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bixtya [17]
3 years ago
12

What is the frequency of this wave? can someone pls explain and answer ​

Physics
1 answer:
Alenkasestr [34]3 years ago
6 0

Answer:

f = 1 Hz

Explanation:

From the attached figure, we find that the time period of the wave is 1 second.

It is a longitudinal wave. It travel in the form of compression and rarefaction. The relation between time period T and frequency f is given by :

f=\dfrac{1}{T}\\\\\text{As T = 1 s}\\\\f=1\ s^{-1}\\\\=1\ Hz

Hence, the frequency of this wave is 1 Hz.

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A coin of mass m rests on a turntable a distance r from the axis of rotation. The turntable rotates with a frequency of f. What
Crank

Answer:

\mu = \frac{r (2\pi f)^{2}}{g}

Explanation:

N = normal force acting on the coin

Normal force in the upward direction balances the weight of the coin, hence

N = mg

f = frequency of rotation

Angular velocity of turntable is hence given as

w = 2\pi f

r = distance from the axis of rotation

\mu = minimum coefficient of static friction

static frictional force is given as

f = \mu N\\f = \mu mg

The  static frictional force provides the necessary centripetal force , hence

Centripetal force = Static frictional force

m r w^{2} = \mu mg\\r w^{2} = \mu g\\\\\mu = \frac{r w^{2}}{g} \\\mu = \frac{r (2\pi f)^{2}}{g}

3 0
3 years ago
You move a 25 n object 4 meters. find the work you did
d1i1m1o1n [39]
In physics, "work<span>" is when a force applied to an object moves the object in the same direction as the force. If someone pushes against a wall, no </span>work<span> is done on the system. It is calculated as follows:

Work = Force x distance
Work = 25 N x 4 meters
Work = 100 N.m</span>
5 0
3 years ago
Read 2 more answers
Which of the following is NOT an argument for showing that the Earth must be round: a. during an eclipse of the Moon, the shadow
Hitman42 [59]

Answer: The correct option is option E (the Sun is seen blocking different constellations in the course of a year.

Explanation:

The earth, which is one of the planets of the solar system that supports life, is shperical in shape. The spherical ( round) shape of the earth is marked by the intervening highlands and oceans on its surface.

Evidence to show that the earth is shperical are:

--> The Lunar eclipse: During an eclipse of the Moon, the shadow of the Earth is always seen to be round.

--> Ships Visibility: When ships travel a large distance away, we see their hulls disappear first and their masts disappear last.

-->Altitude of Polaris (North Star): The height of the North Star changes as we travel to different latitudes. That is ,increases as you move toward the North pole, or decreases as you move toward the equator.

--> Aerial photographs: Photographs of the Earth from space always show a round body.

The statement that doesn't prove that the earth is spherical in shape is (the Sun is seen blocking different constellations in the course of a year). The sun is seen in front of stars blocking different constellation in a year because the earth orbits round the sun in a year and not that it is shperical in shape.

4 0
3 years ago
A particle with charge -40.0nC is on the x axis at the point with coordinate x=0 . A second particle, with charge -20.0 nC , is
ICE Princess25 [194]

A particle with charge -40.0nC is on the x axis at the point with coordinate x=0 . A second particle, with charge -20.0 nC, is on the x axis at x=0.500 m.

No, there is no point at a finite distance where the electric potential is zero.

Hence, Option D) is correct.

What is electric potential?

Electric potential is the capacity for doing work. In the electrical case, a charge will exert a force on some other charge and the potential energy arises. For example, if a positive charge Q is fixed at some point in space, any other positive charge when brought close to it will experience a repulsive force and will therefore have potential energy.

It is also defined as the amount of work required to move a unit charge from a reference point to a specific point against an electric field.

To learn more about electric potential, refer to:

brainly.com/question/15764612

#SPJ4

4 0
1 year ago
A spaceship hovering over the surface of Venus drops an object from a height of 17 m. How much longer does it take to reach the
Paraphin [41]

1.96s and 1.86s. The time it takes to a spaceship hovering the surface of Venus to drop an object from a height of 17m is 1.96s, and the time it takes to the same spaceship hovering the surface of the Earth to drop and object from the same height is 1.86s.

In order to solve this problem, we are going to use the motion equation to calculate the time of flight of an object on Venus surface and the Earth. There is an equation of motion  that relates the height as follow:

h=v_{0} t+\frac{gt^{2}}{2}

The initial velocity of the object before the dropping is 0, so we can reduce the equation to:

h=\frac{gt^{2}}{2}

We know the height h of the spaceship hovering, and the gravity of Venus is g=8.87\frac{m}{s^{2}}. Substituting this values in the equation h=\frac{gt^{2}}{2}:

17m=\frac{8.87\frac{m}{s^{2} } t^{2}}{2}

To calculate the time it takes to an object to reach the surface of Venus dropped by a spaceship hovering from a height of 17m, we have to clear t from the equation above, resulting:

t=\sqrt{\frac{2(17m)}{8.87\frac{m}{s^{2} } }} =\sqrt{\frac{34m}{8.87\frac{m}{s^{2} } } }=1.96s

Similarly, to calculate the time it takes to an object to reach the surface of the Earth dropped by a spaceship hovering from a height of 17m, and the gravity of the Earth g=9.81\frac{m}{s^{2}}.

t=\sqrt{\frac{2(17m)}{9.81\frac{m}{s^{2} } }} =\sqrt{\frac{34m}{9.81\frac{m}{s^{2} } } }=1.86s

8 0
3 years ago
Read 2 more answers
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