Answer:
<em>1,378.9ms²</em>
Explanation:
Given the following
Distance S = 70.6m
Time t = 0.32secs
Initial velocity = 0m/s
Required
Acceleration
Using the equation of motion
S = ut+1/2at²
Substitute
70.6 = 0+1/2a(0.32)²
70.6 = 0.0512a
a = 70.6/0.0512
a = 1,378.9
<em>Hence the acceleration is 1,378.9ms²</em>
Decreases. Air resistance will slow a falling object to its terminal velocity, placing a limit on its acceleration.
A mid ocean ridge is a under water mountain range, knowing this what would you say it is most similar to on land? hope this helped!
Answer:
ΔU = 2 mg h
Explanation:
In a spring mass system the potential energy is U = m g h
where h is measured from the equilibrium point of the spring
the potential energy at the highest point is
U₁ = m g h
the potential energy at the lowest point is
U₂ = m g (-h)
instead in this energy it is
ΔU = 2 mg h
In this two points the kinetic energy is zero, but there is elastic potential energy that has the same value in the two points, so its change is zero
Answer:
5.5 km
Explanation:
First, we convert the distance from km/h to m/s
910 * 1000/3600
= 252.78 m/s
Now, we use the formula v²/r = gtanθ to get our needed radius
making r the subject of the formula, we have
r = v²/gtanθ, where
r = radius of curvature needed
g = acceleration due to gravity
θ = angle of banking
r = 252.78² / (9.8 * tan 50)
r = 63897.73 / (9.8 * 1.19)
r = 63897.73 / 11.662
r = 5479 m or 5.5 km
Thus, we conclude that the minimum curvature radius needed for the turn is 5.5 km
