Suppose car A is moving with a velocity Va, and car b with a velocity Vb,
According the principle of conservation of momentum:
Va x Ma + Vb x Mb = (Ma + Mb) V
V = (Va x Ma + Vb x Mb)/(Ma +Mb)
V = speed of cars after coupling
V = (Va x 20 mg + Vb x 15 mg)/(20 mg + 15 mg)
Put in the values of Va and Vb, and get the V
Answer:
Total displacement will be 47 meter
Total distance will be 83 meters
Explanation:
We have given that first the student go eastward towards bus stop 20 meters
But he realizes that she dropped his physics notebook and so h=she turns back along the same way up to 18 meters
So displacement = 20-18 = 2 meters
And he travel 45 meters in east along the bus stop so total displacement = 45+2 = 47 meters
Total distance traveled by the student = 20+18+45 = 83 meters
The correct answer should be c.The kinetic energy of the water molecules decreases.
If the temperature drops that means that the molecules are coming together. If the temperature rises then it means that the molecules are spreading. If the kinetic energy falls down that means that they are slower which means that they are cooler.
Here is my step-by-step-work. Let me know if you have any questions! :)
Answer:
189 m/s
Explanation:
The pilot will experience weightlessness when the centrifugal force, F equals his weight, W.
So, F = W
mv²/r = mg
v² = gr
v = √gr where v = velocity, g = acceleration due to gravity = 9.8 m/s² and r = radius of loop = 3.63 × 10³ m
So, v = √gr
v = √(9.8 m/s² × 3.63 × 10³ m)
v = √(35.574 × 10³ m²/s²)
v = √(3.5574 × 10⁴ m²/s²)
v = 1.89 × 10² m/s
v = 189 m/s
