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Andru [333]
3 years ago
10

How many nodes are in the standing wave along the wire in the given figure?

Physics
2 answers:
dalvyx [7]3 years ago
6 0

sorry figure cannot there

Softa [21]3 years ago
3 0

Answer:

3

Explanation:

3 for plato users at every cross

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I need help fast!!!
Fantom [35]

You could easily help yourSELF if you simply follow the instructions in the question and "Use Ohm's law" to calculate the resistance.

Ohm's Law:  Resistance = (voltage) / (current)

Resistance = (10 v) / (5 A)

<em>Resistance = 2 ohms</em>

8 0
3 years ago
You find an unmarked blue laser on your way to physics class. When you get to class you realize you can determine the wavelength
Afina-wow [57]

Answer:

λ = 451.7 nm

Explanation:

The expression for the constructive interference of the double diffraction experiment is

          d sin θ = m λ

let's use trigonometry

          tan θ = y / L

   

how the experiment occurs at very small angles

          tan θ = sin θ / cos θ = sin θ

          sin θ = y / L

we substitute

         d y / L = m λ  

         λ = \frac{d \ y}{m \ L}

let's calculate

          λ = \frac{0.342 \ 10^{-3} \ 2.80 \ 10^{-3} }{1 \ 2.12 }

          λ = 4.51699 10⁻⁷ m

          λ = 4.517 10⁻⁷ m (109 nm / 1m)

          λ = 451.7 nm

5 0
3 years ago
A 100-W lightbulb is placed in a cylinder equipped with a moveable piston. The lightbulb is turned on for 0.010 hour, and the as
Taya2010 [7]

Answer:

w =  - 508.53 joules

q = - 3091.47 joules

Explanation:

Let us convert the time in hours into seconds

0.010* 3600\\= 36

Change in internal energy

\delta E = p * \delta t

where E is the internal energy in Joules

p is the power in watts

and t is the time in seconds

\delta E = - 100 * 36\\

\delta E = - 3600 Joules

Amount of work done by the system

w = - P * \delta V

where P is the pressure and V is the volume

Substituting the given values in above equation, we get -

w = - 1 * ( 5.92 -0.90)\\

w = -5.02 liter-atmospheres

Work done in Joules

- 5.02 * 101.3\\= 508.53Joules

q = \delta E - w\\

Substituting the given values we get -

q = - 3600 - (-508.53)\\q = - 3091.47

Thus

w =  - 508.53 joules

q = - 3091.47 joules

7 0
3 years ago
2 * 1.5 * (.850/2)^2A small ball with mass 1.50 kg is mounted on one end of a rod 0.850 m long and of negligible mass. The syste
grin007 [14]

Answer

given,

mass of the rod = 1.50 Kg

length of rod = 0.85 m

rotational velocity = 5060 rev/min

now calculating the rotational inertia of the system.

I = m L^2        

where L is the length of road, we will take whole length of rod because mass is at  the end of it.      

I = 1.5 \times 0.85^2  

I = 1.084 kg.m²                        

hence, the rotational inertia the system is equal to I = 1.084 kg.m²

8 0
3 years ago
What gas law applies to the situation you have described above. Why?
Allushta [10]

If you give us the situation described then I'll be able to help.

6 0
3 years ago
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