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MrMuchimi
1 year ago
14

GP A sinusoidal wave traveling in the negative x direction (to the left) has an amplitude of 20.0 cm, a wavelength of 35.0 cm, a

nd a frequency of 12.0Hz. The transverse position of an element of the medium at t=0, x=0 is y= 3.00cm, and the element has a positive velocity here. We wish to find an expression for the wave function describing this wave.(c) Find the period T from the frequency.
Physics
1 answer:
SIZIF [17.4K]1 year ago
8 0

The period of the wave is determined as 0.083 seconds.

<h3>What is period of a wave?</h3>

The period of a wave is the time taken by a particle of the medium to complete one vibration.

<h3>Period of the wave</h3>

The period of the wave is calculated as follows;

T = 1/f

where;

  • T is the period of the wave
  • f is frequency of the wave

T = 1/12

T = 0.083 seconds

Thus, the period of the wave is determined as 0.083 seconds.

Learn more about period of a wave here: brainly.com/question/18818486

#SPJ4

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A piston raises a weight, then lowers it again to its original height. If the process the system follows as the piston rises is
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Answer:

Negative

Explanation:

First law of thermodynamic also known as the  law of conservation of energy states that the total energy of an isolated system is constant; energy can be transformed from one form to another, but can be neither created nor destroyed.

The first law relates  relates changes in internal energy to heat added to a system and the work done by a system by the conservation of energy.

The first law is mathematically given as ΔU  = U_{F} - U_{0} = Q - W

Where Q  = Quantity of heat

            W = Work done

From the first law The internal energy has the symbol U. Q is positive if heat is added to the system, and negative if heat is removed; W is positive if work is done by the system, and negative if work is done on the system.

Analyzing the pistol when it raises in isothermal and when it falls in  isobaric state.The following can be said:

In the Isothermal compression of a gas there is work done on the system to decrease the volume and increase the pressure. For work to be done on the system it is a negative work done then.

In the Isobaric State An isobaric process occurs at constant pressure. Since the pressure is constant, the force exerted is constant and the work done is given as PΔV.If a gas is to expand at a constant pressure, heat should be transferred into the system at a certain rate.Isobaric is a fuction of heat which is Isothermal Provided the pressure is kept constant.

In Isobaric definition above it can be seen that " Heat should be transferred into the system ata certain rate. For heat to be transferred into the system work is deinitely been done on the system thereby favouring the negative work done.

6 0
3 years ago
What is the value of sin r and sin i in physics​
taurus [48]

Answer:

★The second law of refraction

The ratio of sine of angle of incidence to the sine of angle of refraction is a constant for a light of given colour and for a given pair of media. This law is also called Snell's law of refraction. If 'i' is the angle of incidence and 'r' is the angle of refraction then, Sin i/Sin r = constant

This constant value is called the refractive index of the second medium with respect to the first.

7 0
3 years ago
Carole is surprised when she makes an emergency stop and the tissue box that had been in the back window of her car hits the bac
krek1111 [17]
B An object in motion tends to stay in motion .
4 0
3 years ago
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In this lab you will use a cart and track to explore various aspects of motion. You will measure and record the time it takes th
Keith_Richards [23]

It is given that by using track and cart we can record the time and the distance travelled and also the speed of the cart can be recorded. With all this data we can solve questions on the laws of motion.  

Like using the first law of motion we can determine the force of gravity acting on the cart that has moved a certain distance and the velocity or the speed of card has already been registered and since time is known putting the values in formula would help us calculate the gravitational pull acting on cart.

3 0
3 years ago
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A gasoline tank has the shape of an inverted right circular cone with base radius 4 meters and height 5 meters. Gasoline is bein
RSB [31]

Answer:

h'=0.25m/s

Explanation:

In order to solve this problem, we need to start by drawing a diagram of the given situation. (See attached image).

So, the problem talks about an inverted circular cone with a given height and radius. The problem also tells us that water is being pumped into the tank at a rate of 8m^{3}/s. As you  may see, the problem is talking about a rate of volume over time. So we need to relate the volume, with the height of the cone with its radius. This relation is found on the volume of a cone formula:

V_{cone}=\frac{1}{3} \pi r^{2}h

notie the volume formula has two unknowns or variables, so we need to relate the radius with the height with an equation we can use to rewrite our volume formula in terms of either the radius or the height. Since in this case the problem wants us to find the rate of change over time of the height of the gasoline tank, we will need to rewrite our formula in terms of the height h.

If we take a look at a cross section of the cone, we can see that we can use similar triangles to find the equation we are looking for. When using similar triangles we get:

\frac {r}{h}=\frac{4}{5}

When solving for r, we get:

r=\frac{4}{5}h

so we can substitute this into our volume of a cone formula:

V_{cone}=\frac{1}{3} \pi (\frac{4}{5}h)^{2}h

which simplifies to:

V_{cone}=\frac{1}{3} \pi (\frac{16}{25}h^{2})h

V_{cone}=\frac{16}{75} \pi h^{3}

So now we can proceed and find the partial derivative over time of each of the sides of the equation, so we get:

\frac{dV}{dt}= \frac{16}{75} \pi (3)h^{2} \frac{dh}{dt}

Which simplifies to:

\frac{dV}{dt}= \frac{16}{25} \pi h^{2} \frac{dh}{dt}

So now I can solve the equation for dh/dt (the rate of height over time, the velocity at which height is increasing)

So we get:

\frac{dh}{dt}= \frac{(dV/dt)(25)}{16 \pi h^{2}}

Now we can substitute the provided values into our equation. So we get:

\frac{dh}{dt}= \frac{(8m^{3}/s)(25)}{16 \pi (4m)^{2}}

so:

\frac{dh}{dt}=0.25m/s

3 0
3 years ago
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