Why was Dalton's atomic theory difficult for other scientists to accept? please help?!!!!!

A simple generator is used to generate a peak output voltage of 19.0 V . The square armature consists of windings that are 6.65

salantis [7]

One of the efficient concepts that can help us find the number of turns of the cable is through the concept of induced voltage or electromotive force given by Faraday's law. The electromotive force or emf can be described as,

$\epsilon = NBA\omega$

Where,

N = Number of loops

B = Magnetic Field

A = Cross-sectional Area

$\omega$ = Angular velocity

Re-arrange to find N,

$N = \frac{\epsilon}{BA\omega}$

Our values are given as,

$\epsilon = 19V$

$B = 0.434T$

$\omega = 49.8\frac{rev}{s} (\frac{2\pi rad}{1 rev}) = 99.6\pi rad/s$

$A = (6.65*10^{-2})^2 m^2$

Replacing at our equation we have:

$N = \frac{\epsilon}{( 0.434)A\omega}$

$N = \frac{19}{( 0.434)((6.65*10^{-2})^2)(99.6\pi)}$

$N = 31.63 \approx 32$

Therefore the number of loops of wire should be wound on the square armature is 32 loops

6 0

3 years ago

When the atmosphere is not quite clear, one may sometimes see colored circles concentric with the Sun or the Moon. These are gen

stealth61 [152]

Answer:

D) diffraction

Explanation:

Corona is an optical phenomenon produced by the diffraction of sunlight or moonlight, as light moves through water droplets in the atmosphere.

This phenomenon produces one or more diffuse concentric rings of light around the Sun or Moon, usually seen as colored circles.

Therefore, the explanation for these phenomena of colored concentric circles, sometimes seen with the Sun or the Moon involves diffraction.

5 0

3 years ago

A rope of negligible mass passes over a uniform cylindrical pulley of 1.50kg massand 0.090m radius. The bearings of the pulley h

boyakko [2]

Answer:

Explanation:

mass of pulley, m3 = 1.5 kg

Radius of pulley, R = 0.09 m

mass of monkey, m2 = 4.5 kg

mass of banana bunch, m1 = 3 kg

Let a is teh acceleration ans T1 and T2 be the tension in the rope.

The moment of inertia of the pulley

I = 0.5 x m3 x R² = 0.5 x 1.5 x 0.09 x 0.09 = 0.006075 kgm²

According to Newton's second law

T1 - m1 g = m1 x a .... (1)

m2 g - T2 = m2 x a ..... (2)

(T2 - T1 ) x R = I x α

where, α is the angular acceleration

α = a / R

(T2 - T1)R = 0.5 x m3 x R² x a / R

T2 - T1 = 0.5 x m3 x a ..... (3)

from (1), (2) and (3)

$a = \frac{m_{2}-m_{1}}{m_{1}+m_{2}+\frac{m_{3}}{2}}\times g$

$a = \frac{4.5-3}{3+4.5+0.75}\times 9.8$

a = 1.78 m/s²

from equation (1)

T1 = m1 ( g + a) = 3 ( 9.8 + 1.78) = 34.77 N

from equation (2)

T2 = m2 (g - a) = 4.5 (9.8 - 1.78) = 36.13 N

4 0

3 years ago

SHOW WORK

Helga [31]

Answer:

Follows are the solution to the given question:

Explanation:

For point a:

$T= 2\pi \sqrt{\frac{m}{k}}\\\\k = \frac{4 \pi^2 m}{T^2}\\\\= \frac{4 \times (3.14)^2 \times 3}{2^2}\\\\=29.578 \ \frac{N}{m}\\\\$

For Point b:

$E=\frac{1}{2} m a^2 w^2\\\\$

$=\frac{1}{2} \times m \times a^2 \times \frac{4\pi^2}{T^2}\\\\=\frac{1}{2} \times 3 \times (0.15)^2 \times \frac{4\times 3.14^2}{2^2}\\\\=0.332 \ J$

For Point C:

$V_{max}= a w$

$= (0.15) \times \frac{2\pi}{T}\\\\= (0.15) \times \frac{2\times 3.14}{2}\\\\=0.471 \frac{m}{s}$

For point D:

$X= a \sin (wt+ \phi)\\\\0.91=0.15 \sin(\frac{2\pi}{T} \times t+\phi)\\\\0.91=0.15 \sin(\frac{2\times 3.14}{2} \times 0.5+\phi)\\\\0.60 = \sin(3.14 \times 0.5+\phi)\\\\0.60 = \sin(1.57+\phi)\\\\1.57 +\phi =\sin^{-1} 60^{\circ}\\\\1.57 +\phi = 36.86^{\circ}\\\\=35.29^{\circ}\\\\So, X=15 \sin(3.14t+35.29^{\circ}) \ cm$

5 0

3 years ago

With the frequency set at the mid-point of the slider and the amplitude set at the mid-point of the slider, approximately how ma

Alenkinab [10]

Answer:

The wavelength stays the same.

Explanation:

When the amplitude is increased, the wavelength stays the same.

Here the wavelength doesn't depend upon the amplitude.

4 0

3 years ago