If the probes are identical, then the one that feels a larger gravitational
force is orbiting closer to Jupiter than the other one is.
If they're not identical, then the one with greater mass will feel more
gravitational force than the one with less mass, even if they're both
the same distance from Jupiter. (We know this from the experimental
observation that fatter people weigh more, even on Earth.)
Answer:
B. ) 0.34 m
I definitely guessed and got the right answer so :))
The value of g at sea level is 9.81 ms^-2.
The boy's mass is constant wherever he is in the universe but his weight will depend on the strength gravity where he is.
By proportion its value on the mountain peak is (360 /400) * 9.81
= 0.9 * 9.81 = 8.83 ms^-2 to nearest hundredth, (answer).
Answer:
false
Explanation:
since it has seven valence electrons it means it's a non metal and non metals always gain electrons.
Answer:
The resultant force would (still) be zero.
Explanation:
Before the 600-N force is removed, the crate is not moving (relative to the surface.) Its velocity would be zero. Since its velocity isn't changing, its acceleration would also be zero.
In effect, the 600-N force to the left and 200-N force to the right combines and acts like a 400-N force to the left.
By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.
When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.
However, recall that before the 600-N force is removed, the friction on the crate is 400 N. The normal force on the crate (which is in the vertical direction) did not change. As a result, one can hence be assured that the maximum friction would be at least 400 N. That's sufficient for balancing the 200-N force to the right. Hence, the resultant force on the crate would still be zero, and the crate won't move.
