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3 years ago

How does frequency change as wavelength increases?

1 answer:

5 0

"Frequency decreases" is the one way among the following choices given in the question that frequency change as wavelength increases. The correct option among all the options that are given in the question is the second option. I hope that this is the answer that has actually come to your desired help.

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An amusement park ride called the Rotor debuted in 1955 in Germany. Passengers stand in the cylindrical drum of the Rotor as it

steposvetlana [31]

Answer:

μs = 0.36

Explanation:

While the drum is rotating, the riders, in order to keep in a circular movement, are accelerated towards the center of the drum.
This acceleration is produced by the centripetal force.
Now, this force is not a different type of force, is the net force acting on the riders in this direction.
Since the riders have their backs against the wall, and the normal force between the riders and the wall is perpendicular to the wall and aiming out of it, it is easily seen that this normal force is the same centripetal force.
In the vertical direction, we have two forces acting on the riders: the force of gravity (which we call weight) downward, and the friction force, that will oppose to the relative movement between the riders and the wall, going upward.
When this force be equal to the weight, it will have the maximum possible value, which can be written as follows:

$F_{frmax} = \mu_{s}* F_{n} = m * g (1)$

where μs= coefficient of static friction (our unknown)

As we have already said Fn = Fc.
The value of the centripetal force, is related with the angular velocity ω and the radius of the drum r, as follows:

$F_{n} = m* \omega^{2} * r (2)$

Replacing (2) in (1), simplifying and rearranging terms, we can solve for μs, as follows:

$\mu_{s} = \frac{g}{\omega^{2} r} (3)$

Prior to replace ω for its value, is convenient to convert it from rev/min to rad/sec, as follows:

$\omega = 26.0 \frac{rev}{min} * \frac{1min}{60 sec} *\frac{2*\pi rad }{1 rev} = 2.72 rad/sec (4)$

Replacing g, ω and r in (3):
$\mu_{s} = \frac{g}{\omega^{2} r} = \frac{9.8m/s2}{(2.72rad/sec)^{2} *3.7 m} = 0.36 (5)$

3 0

3 years ago

Estimate how far apart the rays of deepest red and deepest violet light are as they exit the bottom surface. assume nred = 1.57

Harlamova29_29 [7]

We begin by noting that the angle of incidence is the one that's taken with respect to the normal to the surface in question. In this case the angle of incidence is 30. The material is Flint Glass according to the original question. The refractive indez of air n1=1, the refractive index of red in flint glass is nred=1.57, finally for violet in the glass medium is nviolet=1.60. Snell's Law dictates:
$n_1sin(\theta_1)=n_2sin(\theta_2)$
Where $\theta_2$ differs for each wavelenght, that means violet and red will have different refractive indices in the glass.
In the second figure provided details are given on which are the angles in question, $\Delta x$ is the distance between both rays.
$\theta_{2red}=Asin(\frac{sin(30)}{1.57})\approx 18.5705$
$\theta_{2violet}=Asin(\frac{sin(30)}{1.60})\approx 18.21$
At what distance d from the incidence normal will the beams land at the bottom?
For violet we have:
$d_{violet}=h.tan(\theta_{2violet})\approx 0.0132m$
For red we have:
$d_{red}=h.tan(\theta_{2red})\approx 0.0134m$
We finally have:
$\Delta x=d_{red}-d_{violet}\approx2.8\times10^{-4}m$

6 0

3 years ago

During an adiabatic process an object does 100 J of work and its temperature decreases by 5 K. During another process it does 25

Liono4ka [1.6K]

Answer:

The heat capacity for the second process is 15 J/K.

Explanation:

Given that,

Work = 100 J

Change temperature = 5 k

For adiabatic process,

The heat energy always same.

$dQ=0$

$dU=-dW$

We need to calculate the number of moles and specific heat

Using formula of heat

$dU=nC_{v}dT$

$nC_{v}=\dfrac{dU}{dT}$

Put the value into the formula

$nC_{v}=\dfrac{-100}{5}$

$nC_{v}=-20\ J/K$

We need to calculate the heat

Using formula of heat

$dQ=nC_{v}(dT_{1})+dW_{1}$

Put the value into the formula

$dQ=-20\times5+25$

$dQ=-75\ J$

We need to calculate the heat capacity for the second process

Using formula of heat

$dQ=nC_{v}(dT_{1})$

Put the value into the formula

$-75=nC_{v}\times(-5)$

$nC_{v}=\dfrac{-75}{-5}$

$nC_{v}=15\ J/K$

Hence, The heat capacity for the second process is 15 J/K.

5 0

3 years ago

How long would it take a drag racer to increase her speed from 10.m/s to 20 m/s if her car accelerates at a uniform rate of 15 m

garik1379 [7]

It would take about 2 thirds of a second or .66666666 repeating of a second. please give brainliest?

5 0

3 years ago

A truck covered 2/7 of a journey at an average speed of 40

sertanlavr [38]

Answer:

The amount of time for the whole journey is 8 hours.

Explanation:

A truck covered 2/7 of a journey at an average speed of 40 mph. Representing 1 the total of the trip traveled, then the rest of the distance traveled is calculated as: $1-\frac{2}{7} =\frac{5}{7}$

So if the truck covered the remaining 200 miles at $\frac{2}{7}$ , this means that $\frac{5}{7}$ of the trip represents the 200 miles. So, to calculate the total distance traveled by the truck, you apply the following rule of three: if $\frac{5}{7}$ of the route represents 200 miles, the integer 1 (which represents the total of the route), how many miles are they?

$miles=\frac{1*200miles}{\frac{5}{7} } =\frac{7}{5} *200 miles$

miles= 280

So the total distance traveled is 280 miles. Since speed is the relationship between the space traveled by an object and the time used for it ( $speed=\frac{distance}{time}$ ), then if the average of the entire trip was 35 mph and the distance traveled 280 miles, the time is calculated as:

$time=\frac{distance}{speed}=\frac{280 miles}{35 mph}$

time= 8 h

 The amount of time for the whole journey is 8 hours.

6 0

3 years ago