A car speeding down a 20 foot hill
The difference in frequency of the two signals is
.
The given parameters;
- <em>frequency of the 13 C signal = 201.16 MHz</em>
The energy of the 13 C signal located at 20 ppm is calculated as follows;

The energy of the 13 C signal located at 179 ppm is calculated as follows;

The difference in frequency of the two signals is calculated as follows;

Thus, the difference in frequency of the two signals is
.
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Answer:
F = 1.2×10⁻³ N
Explanation:
From the question,
Applying newton's second law of motion,
F = m(v-u)/t................... Equation 1
Given: F = magnitude of the average force exerted on the ball, m = mass of the ball, v = final velocity, u = initial velocity, t = time of contact.
Note: let downward be negative and upward be positive.
Given: m = 48 g = 48/1000 = 0.048 kg, v = 17 m/s, u = -28 m/s (downward),
t = 1800 s
Substitute into equation 1
F = 0.048(17-[28])/1800
F = 1.2×10⁻³ N