T=5s
S=5m
U=0s/m
Acceleration=m/s^2
S=ut+(1/2)(a)(5^2)
A=(25/2)=5
A=5/(25/2)
A=10/25
A=2/5
A=0.4
A=0.4m/s^2.
Answer:
Her angular velocity when tucked is greater than when straight by a factor of 0.23
Explanation:
Moment of inertia (I) = mr^2 = mv^2/w^2
m is mass of the diver
v is diver's linear velocity
w is her angular velocity
When straight, I = 14 kg.m^2
mv^2/w^2 = 14
w^2 = mv^2/14
w = sqrt(mv^2/14) = 0.27sqrt(mv^2)
When tucked, I = 4 kg.m^2
w^2 = mv^2/4
w = sqrt(mv^2/4) = 0.5sqrt(mv^2)
Her angular velocity when tucked is greater than when straight by 0.23 (0.5 - 0.27 = 0.23)
The distance of the airplane from the radar will be 1.08×10⁶ m. Distance can refer to a physical length.
<h3>What is distance?</h3>
Distance is a numerical representation of the distance between two objects or locations.
The distance of the airplane from the radar is found by the formula;
Distance = velocity × time
Distance = 3.0 x 10^8 m/s× 0.0036 s,
Distance = 1.08×10⁶ m.
Hence, the distance of the airplane from the radar will be 1.08×10⁶ m.
To learn more about the distance, refer to the link;
brainly.com/question/26711747
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Answer:
a. no lift
Explanation:
To determine if the forces acting upon an object are balanced or unbalanced, an analysis must first be conducted to determine what forces are acting upon the object and in what direction. If two individual forces are of equal magnitude and opposite direction, then the forces are said to be balanced.
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