Answer:
3 seconds
Explanation:
Applying,
Applying,
v = u±gt................ Equation 1
Where v = final velocity, u = initial velocity, t = time, g = acceleration due to gravity.
From the question,
Given: v = 0 m/s ( at the maximum height), u = 30 m/s
Constant: g = -10 m/s
Substitute these values into equation 1
0 = 30-10t
10t = 30
t = 30/10
t = 3 seconds
a) we can answer the first part of this by recognizing the player rises 0.76m, reaches the apex of motion, and then falls back to the ground we can ask how
long it takes to fall 0.13 m from rest: dist = 1/2 gt^2 or t=sqrt[2d/g] t=0.175
s this is the time to fall from the top; it would take the same time to travel
upward the final 0.13 m, so the total time spent in the upper 0.15 m is 2x0.175
= 0.35s
b) there are a couple of ways of finding thetime it takes to travel the bottom 0.13m first way: we can use d=1/2gt^2 twice
to solve this problem the time it takes to fall the final 0.13 m is: time it
takes to fall 0.76 m - time it takes to fall 0.63 m t = sqrt[2d/g] = 0.399 s to
fall 0.76 m, and this equation yields it takes 0.359 s to fall 0.63 m, so it
takes 0.04 s to fall the final 0.13 m. The total time spent in the lower 0.13 m
is then twice this, or 0.08s
Answer:
The answer is given in the attachment
Explanation:
Answer:
T₂ = 20.06 ° C
Explanation:
Given
P = 90 kg, T₁ = 20 ° C, h = 30 m, c = 1.82 kJ / Kg * ° C
Using the formula to determine the final temperature of the water
T₂ = T₁ * P * h / Eₐ * c
The work done of the person to the water
Eₐ = 1000 kg / m³ * 5 m³ * 9.8 m / s²
Eₐ = 49000 N
T₂ = 20 ° C +[ (90 kg * 30m) / (49000 N * 1.82) ]
T₂ = 20.06 ° C
