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Gennadij [26K]
3 years ago
11

Why would a positive current flow from the positive end of a wire towards its negative end?

Physics
1 answer:
irinina [24]3 years ago
5 0

because positive and negative attract

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A car traveling at 60 mph has how much more energy than a car going at 15 mph?
VikaD [51]

Answer:

No, if a car is going faster. The RPM is obviously higher. If that is higher, you can burn through gas and energy much faster. A car going at 15mph would be cruising and wouldn't have to worry too much about burning our your vehicle.

Explanation:

7 0
3 years ago
17. (16 pts) You set up a two-slit experiment with a laser that produces light with a
Natali [406]

when the two waves interfere with eachother to make a dark spot the periodic difference of the two waves is π . the wave length for 2π is 600nm

. ie. for π difference it is 300nm

4 0
3 years ago
A particle R moves with a velocity of 6 m/s due East and another particle S moves at 8 m/s due South. What is the magnitude of t
cluponka [151]

Answer:

10

Explanation:

We can look at this problem as a triangle, r is the hypotenuse so if we take the square root of 6^2+8^2 we get 10

6 0
2 years ago
Consider the interference pattern produced by two parallel slits of width a and separation d, in which d = 3a. The slits are ill
laila [671]

Answer:

a)   m =1  θ = sin⁻¹  λ  / d,  m = 2        θ = sin⁻¹ ( λ  / 2d) ,   c)     m = 3

Explanation:

a) In the interference phenomenon the maxima are given by the expression

         d sin θ = m λ

the maximum for m = 1 is at the angle

          θ = sin⁻¹  λ  / d

the second maximum m = 2

          θ = sin⁻¹ ( λ  / 2d)

the third maximum m = 3

        θ = sin⁻¹ ( λ  / 3d)

the fourth maximum m = 4

       θ = sin⁻¹ ( λ  / 4d)

b) If we take into account the effect of diffraction, the intensity of the maximums is modulated by the envelope of the diffraction of each slit.

       I = I₀ cos² (Ф) (sin x / x)²

       Ф = π d sin θ /λ

       x = pi a sin θ /λ

where a is the width of the slits

with the values ​​of part a are introduced in the expression and we can calculate intensity of each maximum

c) The interference phenomenon gives us maximums of equal intensity and is modulated by the diffraction phenomenon that presents a minimum, when the interference reaches this minimum and is no longer present

maximum interference       d sin θ = m λ

first diffraction minimum    a sin θ = λ

we divide the two expressions

                       d / a = m

In our case

                   3a / a = m

                    m = 3

order three is no longer visible

7 0
2 years ago
In a transverse wave the particles Name
7nadin3 [17]

Answer:

The direction a wave propagates is perpendicular to the direction it oscillates for transverse waves. A wave does not move mass in the direction of propagation; it transfers energy.

Explanation:

3 0
2 years ago
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