Asked by a friend
Objects with masses of 105 kg and 528 kg
are separated by 0.482 m. A 41.1 kg mass is
placed midway between them.
0.482 m
b b
105 kg
41.1 kg
528 kg
Find the magnitude of the net gravitational
force exerted by the two larger masses on the
41.1 kg mass. The value of the universal gravitational constant is 6.672 × 10
−11
N · m2
/kg
2
.
Answer in units of N
Objects with masses of 105 kg and 528 kg
are separated by 0.482 m. A 41.1 kg mass is
placed midway between them.
0.482 m
b b
105 kg
41.1 kg
528 kg
Find the magnitude of the net gravitational
force exerted by the two larger masses on the
41.1 kg mass. The value of the universal gravitational constant is 6.672 × 10
−11
N · m2
/kg
2
.
Answer in units of N
are separated by 0.482 m. A 41.1 kg mass is
placed midway between them.
0.482 m
b b
105 kg
41.1 kg
528 kg
Find the magnitude of the net gravitational
force exerted by the two larger masses on the
41.1 kg mass. The value of the universal gravitational constant is 6.672 × 10
−11
N · m2
/kg
2
.
Answer in units of N
Objects with masses of 105 kg and 528 kg
are separated by 0.482 m. A 41.1 kg mass is
placed midway between them.
0.482 m
b b
105 kg
41.1 kg
528 kg
Find the magnitude of the net gravitational
force exerted by the two larger masses on the
41.1 kg mass. The value of the universal gravitational constant is 6.672 × 10
−11
N · m2
/kg
2
.
Answer in units of N
Answers
Answered by
Elena
F =G•m1•m2/R²
the gravitational constant G =6.67•10⁻¹¹ N•m²/kg²,
R= 0.241 m
F13= G•m1•m3/R²
F23= G•m2•m3/R²
F23-F13=
=G •m3• (m2-m1)/R²=
=6.67•10⁻¹¹•41.1•(528-105)/0.241²=…
the gravitational constant G =6.67•10⁻¹¹ N•m²/kg²,
R= 0.241 m
F13= G•m1•m3/R²
F23= G•m2•m3/R²
F23-F13=
=G •m3• (m2-m1)/R²=
=6.67•10⁻¹¹•41.1•(528-105)/0.241²=…
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