Answer:
a) 3.78 m/s
b) 3.78 m/s
Explanation:
a )From the equations of kinematics we know that
Vf - Vi = at
since Initial speed Vi = 0
acceleration = 4.2 m/s2
so we have
Vf = a t
= (4.2) (0.9)
= 3.78 m/s
velocity at t = 0.9 s. m/s is 3.78 m/s
b) If the sprinter maintains constant velocity then acceleration becomes zero.
So velocity is 3.78 m/s
Centripetal acceleration ac is the acceleration experienced while in uniform circular motion. It always points toward the center of rotation. It is perpendicular to the linear velocity (tangential speed) v and has the magnitude:
Solving for v:
Answer:
S=d/t.
distance is= speed multiplied by the time
<em>D</em><em>=</em><em> </em><em>1</em><em>1</em><em>5</em> multiplied by 20
the answer is 2300km
From Newton's Three Laws of Motion, derived formulas are already conveniently presented for a rectilinear motion at constant acceleration. One of its equations is
y = (Vf² - Vi²)/2a, where
y is the vertical height travelled by the object
Vf is the final velocity
Vi is the initial velocity
a is the acceleration
Now, when a man jumps, the only force acting on him is gravity pulling him down. When he reaches his maximum height, eventually his velocity will reach zero. So, Vf = 0. Suppose all parameters with subscript 1 refers to man jumping on Earth and those with subscript 2 refers to the man jumping on moon. Since initial velocity and angle is said to be the same, when we find the ratio of x₂/x₁, the terms (Vf²-Vi²) cancels out leaving us with
x₂/x₁ = a₂/a₁
It is common knowledge that gravity on Earth is 9.81 m/s². According to literature, the gravity on the moon is 1.62 m/s². Thus,
x₂/x₁ = a₁/a₂ = 9.81/1.62 = 6
x₂ = 6x₁
Therefore, the man jumping on the moon can reach 6 times higher than in Earth.
