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3 years ago

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A cannonball is fired at a 45.0° angle and an initial velocity of 670 m/s. Assume no air resistance. What is the vertical compon

ent of the cannonball’s velocity? What is the horizontal component of the cannonball’s velocity?
473.8 m/s; 473.8 m/s
-525.2 m/s; 435.5 m/s
0 m/s; 670 m/s
-378 m/s; 378 m/s

1 answer:

elixir [45]3 years ago

6 0

Answer:

<h3>473.8 m/s; 473.8 m/s</h3>

Explanation:

Given the initial velocity U = 670m/s

Horizontal velocity Ux = Ucos theta

Vertical component of the cannon velocity Uy = Usin theta

Given

U = 670m/s

theta = 45°

horizontal component of the cannonball’s velocity = 670 cos 45

horizontal component of the cannonball’s velocity = 670(0.7071)

horizontal component of the cannonball’s velocity = 473.757m/s

Vertical component of the cannonball’s velocity = 670 sin 45

Vertical component of the cannonball’s velocity = 670 (0.7071)

Vertical component of the cannonball’s velocity = 473.757m/s

Hence pair of answer is 473.8 m/s; 473.8 m/s

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Write equations for both the electric and magnetic fields for an electromagnetic wave in the red part of the visible spectrum th

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Answer:

$E=3.5(8.98*10^{6}x-2.69*10^{15}t)$

$B=1.17*10^{-8}(8.98*10^{6}x-2.69*10^{15}t)$

Explanation:

The electric field equation of a electromagnetic wave is given by:

$E=E_{max}(kx-\omega t)$ (1)

E(max) is the maximun value of E, it means the amplitude of the wave.
k is the wave number
ω is the angular frequency

We know that the wave length is λ = 700 nm and the peak electric field magnitude of 3.5 V/m, this value is correspond a E(max).

By definition:

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

$k=8.98*10^{6} [rad/m]$

And the relation between λ and f is:

$c=\lambda f$

$f=\frac{c}{\lambda}$

$f=\frac{3*10^{8}}{700*10^{-9}}$

$f=4.28*10^{14}$

The angular frequency equation is:

$\omega=2\pi f$

$\omega=2\pi*4.28*10^{14}$

$\omega=2.69*10^{15} [rad/s]$

Therefore, the E equation, suing (1), will be:

$E=3.5(8.98*10^{6}x-2.69*10^{15}t)$ (2)

For the magnetic field we have the next equation:

$B=B_{max}(kx-\omega t)$ (3)

It is the same as E. Here we just need to find B(max).

We can use this equation:

$E_{max}=cB_{max}$

$B_{max}=\frac{E_{max}}{c}=\frac{3.5}{3*10^{8}}$

$B_{max}=1.17*10^{-8}T$

Putting this in (3), finally we will have:

$B=1.17*10^{-8}(8.98*10^{6}x-2.69*10^{15}t)$ (4)

I hope it helps you!

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<u>Option A is correct.</u>

Given, Frequency = $1\times 10^{18}\ Hz$

Thus, Wavelength is:

$Wavelength=\frac{c}{Frequency}$

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Answer

given,

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From the above expression we can clearly see that tension is directly proportional to velocity and inversely proportional to radius.

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Answer:

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