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Lunna [17]
3 years ago
10

A wave is moving at the rate of 40 cm/s. Its wavelength is 8 cm. What is the frequency of the wave?

Physics
1 answer:
IgorC [24]3 years ago
5 0
Speed = frequency times wavelength.
40=8xf
f = 5 Hz or cycles per second.
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OleMash [197]
<span>experimental? I hope that helps :)</span>
4 0
3 years ago
Read 2 more answers
Ben franklin showed that 1 teaspoon of oil would cover about 0.50 acre of still water. if you know that 1.0 x 104m2 = 2.47 acres
Dmitry_Shevchenko [17]

To solve for the thickness of the latter of oil, we can simply divide the volume by the total surface area:

Thickness = Volume / Surface Area

We are both given the volume and the surface area, all we have to do now is to convert the units in like terms so that we can cancel those out.

Surface Area = 0.50 acres (1.0 x 10^4 m^2 / 2.47 acres)

Surface Area = 2,024.29 m^2

Further converting this into cm:

Surface Area = 2,024.29 m^2 (100 cm / m)^2

Surface Area = 20,242,914.98 cm^2

 

Therefore the thickness is:

Thickness = 5.0 cm^3 / 20,242,914.98 cm^2

Thickness = 2.47 x 10^-7 cm

or

<span>Thickness = 2.47 x 10^-9 m = 2.47 nm = 24.7 A</span>

3 0
3 years ago
A zone reconnaissance involves a directed effort to obtain detailed information on all routes, obstacles, terrain, enemy forces,
son4ous [18]

Answer:

It is True

Explanation:.

A  commander assigns a zone reconnaissance mission when he seeks additional information on a zone before committing other forces in the zone. It is appropriate when the enemy situation is vague,  existing knowledge of the terrain is limited, or combat operations have altered the terrain. A zone  reconnaissance could include several route or area reconnaissance missions assigned to subordinate units.

7 0
2 years ago
The space station is 4.41 x 10^5 kg and orbits the earth 6.78 x 10^6 m from the center of earth. The mass of earth is 5.97 x 10^
allochka39001 [22]

Answer:

3 820 885 N

Explanation:

Gravitational equation

   F = G  m1 m2 / r^2    

         G = gravitational constant = 6.6713 x 10^-11 m^3/kg-s^2

F = 6.6713 x 10^-11   *   4.41 x 10^5  * 5.97 x 10^24  / ( 6.78x 10^6)^2

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6 0
2 years ago
Consider a car travelling at 60 km/hr. If the radius of a tire is 25 cm, calculate the angular speed of a point on the outer edg
vlabodo [156]

To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.

From the perspective of angular movement, we find the relationship with the tangential movement of velocity through

\omega = \frac{v}{R}

Where,

\omega =Angular velocity

v = Lineal Velocity

R = Radius

At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is

\alpha = \frac{\omega}{t}

Where

\alpha =Angular acceleration

\omega = Angular velocity

t = Time

Our values are

v = 60\frac{km}{h} (\frac{1h}{3600s})(\frac{1000m}{1km})

v = 16.67m/s

r = 0.25m

t=6s

Replacing at the previous equation we have that the angular velocity is

\omega = \frac{v}{R}

\omega = \frac{ 16.67}{0.25}

\omega = 66.67rad/s

Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s

At the same time the angular acceleration would be

\alpha = \frac{\omega}{t}

\alpha = \frac{66.67}{6}

\alpha = 11.11rad/s^2

Therefore the angular acceleration of a point on the outer edge of the tires is 11.11rad/s^2

5 0
2 years ago
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