Answer:
As Per Provided Information
Moving body has 2m/s² acceleration
Time taken by body is 4 second
We are asked to find the 'change in velocity' ( ∆V) by the body.
<u>Formula Used here</u>
<u>Substituting </u><u>the </u><u>given </u><u>value</u>
<u></u>
<u>Therefore</u><u>,</u>
- <u>Change </u><u>in </u><u>velocity </u><u>is </u><u>8</u><u> </u><u>m/</u><u>s</u>
In order to catch up with the cockroach, the time it took for it to reach just before under the counter must be the same time for me to travel its covered distance. Thus,
time of cockroach = time of my motion
Time of cockroach = Distance/Constant Speed = 1.30 m/1.50 m/s = 0.867 seconds
We use this for the equation below:
x = v₀t + 0.5at²
0.75 m = (0.7 m/s)(0.867 s) + 0.5(a)(0.867 s)²
Solving for a,
a = 0.381 m/s²
<span>Downhill walking or walking down stairs causes D. Eccentric contractions of the leg muscles.</span>
Answer:
Explanation:
As we know that the block of wood is suspended by spring
Now at equilibrium position we have net force balanced on it
so we have
so we will have
now the time period of the spring block system for one complete oscillation is given as
now plug in all values in it
Now total time to complete 3 cycles is given as
Answer:
It would take the object 5.4 s to reach the ground.
Explanation:
Hi there!
The equation of the height of a free-falling object at any given time, neglecting air resistance, is the following:
h = h0 + v0 · t + 1/2 · g · t²
Where:
h = height of the object at time t.
h0 = initial height.
v0 = initial velocity.
g = acceleration due to gravity (-32.2 ft/s² considering the upward direction as positive).
t = time
Let´s supose that the object is dropped and not thrown so that v0 = 0. Then:
h = h0 + 1/2 · g · t²
We have to find the time at which h = 0:
0 = 470 ft - 1/2 · 32.2 ft/s² · t²
Solving for t:
-470 ft = -16.1 ft/s² · t²
-470 ft / -16.1 ft/s² = t²
t = 5.4 s