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2 years ago

6

Fig. 19-4 shows two wires, in cross section, carrying 7.0 Amperes out of the page. Determine the total magnetic field, due to bo

th wires, at point X midway between the wires.

1 answer:

mel-nik [20]2 years ago

6 0

I habe no idea i just need points sorry

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Cell phone conversations are transmitted by high-frequency radio waves. Suppose the signal has wavelength 36.5 cm while travelin

tia_tia [17]

Answer:

f=8.219*10^{8}Hz

Explanation:

We are going to use the formula v=fλ

Where v= velocity of radio waves

f= frequency

λ= wavelength of wave

radio waves are electromagnetic waves and as such they have the speed of light which is 3*10^{8}m/s.

also when a wave travels from one medium to another, the wavelength changes while the frequency remains the same.

calculating for the frequency of the wave in air also gives us the frequency in the window glass.

f=\frac{v}{λ}

v=3*10^{8}m/s.

λ=36.5 cm = 36.5/100= 0.365m

f=\frac{3*10^{8}m/s.}{0.365m}

f=8.219*10^{8}Hz

7 0

2 years ago

An object executes simple harmonic motion with an amplitude A. (Use any variable or symbol stated above as necessary.) (a) At wh

valentina_108 [34]

Answer:

(a) x=ASin(ωt+Ф₀)=±(√3)A/2

(b) x=±(√2)A/2

Explanation:

For part (a)

V=AωCos(ωt+Ф₀)⇒±0.5Aω=AωCos(ωt+Ф₀)

Cos(ωt+Ф₀)=±0.5⇒ωt+Ф₀=π/3,2π/3,4π/3,5π/3

x=ASin(ωt+Ф₀)=±(√3)A/2

For part(b)

U=0.5E and U+K=E→K=0.5E

E=K(Max)

(1/2)mv²=(0.5)(1/2)m(Vmax)²

V=±(√2)Vmax/2→ωt+Ф₀=π/4,3π/4,7π/4

x=±(√2)A/2

7 0

2 years ago

Motion maps for two objects, Y and Z, are shown.

Vesna [10]

Answer:

The answer is B) 3 seconds

Explanation:

I just took the test on 2020 edge and got it right

5 0

2 years ago

When the wind kicks up dust and sand, the dust grains are charged. The small grains tend to get a negative charge, and the large

diamong [38]

Answer:

Explanation:

Small grains are negatively charged by the wind while big grains is positively charged and remains at the ground . This process creates an electric field due to the presence of oppositely charged particles.

When ever electric field exists it is directed from a positive charge to a negative charge so the here electric field is towards an upwards direction.

4 0

2 years ago

A proton is confined in an infinite square well of width 10 fm. (The nuclear potential that binds protons and neutrons in the nu

kvasek [131]

Answer:

First Question

$E = 1.065*10^{-12} \ J$

Second Question

The wavelength is for an X-ray

Explanation:

From the question we are told that

The width of the wall is $w = 10\ fm = 10*10^{-15 }\ m$

The first excited state is $n_1 = 2$

The ground state is $n_0 = 1$

Gnerally the energy (in MeV) of the photon emitted when the proton undergoes a transition is mathematically represented as

$E = \frac{h^2 }{ 8 * m * l^2 [ n_1^2 - n_0 ^2 ] }$

Here h is the Planck's constant with value $h = 6.62607015 * 10^{-34} J \cdot s$

m is the mass of proton with value $m = 1.67 * 10^{-27} \ kg$

So

$E = \frac{( 6.626*10^{-34})^2 }{ 8 * (1.67 *10^{-27}) * (10 *10^{-15})^2 [ 2^2 - 1 ^2 ] }$

=> $E = 1.065*10^{-12} \ J$

Generally the energy of the photon emitted is also mathematically represented as

$E = \frac{h * c }{ \lambda }$

=> $\lambda = \frac{h * c }{E }$

=> $\lambda = \frac{6.62607015 * 10^{-34} * 3.0 *10^{8} }{ 1.065 *10^{-15 } }$

=> $\lambda = 1.87*10^{-10} \ m$

Generally the range of wavelength of X-ray is $10^{-8} \to 1)^{-12}$

So this wavelength is for an X-ray.

8 0

2 years ago