The best answer is D. field lines should always be crossing each other.
Answer:
The weight of the landing craft in the vicinity of Callisto's surface is 3480 N.
Explanation:
The engine of the craft provides an upward thrust of 
so that the space craft descends at a constant speed.
This implies that the net force on the space craft is zero.
The upward thrust will be equal to the downward gravitational pull by Callisto.
So the weight of the craft near the vicinity will be 3480 N.
Buhrs atomic model differed from ruthofords because it explained that electrons exist in specified energy levels surrounding the nucleus. This means that, Ruthoford believed that electrons can't do very much. However, Buhrs' model showed that electrons are much more powerful than anyone else believes they can be.
Answer:
100,048
Explanation:
K.E = 1/2 m (v)^2
K.E = 1^/2 * 74 * (52)^2
K.E = 100,048J =100.048kJ
B4 the tackle:
<span>The linebacker's momentum = 115 x 8.5 = 977.5 kg m/s north </span>
<span>and the halfback's momentum = 89 x 6.7 = 596.3 kg m/s east </span>
<span>After the tackle they move together with a momentum equal to the vector sum of their separate momentums b4 the tackle </span>
<span>The vector triangle is right angled: </span>
<span>magnitude of final momentum = √(977.5² + 596.3²) = 1145.034 kg m/s </span>
<span>so (115 + 89)v(f) = 1145.034 ←←[b/c p = mv] </span>
<span>v(f) = 5.6 m/s (to 2 sig figs) </span>
<span>direction of v(f) is the same as the direction of the final momentum </span>
<span>so direction of v(f) = arctan (596.3 / 977.5) = N 31° E (to 2 sig figs) </span>
<span>so the velocity of the two players after the tackle is 5.6 m/s in the direction N 31° E </span>
<span>btw ... The direction can be given heaps of different ways ... N 31° E is probably the easiest way to express it when using the vector triangle to find it</span>
