C. Is the correct answer for this question
Answer:
v= 300 m/s
Explanation:
Given that
altitude ,h= 4500 m
The mass ,m = 3 kg
Lets take acceleration due to gravity , g= 10 m/s²
The speed before impact at sea level = v
Initial speed ,u = 0 m/s
We know that
v²=u²+2 g h
v=final speed
u=initial speed
h=height
Now by putting the values in the above equation
v² = 0²+ 2 x 10 x 4500
v²=90000
v= 300 m/s
Therefore the speed at sea level will be 300 m/s.
Answer:
Car 1 and Car 2 have the same momentum!
Explanation:
Using the formula of momentum (P=m*v), we get for each car:
Car 1: 5kg*2.2m/s = 11kg*m/s
Car 2: 5.5kg*2m/s = 11kg*m/s
Car 3: 6kg*1.35m/s = 8.1kg*m/s
Car 4: 6.5kg*1.9m/s = 12.35kg*m/s
Car 5: 7kg*1.25m/s = 8.75kg*m/s
Answer:
384.6 m
Explanation:
The length of the runway airport should be long enough to accommodate the aircraft during its acceleration from rest to 161 km/h at rate of 2.6 m/s. We can use the following equation of motion to solve for this:

where v0 = 0 m/s is the initial velocity of the airplane when it start accelerating, v = 161 km/h = 161*1000*(1/60)(1/60) = 44.72 m/s is the airborn speed, a = 2.6 m/s2 is the acceleration, and
is the distance of the runway, which we care looking for


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