to be increased as per Lenz's law . Hence current induced will be anticlockwise so that additional magnetic flux is induced out of the page.

4 0

2 years ago