The radius of Mercury (from the center to just above the atmosphere) is 2440 km

(a)4676m/s

(b) 4227m/s

Explanation:

Let the mass of mercury be Mm = 3×10²³ kg and the mass of the object be M

Also let the distance from the center of mercury be R1 = 2.240×10⁶m let the distance the object is launched to be R

From the energy conservation equation

K1 + U1 = K2 + U2

K1, K2 = initial and final kinetic energies.

U1, U2 = initial and final potential energies.

1/2MV1² + (-GMm×M/R1) = 1/2MV2² + (-GMm×M/R)

Assuming that the very far R approaches infinity then the second term to the right side of the equation above approaches zero

So dividing through by M since M is common to both sides we have that

1/2V1² + (-G×Mm/R1) = 1/2V2²

Rearranging,

V1² = 2×( 1/2V2²+ GMm/R1 )

Given V2 = 2000m/s and also the constant value of G = 6.67×10‐¹¹ N•m²/kg²

Substituting,

V1² =2×( 1/2×2000² + 6.67×10-¹¹ ×3×10²³ /(2.24×10⁶))

V1² = 21866071.43

V1 = 4676m/s

(b) this is still similar to the case described above but this time V2 = 0m/s

V1² = 2×( 1/2V2²+ GMm/R1 ) = 2×(1/2×0+ GMm/R1) = 2GMm/R1

V1² = 2×6.67×10-¹¹ × 3×10²³/(2.24×10⁶)

= 17866071.43

V1 = 4227m/s

8 Hz, 48 Hz

Explanation:

The standing waves on a string (or inside a pipe, for instance) have different modes of vibrations, depending on how many segments of the string are vibrating.

The fundamental frequency of a standing wave is the frequency of the fundamental mode of vibration; then, the higher modes of vibration are called harmonics. The frequency of the n-th harmonic is given by

where

is the fundamental frequency

In this problem, we know that the wave's third harmonic has a frequency of

This means this is the frequency for n = 3. Therefore, we can find the fundamental frequency as:

Now we can also find the frequency of the 6-th harmonic using n = 6: