Question

A lunar excursion module (LEM) weights 1500 kg, on earth where g = 9.75 mps2.

What will be its weight on the surface of the moon where gm = 1.70 mps2. On the surface

of the moon, what will be the force in kgf, and in newtons required to accelerate the

module at 10 mps2?​

Answer 1

Who knows the ans yaar

Answer 2

Given:

$W=1500kg$

$g=9.75mps^{2}$

$g_{m}=1.70\frac{m}{s^{2}}$

To find:

The surface of the moon where on the surface and newtons required to accelerate the model.

Explanation:

$F=\frac{mg}{k}$

$W=\frac{mg}{k}$

$M=\frac{wk}{g}$

$m=\frac{1500*9.8066 }{9.75} }$

$m=1508.7077kg_{m}$

$w=\frac{1508.7077*1.70}{9.8066}$

$w=261.5385kg_{f}$

$f=\frac{1508.7077*10}{9.8066}$

$f=1538.4615kg_{f}$

$f=\frac{1508.7077*10}{1}$

$f=15087.077N$

Answer:

Therefore, he surface the moon the surface  $w=261.5385kg_{f}$

The force in newtons required to accelerate the $f=15087.077N$.