2. Choose one food chain in the web. Name all the organisms in that chain. Start with the producer and end with the top-level co
1 answer:
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Answer:
t = √2y/g
Explanation:
This is a projectile launch exercise
a) The vertical velocity in the initial instants ( = 0) zero, so let's use the equation
y = t -1/2 g t²
y= - ½ g t²
t = √2y/g
b) Let's use this time and the horizontal displacement equation, because the constant horizontal velocity
x = vox t
x = v₀ₓ √2y/g
c) Speeds before touching the ground
vₓ = vox = constant
=
- gt
= 0 - g √2y/g
= - √2gy
tan θ = Vy / vx
θ = tan⁻¹ (vy / vx)
θ = tan⁻¹ (√2gy / vox)
d) The projectile is higher than the cliff because it is a horizontal launch
Answer:
The maximum safe speed of the car is 30.82 m/s.
Explanation:
It is given that,
The formula that models the maximum safe speed, v, in miles per hour, at which a car can travel on a curved road with radius of curvature r r, is in feet is given by :
.........(1)
A highway crew measures the radius of curvature at an exit ramp on a highway as 380 feet, r = 380 feet
Put the value of r in equation (1) as :
v = 30.82 m/s
So, the maximum safe speed of the car is 30.82 m/s. Hence, this is the required solution.