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3 years ago

All electromagnetic waves travel in vacuum at the speed of c=3×10^8 m/s. Find the wavelength of microwaves of frequency 10^10 Hz

1 answer:

5 0

Recall the wave equation, $c=f\lambda$ where c is the speed of the wave (m/s), f is the frequency of the wave (Hz) and λ is the wavelength of the wave (m).

$c=f\lambda \Rightarrow \lambda = \frac{c}{f}$ so $\lambda = \frac{3 \times 10^8}{10^{10}} = 0.03 \text{m}$

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A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a constant angular acceler

ki77a [65]

Answer:

a) $t_1=\frac{w_1-w_o}{\alpha}=\frac{w_1}{\alpha}sec$

b) $\theta_1=\frac{w_1^2}{2\alpha}rad$

c) $t_2=\frac{\alpha t_1}{5\alpha}=\frac{t_1}{5}sec$

Explanation:

1) Basic concepts

Angular displacement is defined as the angle changed by an object. The units are rad/s.

Angular velocity is defined as the rate of change of angular displacement respect to the change of time, given by this formula:

$w=\frac{\Delat \theta}{\Delta t}$

Angular acceleration is the rate of change of the angular velocity respect to the time

$\alpha=\frac{dw}{dt}$

2) Part a

We can define some notation

$w_o=0\frac{rad}{s}$ ,represent the initial angular velocity of the wheel

$w_1=?\frac{rad}{s}$ , represent the final angular velocity of the wheel

$\alpha$ , represent the angular acceleration of the flywheel

$t_1$ time taken in order to reach the final angular velocity

So we can apply this formula from kinematics:

$w_1=w_o +\alpha t_1$

And solving for t1 we got:

$t_1=\frac{w_1-w_o}{\alpha}=\frac{w_1}{\alpha}sec$

3) Part b

We can use other formula from kinematics in order to find the angular displacement, on this case the following:

$\Delta \theta=wt+\frac{1}{2}\alpha t^2$

Replacing the values for our case we got:

$\Delta \theta=w_o t+\frac{1}{2}\alpha t_1^2$

And we can replace $t_1$ from the result for part a, like this:

$\theta_1-\theta_o=w_o t+\frac{1}{2}\alpha (\frac{w_1}{\alpha})^2$

Since $\theta_o=0$ and $w_o=0$ then we have:

$\theta_1=\frac{1}{2}\alpha \frac{w_1^2}{\alpha^2}$

And simplifying:

$\theta_1=\frac{w_1^2}{2\alpha}rad$

4) Part c

For this case we can assume that the angular acceleration in order to stop applied on the wheel is $\alpha_1 =-5\alpha \frac{rad}{s}$

We have an initial angular velocity $w_1$ , and since at the end stops we have that $w_2 =0$

Assuming that $t_2$ represent the time in order to stop the wheel, we cna use the following formula

$w_2 =w_1 +\alpha_1 t_2$

Since $w_2=0$ if we solve for $t_2$ we got

$t_2=\frac{0-w_1}{\alpha_1}=\frac{-w_1}{-5\alpha}$

And from part a) we can see that $w_1=\alpha t_1$ , and replacing into the last equation we got:

$t_2=\frac{\alpha t_1}{5\alpha}=\frac{t_1}{5}sec$

5 0

3 years ago

A researcher measures the thickness of a layer of benzene (nn = 1.50) floating on water by shining monochromatic light onto the

baherus [9]

Answer:

The minimum thickness is $t= 8.75*10^{-8} m$

Explanation:

generally the equation for thin film interference is mathematically represented as

$2nt = (m + \frac{1}{2} ) \lambda$

Where t the thickness

m is any integer

n is the refractive index of the film

$\lambda$ is the wavelength of light

Since we are looking for the thickness we make t the subject of the formula

$t = \frac{(m+ \frac{1}{2} ) \lambda}{2n}$

m= 0 cause the thickness is minimum at m=0

Substituting values

$t = \frac{(0 +\frac{1}{2}) 8525*10^{-9} }{2 *1.5}$

$t= 8.75*10^{-8} m$

8 0

3 years ago

Read 2 more answers

An object is projected upward from the surface of the earth with an initial speed of 3.7 km/s. find the maximum height it reache

Mariulka [41]

The kinetic energy of the object just after it starts its motion from the Earth surface is
$K= \frac{1}{2}mv^2$
where m is the object mass and $v=3.7 km/s=3700 m/s$ its initial speed.
When it reaches its maximum height, the object speed is zero and all its kinetic energy converted into gravitational potential energy, which is
$U=mgh$
where $g=9.81 m/s^2$ and h is the maximum height reached by the object.

Since the energy of the object must be conserved, K=U, therefore we can write
$\frac{1}{2}mv^2 =mgh$
and we can solve to find h, the maximum height:
$h= \frac{v^2}{2g}= \frac{(3700 m/s)^2}{2\cdot 9.81 m/s^2}= 6.98 \cdot 10^5 m=698 km$

5 0

4 years ago

Which of the following is a trend in the health sciences

forsale [732]

Answer

more patient-centered care / advancing technology / integrative medicine

5 0

3 years ago

Read 2 more answers

As you travel from Detroit in a certain direction, the outside temperature, T (in degrees), depends on your distance, d (in mile

Ber [7]

Answer:

a) $\Delta T= 100^{\circ}C$

b) $\bigtriangledown T=1^{\circ}C.mile^{-1}$

c) $\bigtriangledown T_4=1^{\circ}C.mile^{-1}$

d) $\bigtriangledown T_4=1^{\circ}C.mile^{-1}$

Explanation:

Given is the data of variation of temperature with respect to the distance traveled:

Temperature T as a function of distance d:

$T=(d+30) ^{\circ}C$ ...................................(1)

(a)

Total change in temperature from the start till the end of the journey:

$\Delta T= T_f-T_i$ ..............................(2)

where:

$T_f$ = final temperature

$T_i$ = initial temperature

∵In the start of the journey d = 0 miles & at the end of the journey d = 100 miles.

So, correspondingly we have the eq. (2) & (1) as:

$\Delta T= (100+30)-(0+30)$

$\Delta T= 100^{\circ}C$

(b)

Now, the average rate of change of the temperature, with respect to distance, from the beginning of the trip to the end of the trip be calculated as:

$\bigtriangledown T=\frac{\Delta T}{\Delta d}$ ......................(3)

where:

$\Delta d$ = change in distance

$\bigtriangledown T=$ change in temperature with respect to distance

putting the respective values in eq. (3)

$\bigtriangledown T=\frac{100}{100}$

$\bigtriangledown T=1^{\circ}C.mile^{-1}$

(c)

comparing the given function of the temperature with the general equation of a straight line:

$y=m.x+c$

We find that we have the slope of the equation as 1 throughout the journey and therefore the rate of change in temperature with respect to distance remains constant.

$\bigtriangledown T_4=1^{\circ}C.mile^{-1}$

(d)

comparing the given function of the temperature with the general equation of a straight line:

$y=m.x+c$

We find that we have the slope of the equation as 1 throughout the journey and therefore the rate of change in temperature with respect to distance remains constant.

$\bigtriangledown T_4=1^{\circ}C.mile^{-1}$

4 0

4 years ago