The right<span> at +20.0 </span>cm/s makes<span> an </span>elastic head<span>-on </span>collision<span> with a 10.0 </span>g object<span> that </span>makes<span> an</span>elastic head<span>-on </span>collision<span> with a 10.0 </span>g object<span> that is </span>initially<span> at </span>rest<span>.(b) Find the fraction of the </span>initial<span>kinetic energy transferred to the 10.0 </span>g object<span>.of small </span>mass<span> before and </span>after collision; V=velocity<span> of big </span>mass after collision<span>.</span>
Answer:
gravitational potential energy
The amount of energy is ALWAYS conserved based on the law of conservation of energy. Therefore, the amount of energy will remain the same.
Answer:
0.05 m
Explanation:
From the question given above, the following data were obtained:
Mass of first object (M1) = 9900 kg
Gravitational force (F) = 12 N
Mass of second object (M2) = 52000 kg
Distance apart (r) =?
Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
Thus, we can obtain the distance between the two objects as shown below:
F = GM1M2/r²
12 = 6.67×10¯¹¹ × 9900 × 52000 /r²
Cross multiply
12 × r² = 6.67×10¯¹¹ × 9900 × 52000
Divide both side by 12
r² = (6.67×10¯¹¹ × 9900 × 52000)/12
Take the square root of both side
r = √[(6.67×10¯¹¹ × 9900 × 52000)/12]
r = 0.05 m
Therefore, the distance between the two objects is 0.05 m
