A 2.00-m rod of negligible mass connects two very small objects at its ends. The mass of one object is 1.00 kg and the mass of t
1 answer:
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Answer: hello options related to your question is missing attached below is the missing part of your question
answer: No charge of the length of the bonds expected because the rod did not touch the charge source ( option A )
Explanation:
When the Charge is first, Furthest away and second and closest to the source charge. <em>The spring like bonds can be said to have No charge of the length of the bonds expected because the rod did not touch the charge source </em><em>when Furthest away the bond with charge will be less effective </em>
Explanation:
PRIMERO HACES EL RECUENTO DEL TIEMPO Y LO CONVIERTES EN
SEGUNDOS Y ENTONCES
<em>t</em> = 227 s = 227 S - 38 s = 189 s
= 38 s
LUEGO USANDO LA ECUACIÓN DE GALILEO GALILEI SSUPONIENDO
QUE EL MOVIL VIAJA A VELOCIDAD CONSTANTE
<em>v</em> = 3.50 m/189 s = 0.0185 m/s
PARA LA DISTANCIA NTRE B Y C
= 0.0185 m/S( 38 s) = 0.703 m
LA HORA EN QUE EL MOVIL PASA POR A ES
11:43:15 - 38 s - 189 s = 11:39:29
Answer: Decreasing the distance of the space shuttle from Earth .
Explanation:
According to expression of gravitational force:
G = gravitational constant
= masses of two objects
r = Distance between the two objects.
F = Gravitational force
From the above expression we can say that gravitational force is inversely proportional to squared of the distance between the two masses.
So, in order to increase the gravitational force on space shuttle distance between the space space shuttle must be decreased.
Hence, the correct answer 'decreasing the distance of the space shuttle from Earth '.