Answer:v=41.23 m/s
Explanation:
Given
mass of heavy object 
distance of
from the axle 
mass of rock 
Length of rod 
distance of
from axle 
Net torque acting is




Work done by
is converted to rock kinetic Energy
thus

Where 





recoils and must be tethered or he's gone.
Answer: 40
Explanation:
= 34 + 5 * 1.2465
= 40.2325
= 40
The number of significant figures in the answer should be the same as the number with the least number of significant figures that any of the digits in the equation have.
32 has 2 significant figures so the answer has to be 2 significant figures which is 40.
Answer:
we can say here that | v² - u² | is the same for upward as for downward and change in the speed is different here so | v - u | same whenever rock travel up, down for same time and not same distances
Explanation:
given data
base = 3.60 m
speed u = 8 m/s
height = 1.70 m
to find out
check change in speed
solution
we know here formula for v that is
v² = u² - 2gh ............1 for upward speed
v² = u² + 2gh ............2 for projected speed
so here put all value and find v with h = 3.60 - 1.70 = 1.9 m
v² = 8² - 2(9.8) 1.9 = 26.76
v² = 8² + 2(9.8) 1.9 = 101.24
v = 5.173 m/s ..............3
v = 10.061 m/s ...................4
so change in speed form 3 and 4 equation
change in speed = v - u = 8 - 5.173 = 2.827 m/s .................5
change in speed = v - u = 10.061 - 8 = 2.061 m/s ..................6
so now we can say here that | v² - u² | is the same for upward as for downward and change in the speed is different here so | v - u | same whenever rock travel up, down for same time and not same distances
The answer is 15 kilometers in 20 minutes.