No.
The acceleration of gravity on or near Earth's surface is 9.8 m/s² ,
not 20 m/s² .
If it were 20 m/s², then you would weigh almost exactly double
what you really weigh now.
Given that:
k = 500 n/m,
work (W) = 704 J
spring extension (x) = ?
we know that,
Work = (1/2) k x²
704 = (1/2) × 500 × x²
x = 1.67 m
A spring stretched for 1.67 m distance.
Answer:
4.17 m/s²
Explanation:
We are told the reaction time is 0.2 s. Now, during this reaction time the car is going to travel an additional distance of
: x = u × t = 40 × 0.2 = 8 m
where u is the initial velocity of the car which is 40.0 m/s.
We are told that he had 200 m to stop before applying brakes. Thus, after applying brakes, he now has a distance to cover of; s = 200 - 8 = 192 m
Since vehicle is coming to rest acceleration would be negative, thus using Newton's equation of motion, we have;
v
² = u² - 2as
v = 0 m/s since it's coming to rest
u = 40 m/s
s = 192 m
Thus;
0² = 40² - 2(a)(192)
0² = 1600 - 384a
a = 1600/384
a = 4.17 m/s²
I Think Its True My Dude Or Dudette
.
Hope this helps
.
Zane
Answer:
Explanation:
Using the principle of moment, assuming the rod is uniform rod of mass 1 kg
the center of mass of the rod will be at 1 m
assuming the system is in equilibrium,
clockwise moment = anticlockwise moment
let the distance of the man shoulder be x from the center of gravity and also is the pivot point
total mass of bucket + mass of honey = 2kg + 3 kg = 5 kg for rear bucket and
2kg + 5 kg = 7 kg for front bucket
( 5kg × ( 1+x)) + ( 1 kg × x) = 7 kg × ( 1 - x)
5 + 5 x + x = 7 - 7x
5 + 6x = 7 - 7x
6x + 7x = 7 - 5
13x = 2
x = 2 / 13 = 0.154 m
the honeybucket man's shoulder is 0.154 m from the center of the pole ( forward ).
