Two colours of light which mix to make white light are called

Frequency refers to the number of wavelengths that pass a fixed point in a minute. true or false

Art [367]

Frequency refers to the number of wavelengths that pass a fixed point in a minute. true or false

Answer:True

7 0

4 years ago

The tape in a videotape cassette has a total

xxMikexx [17]

Answer:

$w=19.76 \ rad/s$

Explanation:

<u>Circular Motion </u>

Suppose there is an object describing a circle of radius r around a fixed point. If the relation between the angle of rotation by the time taken is constant, then the angular speed is also constant. If that relation increases or decreases at a constant rate, the angular speed is given by:

$w=w_o+\alpha \ t$

Where $\alpha$ is the angular acceleration and t is the time. If the object was instantly released from the circular path, it would have a tangent speed of:

$\displaystyle v_t=w.r$

We have two reels: one loaded with the tape to play and the other one empty and starting to fill with tape. They both rotate at different angular speeds, one is increasing and the other is decreasing as the tape goes from one to the other. We'll assume the tangent speed is constant for both (so the tape can play correctly). Let's call $w_1$ the angular speed of the loaded reel and $w_2$ that from the empty reel. We have

$w_1=w_{o1}+\alpha_1 \ t$

$w_2=w_{o2}+\alpha_2 \ t$

If $r_f=35\ mm=0.035\ m$ is the radius of the reel when it's full of tape, the angular speed for the loaded reel is computed by

$\displaystyle w_{01}=\frac{v_t}{r_f}$

The tangent speed is computed by knowing the length of the tape and the time needed to fully play it.

$t=1.8\ h=1.8*3600=6480\ sec$

$\displaystyle v_t=\frac{x}{t}=\frac{249\ m}{6480\ sec}=0.0384 \ m/s$

$\displaystyle w_{01}=\frac{0.0384}{0.035}=1,098 \ rad/s$

If $r_e=10\ mm=0.001\ m$ is the radius of the reel when it's empty, the angular speed for the empty reel is computed by

$\displaystyle w_{02}=\frac{0.0384}{0.001}=38.426 \ rad/s$

The full reel goes from $w_{01}$ to $w_{02}$ in 6480 seconds, so we can compute the angular acceleration:

$\displaystyle \alpha_1=\frac{w_{02}-w_{01}}{6480}$

$\displaystyle \alpha_1=\frac{38.426-1.098}{6480}=0.00576 \ rad/sec^2$

The empty reel goes from $w_{02}$ to $w_{01}$ in 6480 seconds, so we can compute the angular acceleration:

$\displaystyle \alpha_2=\frac{1.098-38.426}{6480}=-0.00576 \ rad/sec^2$

So the equations for both reels are

$w_1=1.098+0.00576 \ t$

$w_2=38.426-0.00576 \ t$

They will be the same when

$1.098+0.00576 \ t=38.426-0.00576 \ t$

Solving for t

$\displaystyle t=\frac{38.426-1.098}{0.0115}$

$t=3240 \ sec$

The common angular speed is

$w_1=1.098+0.00576 \ 3240=19.76 \ rad/s$

$w_2=38.426-0.00576 \ 3240=19.76 \ rad/s$

They both result in the same, as expected

$\boxed{w=19.76 \ rad/s}$

6 0

4 years ago

Suppose 8.41 moles of a monatomic ideal gas expand adiabatically, and its temperature decreases from 395 to 279 K. Determine (a)

sergeinik [125]

Answer:

a) W=12166.20876 J

b) U= -12166.20876 J

Explanation:

No. of moles, n = 8.41

Change of temperature, ΔT = T1 - T2

= 395 - 279

= 116 K

For monatomic gas, γ = 5/3

γ -1 = 2 /3

Solution:

(a)

Work done, $W= \frac{nR}{\gamma-1}(T_1-T_2)$

plugging values we get

$W= \frac{8.314\times8.41}{2/3}(116)$

Ans: 12166.20876 J

Work done, W = + 12166.20876 J

(b)

From first law of thermodynamics, dQ = U + W

but, dQ = 0 ( adiabatic process)

Hence, U = - W

= - 12166.20876 J

Ans:

Change in internal energy, U = - 12166.20876 J

8 0

3 years ago

If i took physical science in 7thgrade do i get credit for physics

Anni [7]

Answer:

I don't think so

Explanation:

5 0

2 years ago

Read 2 more answers

The speed of sound in air is 320 ms-1 and in water it is 1600 ms-1. It takes 2.5 s for sound to reach a certain distance from th

Nonamiya [84]

Answer:

Distance covered by the sound in air is 800 meter and the time taken by the sound in water for the same distance is 0.5 seconds.

Explanation:

Given:

Speed of sound in air = 320 m/s

Speed of sound in water = 1600 m/s

Time taken to reach certain distance in air = 2.5 sec

a.

We have to find the distance traveled by sound in air.

Distance = Product of speed and time.

⇒ $Distance = Speed\times time\ taken$

⇒ $Distance = 320\times 2.5$

⇒ $Distance = 800$ meters.

b.

Now we have to find how much time the sound will take to travel in water.

⇒ Time = Ratio of distance and speed

⇒ $Time =\frac{distance}{speed}$

⇒ $Time =\frac{800}{1600}$ <em> ...distance = 800 m and speed = 1600 m/s</em>

⇒ $Time =\frac{1}{2}$

⇒ $Time =0.5$ seconds.

Distance covered by the sound in air is 800 meter and the time taken by the sound in water for the same distance is 0.5 seconds.

7 0

3 years ago