T= 24.5 feet per second. That’s the speed it attains just before hitting the ground.
To counteract a 58 mph crosswind, the western component of the trajectory must be accounted for. Consequently, directing towards the northwest creates a 45-degree angle, aligning with the destination. This triangle's third vertex is located at the destination, with the right angle positioned there. The western aspect of their flight represents the triangle's base, while the vertical side reflects the resultant path, and the hypotenuse indicates the actual distance traveled. Since the 58 mph crosswind was countered by flying in a northwest direction, the distance from the starting point to the destination should equal the westward segment of their journey. The hypotenuse can be determined via the square root of twice the dimension of the identical sides.
c = sqrt (58^2 + 58^2) = sqrt (6728) = 82.02
An alternative method:
c = sqrt (2) * 58 = 1.414 * 58 = 82.02
Thus, they must fly at 82.02 mph.
<span>E = h x f </span>
<span>Thus: </span>
<span>f = E / h </span>
<span>f = 4.41•10^-19 / 6.62•10^-34 </span>
<span>f = 6.66•10^14 Hz (s^-1) </span>
<span>b/ What is the wavelength of this light? </span>
<span>------------------------------ </span>
<span>λ = c / f </span>
<span>λ = 3•10^8 / 6.66•10^14 </span>
<span>λ = 4.50•10^-7 m </span>
For a string that is secured at both ends, the oscillation amplitude must be zero at both extremities of the string. Refer to the attached images for clarity. It is evident that our fundamental harmonic will have the following wavelength: All subsequent harmonics are simply multiples of the fundamental. The three longest wavelengths are as follows:
