Can any kind soul help me

2917.4 m/s

Explanation:

From the question given above, the following data were:

Gravitational acceleration of the Moon (g) = 0.25 times the gravitational acceleration of the earth

Radius (r) of the Moon = 1737 Km

Escape velocity (v) =?

Next, we shall determine the gravitational acceleration of the Moon. This can be obtained as follow:

Gravitational acceleration of the earth = 9.8 m/s²

Gravitational acceleration of the Moon (g) = 0.25 times the gravitational acceleration of the earth

= 0.25 × 9.8 = 2.45 m/s²

Next, we shall convert 1737 Km to metres (m). This can be obtained as follow:

1 Km = 1000 m

Therefore,

1737 Km = 1737 Km × 1000 m / 1 Km

1737 Km = 1737000 m

Thus, 1737 Km is equivalent to 1737000 m

Finally, we shall determine the escape velocity of the rocket as shown below:

Gravitational acceleration of the Moon (g) = 2.45 m/s²

Radius (r) of the moon = 1737000 m

Escape velocity (v) =?

v = √2gr

v = √(2 × 2.45 × 1737000)

v = √8511300

v = 2917.4 m/s

Thus, the escape velocity is 2917.4 m/s

explanation:

considering the bohr theory, the type of connection that requires the transfer the answer conuldn't be right