Answer:
x' = 1.01 m
Explanation:
given,
mass suspended on the spring, m = 0.40 Kg
stretches to distance, x = 10 cm = 0. 1 m
now,
we know
m g = k x
where k is spring constant
0.4 x 9.8 = k x 0.1
k = 39.2 N/m
now, when second mass is attached to the spring work is equal to 20 J
work done by the spring is equal to
x'² = 1.0204
x' = 1.01 m
hence, the spring is stretched to 1.01 m from the second mass.
in this since your volume remains at a constant you'll need to use Gay-Lussacs law, p1/t1=p2/t2.
your temp should be converted in kelvin
variables:
p1=3.0×10^6 n/m^2
t1= 270k
just add 273 to your celcius
p2= ? your solving for this
t2= 315k
then you set up the equation
(3.0×10^6)/270= (x)(315)
you then cross multiply
(3.0×10^6)315=270x
distribute the 315 to the pressure.
9.45×10^8=270x then you divide 270 o both sides to get
answer
3.5×10^6 n/m^2
Answer: 10 m/s
We're told the speed is constant, so it's not changing throughout the time period given to us. So throughout the entire interval, the speed is 10 m/s.
1- interaction between 2 objects
2- action- reaction force pairs
<span>Density can be calculated and found by dividing the sample's mass by its volume. D=m/v</span>
