Answer:
3.6ft
Explanation:
Using= 2*π*sqrt(L/32)
To solve for L, first move 2*n over:
T/(2*π) = sqrt(L/32)
Next,eliminate the square root by squaring both sides
(T/(2*π))2 = L/32
or
T2/(4π2) = L/32
Lastly, multiply both sides by 32 to yield:
32T2/(4π2) = L
and simplify:
8T²/π²= L
Hence, L(T) = 8T²/π²
But T = 2.1
Pi= 3.14
8(2.1)²/3.14²
35.28/9.85
= 3.6feet
I don’t think we can answer this question with the information given. ANY ball thrown with ANY initial velocity v will be observed at a height h twice and with a time interval Δt.
Answer:
<h2> 27m/s</h2>
Explanation:
Given data
initital velocity u=15m/s
deceleration a=3m/s^2
time t= 4 seconds
final velocity v= ?
Applying the expression
v=u+at------1
substituting our data into the expression we have
v=15+3*4
v=15+12
v=27m/s
The velocity after 4 seconds is 27m/s
<span>The weightlifter does no work. Although he has exerted force, work is the product of force over distance. Since he has not moved the wall he has done no work.</span>
Answer:
the pressure per square inch is greater from the smaller feet.
Explanation:
different weight distribution
