An ant carries a morsel of food 4.26 meters along a straight path to his nest. He then turns around and follows the path back to
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Answer:
<em>The acceleration of the racecar is</em>
Explanation:
<u>Uniformly Accelerated Motion</u>
It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.
Following the definition above, the acceleration is defined as:
Where a is the constant acceleration, vo the initial speed, vf the final speed, and t the time.
The racecar goes from vo=18.5 m/s to vf=46.1 m/s in t=2.47 seconds, thus the acceleration is:
The acceleration of the racecar is
Answer:
v = 15.8 m/s
Explanation:
Let's analyze the situation a little, we have a compressed spring so it has an elastic energy that will become part kinetic energy and a potential part for the man to get out of the barrel, in addition there is a friction force that they perform work against the movement. So the variation of mechanical energy is equal to the work of the fictional force
= ΔEm =
-Em₀
Let's write the mechanical energy at each point
Initial
Em₀ = Ke = ½ k x²
Final
= K + U = ½ m v² + mg y
Let's use Hooke's law to find compression
F = - k x
x = -F / k
x = 4400/1100
x = - 4 m
Let's write the energy equation
fr d = ½ m v² + mgy - ½ k x²
Let's clear the speed
v² = (fr d + ½ kx² - mg y) 2 / m
v² = (40 4.00 + ½ 1100 4² - 60.0 9.8 2.50) 2/60.0
v² = (160 + 8800 - 1470) / 30
v = √ (229.66)
v = 15.8 m/s
Answer:
(A) The maximum height of the ball is 40.57 m
(B) Time spent by the ball on air is 5.76 s
(C) at 33.23 m the speed will be 12 m/s
Explanation:
Given;
initial velocity of the ball, u = 28.2 m/s
(A) The maximum height
At maximum height, the final velocity, v = 0
v² = u² -2gh
u² = 2gh
(B) Time spent by the ball on air
Time of flight = Time to reach maximum height + time to hit ground.
Time to reach maximum height = time to hit ground.
Time to reach maximum height is given by;
v = u - gt
u = gt
Time of flight, T = 2t
(C) the position of the ball at 12 m/s
As the ball moves upwards, the speed drops, then the height of the ball when the speed drops to 12m/s will be calculated by applying the equation below.
v² = u² - 2gh
12² = 28.2² - 2(9.8)h
12² - 28.2² = - 2(9.8)h
-651.24 = -19.6h
h = 651.24 / 19.6
h = 33.23 m
Thus, at 33.23 m the speed will be 12 m/s