vf = vi + at
vf – vi = at<span>
<span>vi= 0, vf=26 and afor nil = 9.8m/s2</span></span>
26 = 9.8t
t =<span> 26 / 9.8 = 2.65 s
Now we know the total time, so we can calculate the time 1
second before it hit the ground.
<span>= 2.65 -1 = 1.65s
<span>Now again using the same equation, vf = vi+at, we can find vf
vi = 0, a = 9.8 t=1.65</span></span></span>
vf = 0 + 9.8(1.65) =
16.17 m/s<span>
</span><span>So,
the nail is traveling with the speed of 16.17m/s 1 second before it hits the
ground.</span>
The unit used<span> to measure weight in the metric system is the gram.</span>
Answer:
The time it took the bobsled to come to rest is 10 s.
Explanation:
Given;
initial velocity of the bobsled, u = 50 m/s
deceleration of the bobsled, a = - 5 m/s²
distance traveled, s = 250 m
Apply the following kinematic equation to determine the time of motion of the bobsled;
s = ut + ¹/₂at²
250 = 50t + ¹/₂(-5)t²
250 = 50t - ⁵/₂t²
500 = 100t - 5t²
100 = 20t -t²
t² - 20t + 100 = 0
t² -10t - 10t + 100 = 0
t (t - 10) - 10(t - 10) = 0
(t - 10)(t - 10) = 0
t = 10 s
Therefore, the time it took the bobsled to come to rest is 10 s.
Answer:
8.829 m/s²
Explanation:
M = Mass of Earth
m = Mass of Exoplanet
= Acceleration due to gravity on Earth = 9.81 m/s²
g = Acceleration due to gravity on Exoplanet



Dividing the equations we get

Acceleration due to gravity on the surface of the Exoplanet is 8.829 m/s²
Answer:
A.) 4.81 seconds
B.) 44.6 m/s
Explanation:
He begins his dive by jumping up with a velocity of 5 (m/s).
Let us first calculate the maximum height reached by using third equation of motion
V^2 = U^2 - 2gH
At maximum height, V = 0
0 = 5^5 - 2 × 9.8H
19.6H = 25
H = 25 /19.6
H = 1.28 m
The time taken for the diver to reach the water from the maximum height can be calculated by using second equation of motion.
Where height h = 1.28 + 100 = 101.28 m
h = Ut + 1/2gt^2
As the diver drop from maximum height, U = 0
101.28 = 1/2 × 9.8 × t^2
4.9t^2 = 101.28
t^2 = 101.28/4.9
t^2 = 20.669
t = sqrt ( 20.669)
t = 4.55s
As the diver jumped up, the time taken to reach the maximum height will be
Time = 1.28 / 5 = 0.256
The time taken for him to hit the water below will be 0.256 + 4.55 = 4.81 seconds
B.) Velocity right before he hits the water will be
V^2 = U^2 + 2gH
But U = 0
V^2 = 2 × 9.8 × 101.28
V^2 = 1985.09
V = 44.6 m/s