Significant
Explanation:
All the digits shown on a measuring device and one estimated digit are all considered to be significant.
A significant digit is a set of values that shows how precise a measurement is reported.
Measuring devices such as calculators gives their values in significant digits.
- Non- zero digits in a measurement are always significant.
- Zero's before a decimal are not significant. Those after a number in a decimal are significant.
- Zero's between digits are significant.
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Significant digits brainly.com/question/2743055
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Explanation:
Given
initial velocity(v_0)=1.72 m/s
using 
Where v=final velocity (Here v=0)
u=initial velocity(1.72 m/s)
a=acceleration 
s=distance traveled
s=0.214 m
(b)time taken to travel 0.214 m
v=u+at
t=0.249 s
(c)Speed of the block at bottom
Here u=0 as it started coming downward
v=1.72 m/s
M=meter, km=kilometer, mm=millimeter, mg=micrometer, cm=centimeter
Answer:
V= 14.2m/s
Explanation:
We know that acceleration= dv/dt
So 16m/s²=dv/ dt = v dv/ds
So this wil be
Integral of 16m/s² ds [at 2,2]= integral of v dv at[ 0, v]
So 16[s (3/2)/3/2] at ( s,3) = v²/2
At s= 6m
So v² = 64/3( 6^1.5-3^1.5)
= 14.2m/s
Answer:
42.05 m/s
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 0 m/s
Height (h) = 239 m
Acceleration due to gravity (g) = 3.7 m/s²
Final velocity (v) =?
The velocity with which the camera hits the ground can be obtained as follow:
v² = u² + 2gh
v² = 0² + 2 × 3.7 × 239
v² = 0 + 1768.6
v² = 1768.6
Take the square root of both side
v = √(1768.6)
v = 42.05 m/s
Therefore, the velocity with which the camera hits the ground is 42.05 m/s
