Answer:
Plane will 741.6959 m apart after 1.7 hour
Explanation:
We have given time = 1.7 hr
So if we draw the vectors of a 2d graph we see that the difference in angles is = 
Speed of first plane = 730 m/h
So distance traveled by first plane = 730×1.7 = 1241 m
Speed of second plane = 590 m/hr
So distance traveled by second plane = 590×1.7 = 1003 m
We represent these distances as two sides of the triangle, and the distance between the planes as the side opposing the angle 58.6.
Using the law of cosine, 
representing the distance between the planes, we see that:
r = 741.6959 m
Answer:
The puck moves a vertical height of 2.6 cm before stopping
Explanation:
As the puck is accelerated by the spring, the kinetic energy of the puck equals the elastic potential energy of the spring.
So, 1/2mv² = 1/2kx² where m = mass of puck = 39.2 g = 0.0392 g, v = velocity of puck, k = spring constant = 59 N/m and x = compression of spring = 1.3 cm = 0.013 cm.
Now, since the puck has an initial velocity, v before it slides up the inclined surface, its loss in kinetic energy equals its gain in potential energy before it stops. So
1/2mv² = mgh where h = vertical height puck moves and g = acceleration due to gravity = 9.8 m/s².
Substituting the kinetic energy of the puck for the potential energy of the spring, we have
1/2kx² = mgh
h = kx²/2mg
= 59 N/m × (0.013 m)²/(0.0392 kg × 9.8 m/s²)
= 0.009971 Nm/0.38416 N
= 0.0259 m
= 2.59 cm
≅ 2.6 cm
So the puck moves a vertical height of 2.6 cm before stopping
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