A local FM radio station broadcasts at an energy of 6.04E-29 kJ/photon. Calculate the frequency at which it is broadcasting.

9.11×10⁷ Hz

Explanation:

From the question given above the following data were obtained:

Energy (E) = 6.04×10¯²⁹ KJ

Frequency (f) =?

Next, we shall convert 6.04×10¯²⁹ KJ to joule (J). This can be obtained as follow:

1 KJ = 1000 J

Therefore,

6.04×10¯²⁹ KJ = 6.04×10¯²⁹ KJ × 1000 J / 1 KJ

6.04×10¯²⁹ KJ = 6.04×10¯²⁶ J

Thus, 6.04×10¯²⁹ KJ is equivalent to 6.04×10¯²⁶ J.

Finally, we shall determine the frequency at which the FM radio station is broadcasting as follow:

Energy (E) = 6.04×10¯²⁶ J

Planck's constant (h) = 6.63×10¯³⁴ Js

Frequency (f) =?

E = hf

6.04×10¯²⁶ = 6.63×10¯³⁴ × f

Divide both side by 6.63×10¯³⁴

f = 6.04×10¯²⁶ / 6.63×10¯³⁴

f = 9.11×10⁷ Hz

Therefore, the frequency at which the FM radio station is broadcasting is 9.11×10⁷ Hz

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.

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