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3 years ago

A wave with a period of 0.008 second has a frequency of

2 answers:

8 0

The frequency of a wave is the reciprocal of its period.

A period of 0.008 sec means a frequency of

1 / 0.008 sec = 125 per sec . (125 Hz)

zubka84 [21]3 years ago

4 0

A wave with a period of 0.008 second has a frequency of

1 /.008 = 125 Hz

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A cardinal (Richmondena cardinalis) of mass 3.70×10−2 kg and a baseball of mass 0.144 kg have the same kinetic energy. What is t

Radda [10]

Answer:

$\frac{p_{c}}{p_{b}}\approx 0.507$

Explanation:

Since the cardinal and ball have the same kinetic energy, it is possible to determine the ratio between speeds. (c for cardinal, b for baseball)

$K_{c} = K_{b}$

$\frac{1}{2}\cdot m_{c}\cdot v_{c}^{2}= \frac{1}{2}\cdot m_{b}\cdot v_{b}^{2}$

$\frac{v_{c}}{v_{b}}=\sqrt{\frac{m_{b}}{m_{c}} }$

The ratio is obtained by multiplying each side by $\frac{m_{c}}{m_{b}}$ :

$\frac{p_{c}}{p_{b}}=\frac{m_{c}}{m_{b}}\cdot \sqrt{\frac{m_{b}}{m_{c}} }$

$\frac{p_{c}}{p_{b}}= \sqrt{\frac{m_{c}}{m_{b}} }$

The value of this ratio is:

$\frac{p_{c}}{p_{b}}\approx 0.507$

3 0

3 years ago

Read 2 more answers

A hazard sign has 3 identical

melomori [17]

Answer and Explanation: To know how much tape he will need, we have to calculate the perimeter of each parallelogram-shaped stripe.

Perimeter is the sum of all the sides of a figure.

For a parallelogram:

P = 2*length + 2*width

So, we need to determine width and length of the stripe.

Width is 3 inches. Length is the hypotenuse of the right triangle, whose sides are 6 and 18 inches. Then, length is

$h=\sqrt{18^{2}+6^{2}}$

$h=\sqrt{360}$

h = 19 in

Perimeter of the first stripe is

P = (2*19) + (2*3)

P = 44 inches

The hazard sign has 3 stripes. So total perimeter is

$P_{t}=$ 44 + 44 + 44

$P_{t}=$ 132 inches

To outline the parallelogram-shaped stripes, Charles need a total of 132 inches of tape. Since one roll has 144 inches, he will have enough tape to finish the job.

5 0

3 years ago

A ship maneuvers to within 2.46×10³ m of an island’s 1.80 × 10³ m high mountain peak and fires a projectile at an enemy ship 6.1

Nesterboy [21]

Answer:

The distance close to the peak is 597.4 m.

Explanation:

Given that,

Distance of the first ship from the mountain $d=2.46\times10^{3}\ m$

Height of island $h=1.80\times10^{3}\ m$

Distance of the enemy ship from the mountain $d'=6.10\times10^{2}\ m$

Initial velocity $v=2.55\times10^{2}\ m/s$

Angle = 74.9°

We need to calculate the horizontal component of initial velocity

Using formula of horizontal component

$v_{x}=v\cos\theta$

Put the value into the formula

$v_{x}=2.55\times10^{2}\cos74.9$

$v_{x}=66.42\ m/s$

We need to calculate the vertical component of initial velocity

Using formula of vertical component

$v_{y}=v\sin\theta$

Put the value into the formula

$v_{y}=2.55\times10^{2}\sin74.9$

$v_{y}=246.19\ m/s$

We need to calculate the time

Using formula of time

$t=\dfrac{d}{v_{x}}$

$t=\dfrac{2.46\times10^{3}}{66.42}$

$t=37.03\ sec$

We need to calculate the height of the shell on reaching the mountain

Using equation of motion

$H= v_{y}t-\dfrac{1}{2}gt^2$

Put the value in the equation

$H=246.19\times37.03-\dfrac{1}{2}\times9.8\times(37.03)^2$

$H=2397.4\ m$

We need to calculate the distance close to the peak

Using formula of distance

$H'=H-h$

Put the value into the formula

$H'=2397.4-1800$

$H'=597.4\ m$

Hence, The distance close to the peak is 597.4 m.

6 0

3 years ago

The theory of plate tectonics describes how the Earth’s lithosphere is broken into plates and these plates move over the molten

Mumz [18]

Answer:

Explanation:

A theory can be changed when new evidence is found. A law doesn't change because it is universally a fact. It doesn't need new evidence to support it.

5 0

3 years ago

Light of wavelength 600 nm passes though two slits separated by 0.22mm and is observed on a screen 1.1m behind the slits. The lo

Ne4ueva [31]

Answer:

$y(m=1)=\frac{(1)(600*10^{-9}m)(1.1m)}{0.22m}=3*10^{-6}m\\\\y(m=1)=\frac{(-1)(600*10^{-9}m)(1.1m)}{0.22m}=-3*10^{-6}m$

Explanation:

We have to take into account the expression for the position of the fringes

$dsen\theta=m\lambda\\y=\frac{m\lambda D}{d}$

where m is the number of the maximum, d is the separation of the slits, D is the distance to the screen.

(a) By replacing we obtain

$y(m=1)=\frac{(1)(600*10^{-9}m)(1.1m)}{0.22m}=3*10^{-6}m\\\\y(m=1)=\frac{(-1)(600*10^{-9}m)(1.1m)}{0.22m}=-3*10^{-6}m$

(b) more information is required to solve this point. Please complete the information.

HOPE THIS HELPS!

4 0

3 years ago

Read 2 more answers