The radio waves of a particular AM radio station vibrate 680,000 times per second. What is the wavelength of the wave?
1 answer:
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Answer:
v₀ = 6.64 m / s
Explanation:
This is a projectile throwing exercise
x = v₀ₓ t
y = y₀ + v_{oy} t - ½ g t²
In this case they indicate that y₀ = 1.8 m and the point of the basket is x=4.9m y = 3.0 m
the time to reach the basket is
t = x / v₀ₓ
we substitute
y- y₀ =
y - y₀ = x tan θ -
we substitute the values
3 -1.8 = 3.0 tan 60 -
1.2 = 5.196 - 176.4 1 / v₀²
176.4 1 / v₀² = 3.996
v₀ =
v₀ = 6.64 m / s
Answer:
Explanation:
There's an easy way to answer this and then an easier way. I'll do both since I'm not sure what you're doing this for: physics or calculus. Calculus is the easier way, btw.
Going with the physics version first, here's what we know:
a = -9.8 m/s/s
v₀ = 3.75 m/s
t = ??
That's not a whole lot...at least not enough to directly solve the problem. What we have to remember here is that at the max height of a parabolic path, the final velocity is 0. So we can add that to our info:
v = 0 m/s. Use the one-dimensional equation that utilizes all that info and allows us to solve for time:
v = v₀ +at and filling in:
0 = 3.75 + (-9.8)t and
-3.75 = -9.8t so
t = .38 seconds. This is how long it takes to get to its max height. Another thing we need to remember (which is why calculus is so much easier!) is that at the halfway point of a parabolic path (the max height), the object has traveled half the time it takes to make the whole trip. In other words, if .38 is how long it takes to go halfway, then 2(.38) is how long the whole trip takes:
2(.38) = .76 seconds. Now onto the calculus way:
The position function is
The first derivative of this is the velocity function and, knowing that when the velocity is 0, the time is halfway gone, we will find the velocity function and then set it equal to 0 and solve for t:
v(t) = -9.8t + 3.75 and
0 = -9.8t + 3.75 and
-3.75 = -9.8t so
t = ,38 and multiply that by 2 to find the time the whole trip took:
2(.38) = .76 seconds.