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3 years ago

Two long, straight wires are parallel and 26 cm apart.

Physics

1 answer:

mezya [45]3 years ago

5 0

Answer: 2.49×10^-3 N/m

Explanation: The force per unit length that two wires exerts on each other is defined by the formula below

F/L = (u×i1×i2) / (2πr)

Where F/L = force per meter

u = permeability of free space = 1.256×10^-6 mkg/s^2A^2

i1 = current on first wire = 57A

i2 = current on second wire = 57 A

r = distance between both wires = 26cm = 0.26m

By substituting the parameters, we have that

Force per meter = (1.256×10^-6×57×57)/ 2×3.142 ×0.26

= 4080.744×10^-6/ 1.634

= 4.080×10^-3 / 1.634

= 2.49×10^-3 N/m

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A transverse wave on a string is described by the following wave function.y = (0.090 m) sin (px/11 + 4pt)(a) Determine the trans

alukav5142 [94]

Explanation:

(a) It is known that equation for transverse wave is given as follows.

y = $(0.09 m)sin(\pi \frac{x}{11} + 4 \pi t)$

Now, we will compare above equation with the standard form of transeverse wave equation,

y = $A sin(kx + \omega t)$

where, A is the amplitude = 0.09 m

k is the wave vector = $\frac{\pi}{11}$

$\omega$ is the angular frequency = $4\pi$

x is displacement = 1.40 m

t is the time = 0.16 s

Now, we will differentiate the equation with respect to t as follows.

The speed of the wave will be:

v(t) = $\frac{dy}{dt}$

v(t) = $A \omega cos(kx + \omega t)$

v(t) = $(0.09 m)(4\pi) cos(\frac{\pi \times 1.4}{11} + 4 \pi \times 0.16)$

v(t) = -0.84 m/s

The acceleration of the particle in the location is

a(t) = $\frac{dv}{dt}$

a(t) = $-A \omega 2sin(kx + \omega t)$

a(t) = $-(0.09 m)(4 \pi)2 sin(\frac{\pi \times 1.4}{11} + 4\pi \times 0.16)$

a(t) = -9.49 $m/s^{2}$

Hence, the value of transverse wave is 0.84 m/s and the value of acceleration is 9.49 $m/s^{2}$ .

(b) Wavelength of the wave is given as follows.

$\lambda = \frac{2\pi}{k}$

$\lambda = (frac{2\pi}{\frac{\pi}{11})
$

$\lambda$ = 22 m

The period of the wave is

T = $\frac{2 \pi}{\omega}$

T = $\frac{2 \pi}{4 \pi}$

= 0.5 sec

Now, we will calculate the speed of propagation of wave as follows.

v = $\frac{\lambda}{T}$

= $\frac{22 m}{0.5 s}$

= 44 m/s

therefore, we can conclude that wavelength is 22 m, period is 0.5 sec, and speed of propagation of wave is 44 m/s.

7 0

3 years ago

Plz help this is so confusing

Nata [24]

Answer:

5 Km/h

Explanation:

From the question given above, the following data were obtained:

Distance travelled = 10 Km

Time = 2 hours

Speed =?

Speed is simply defined as the distance travelled per unit time. Mathematically, it can be represented as:

Speed = distance travelled /time.

With the above formula, we can obtain the speed at which the duck is travelling as follow:

Distance travelled = 10 Km

Time = 2 hours

Speed =?

Speed = distance travelled /time.

Speed = 10 / 2

Speed = 5 Km/h

Thus, the duck is travelling at a speed of 5 Km/h

7 0

3 years ago

4.5 billion km, the average separation between the sun and Neptune (report answer in hours). How long does it take light to trav

Liula [17]

Answer:

t = 4.17 hours

Explanation:

given,

The distance between Sun and Neptune, d = 4.5 billion Km

= 4.5 x 10⁹ Km

= 4.5 x 10¹¹ m

The velocity of light, c = 3 x 10⁸ m/s

The velocity is always equal to displacement by the time.

∴ t = d / V

= 4.5 x 10¹¹ m / 3 x 10⁸ m/s

= 15,000 s

= 4.17 h

Hence, the time taken by the light rays to reach the Neptune is, t = 4.17 h

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An object moves along the x axis. The graph shows it’s position x as a function of time t. Find the average velocity of the obje

nadezda [96]

Question is from B to C

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t1=1s, t2=5s
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Time elapsed, ∆t=(5-1)s=4s

So average velocity v =∆x/∆t=6/4=1.5m/s

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ladessa [460]

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