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34kurt
3 years ago
6

A laser beam with a frequency of 180 Hz forms an 8 m standing wave with 10 nodes.

Physics
1 answer:
DIA [1.3K]3 years ago
8 0

Answer:33

Explanation:

F = frequency

N =  Node count

w = wave lenght

v = wave velocity

L = distance wave traveled

First find wave length of laser

w = (2/(N))*(L)

w = (2/(10))*(8)

w = 1.6

then using (w), find velocity

V =  (w)(F)

V = (1.6)*(108)

V = 288

Plug in V and the new frequency to solve for new node count

F = NV/2L

(600) = (N)*(288) / 2 * (8)

(N) = 33.33

there are 33 nodes

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All of the following are kinds of forces except
dangina [55]

Answer:

All of the following are kinds of forces except?

<h2><u><em>b) acceleration is the Answer</em></u></h2>

Explanation:

<h3><u><em>Acceleration is actually produced when a force acts upon a body therefore it is the result after we apply any force.</em></u></h3>
3 0
3 years ago
A pendulum completes 30 cycles every minute. What is the frequency of the pendulum
Strike441 [17]

Answer:

1 cycle every 2 seconds if that is what it is asking. So you have to do 2 x 30 and you will get 60 so ur awnser is 60

Explanation:

8 0
3 years ago
a small negatively charged sphere with a mass of 5.4*10^-5 is suspended between two parallel plates. the potential difference is
labwork [276]
Here, Fe = Fg
q.E = m.g
We have: E = 360 V
m = 5.4 × 10⁻⁵
g = 9.8 m/s²   [ constant value for earth system ]

Substitute their values into the expression:
q (360) = 5.4 × 10⁻⁵ × 9.8
q = 52.92 × 10⁻⁵ / 360
q = -1.47 × 10⁻⁶  [ negative sign represents the nature of charge ] 

So, Your Final answer would be 1.47 × 10⁻⁶

Hope this helps!
3 0
4 years ago
Two horizontal rods are each held up by vertical strings tied to their ends. Rod 1 has length L and mass M; rod 2 has length 2L
antiseptic1488 [7]

Answer:

Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.

Explanation:

For the rod 1 the angular acceleration is

\tau_1 = I_1\alpha _1 \\\\\alpha_1 = \dfrac{\tau_1}{I_1}

Similarly, for rod 2

\alpha_2 = \dfrac{\tau_2}{I_2}.

Now, the moment of inertia for rod 1 is

I_1 = \dfrac{1}{3}ML^2,

and the torque acting on it is (about the center of mass)

\tau_1 = Mg\dfrac{L}{2};

therefore, the angular acceleration of rod 1 is  

\alpha_1 = \dfrac{Mg\dfrac{L}{2}}{\dfrac{1}{3}ML^2},

\boxed{\alpha_1 = \dfrac{3g}{2L} }

Now, for rod 2 the moment of inertia is

I_2 = \dfrac{1}{3}(2M)(2L)^2

I_2 = \dfrac{8}{3} ML^2,

and the torque acting is (about the center of mass)

\tau _2 = (2M)g \dfrac{(2L)}{2}

\tau _2 = 2MgL;

therefore, the angular acceleration \alpha_2 is

\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.

\boxed{\alpha_2 = \dfrac{3g}{4L}}

We see here that

\dfrac{3g}{2L} > \dfrac{3g}{4L}

therefore

\boxed{\alpha_1 > \alpha_2.}

In other words , the initial angular acceleration for rod 1 is greater than for rod 2.

7 0
3 years ago
A car traveling at 12 m/s slows down with an acceleration of -1.5 m/s2. What is the
Alex777 [14]

Answer:

using s=ut +0.5at^2

u=12m/s

a=-1.5m/s2

t=6 s

s=54m

7 0
3 years ago
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