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3 years ago

10

10 PTS. If an electromagnetic wave has a frequency of 8 ⋅ 10^14 H z , what is its wavelength? Use λ=V/F. The speed of light is 3

⋅ 10^8 m/s.

1 answer:

ad-work [718]3 years ago

3 0

Answer :

The wavelength would be 410 m/s

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Investigators are researching the appearance of a strange movement of objects in what the local community considers a haunted ho

wolverine [178]

Answer:

0.18216 T

Explanation:

N = Number of turns = 219

A = Area = $\pi r^2$

r = Radius = 1 cm

$\omega$ = Angular speed = $40\times 2\pi$

Maximum emf is given by

$\epsilon=NBA\omega\\\Rightarrow B=\dfrac{\epsilon}{NA\omega}\\\Rightarrow B=\dfrac{3.15}{219\times \pi 0.01^2\times 40\times 2\pi}\\\Rightarrow B=0.18216\ T$

The strength of the magnetic field is 0.18216 T

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3 years ago

1. In a(n) ______________________ circuit, all parts are connected one after another along one path.

poizon [28]

Answer:

1. A <em>series circuit </em>is a closed circuit which has all loads connected in a row and there is only one path for the current to flow.

2. The <em>Ohm's Law </em>state that a current flow through a resistor is directly proportional to the voltage across it $R = \frac{V}{I}$

3. A <em>parallel circuit </em>is a closed circuit divided into branches that it has two o more paths for the current to flow and the loads are parallel to each other which mean the voltage across them is the same for all loads.

3 0

3 years ago

Read 2 more answers

Advocard [28]

Answer:

$give \\ mass(m) = 1350kg \\ acceleration(a) = 1ms {}^{ - 2} \ \\ sln\ \\ from \: our \: formula \\ \: f = ma \\1350kg \times 1ms {}^{ - 2} \\ f = 1350newton$

the force you applied to your car =1350N

5 0

3 years ago

A baseball hit just above the ground leaves the bat 27 m/s at 45° above the horizontal. A) How far away does the ball strike the

Sedbober [7]

Answer:

A) The ball hits the ground 74.45 m far from the hitting position.

B) Maximum height of the ball = 18.57 m

Explanation:

There are two types of motion in this horizontal and vertical motion.

We have velocity = 27 m/s at 45° above the horizontal

Horizontal velocity = 27cos45 = 19.09 m/s

Vertical velocity = 27sin45 = 19.09 m/s

Time to reach maximum height,

v = u + at

0 = 19.09 - 9.81 t

t = 1.95 s

So total time of flight = 2 x 1.95 = 3.90 s

A) So the ball travels at 19.09 m/s for 3.90 seconds.

Horizontal distance traveled = 19.09 x 3.90 = 74.45 m

So the ball hits the ground 74.45 m far from the hitting position.

B) We have vertical displacement

S = ut + 0.5 at²

H = 19.09 x 1.95 - 0.5 x 9.81 x 1.95² = 18.57 m

Maximum height of the ball = 18.57 m

6 0

3 years ago

A 0.500-nm x-ray photon is deected through 134 in a Compton scattering event. At what angle (with respect to the incident beam)

natita [175]

Answer:

The angle of recoil electron with respect to incident beam of photon is 22.90°.

Explanation:

Compton Scattering is the process of scattering of X-rays by a charge particle like electron.

The angle of the recoiling electron with respect to the incident beam is determine by the relation :

$\cot\phi = (1+\frac{hf}{m_{e}c^{2} })\tan\frac{\theta }{2}$ ....(1)

Here ∅ is angle of recoil electron, θ is the scattered angle, h is Planck's constant, $m_{e}$ is mass of electron, c is speed of light and f is the frequency of the x-ray photon.

We know that, f = c/λ ......(2)

Here λ is wavelength of x-ray photon.

Rearrange equation (1) with the help of equation (1) in terms of λ .

$\cot\phi = (1+\frac{h}{m_{e}c\lambda })\tan\frac{\theta }{2}$

Substitute 6.6 x 10⁻³⁴ m² kg s⁻¹ for h, 9.1 x 10⁻³¹ kg for $m_{e}$ , 3 x 10⁸ m/s for c, 0.500 x 10⁻⁹ m for λ and 134° for θ in the above equation.

$\cot\phi = (1+\frac{6.6\times10^{-34} }{9.1\times10^{-31}\times3\times10^{8}\times0.5\times10^{-9} })\tan\frac{134 }{2}$

$\cot\phi=2.37$

$\phi$ = 22.90°

8 0

4 years ago