An airplane was moving 1200 km per minute traveling in a north-to-south direction when they were spotted by air traffic control.
2 answers:
Answer:
Its velocity was 20 km/s southward
Explanation:
As we know that airplane is moving from North to South
So here we will have
now we know
1 min = 60 s
so we will have
so the velocity of the airplane is towards South moving with speed v = 20 km/s so correct answer will be
Its velocity was 20 km/s southward
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Answer:
<em>The range is 35.35 m</em>
Explanation:
<u>Projectile Motion</u>
It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.
Being vo the initial speed of the object, θ the initial launch angle, and the acceleration of gravity, then the maximum horizontal distance traveled by the object (also called Range) is:
The projectile was launched at an angle of θ=30° with an initial speed vo=20 m/s. Calculating the range:
The range is 35.35 m
a) For the motion of car with uniform velocity we have , , where s is the displacement, u is the initial velocity, t is the time taken a is the acceleration.
In this case s = 520 m, t = 223 seconds, a =0
Substituting
The constant velocity of car a = 2.33 m/s
b) We have
s = 520 m, t = 223 seconds, u =0 m/s
Substituting
Now we have v = u+at, where v is the final velocity
Substituting
v = 0+0.0209*223 = 4.66 m/s
So final velocity of car b = 4.66 m/s
c) Acceleration = 0.0209