Answer:
1966 J
Explanation:
The work done by the hose on the balloon is equal to the elastic potential energy stored in it:
where
k = 130 N/m is the spring constant
x = 5.50 m is the stretching of the hose before it is being released
If we substitute these numbers into the equation, we find:
So, the work done is 1966 J.
Answer:
Carbon 12
Explanation:
I don't 100% know what to put here, but...
When you remove the nucleus from an oxygen atom, almost everything of the base oxygen is essentially stripped away. Since almost everything is made of carbon, and Carbon 12 is one of the most common forms of Carbon, Carbon 12 would be what is left.
The two evidence are the car's engine gets hot as it burns gasoline are the car remains full of gasoline if its engine stays off.
<h3>Which two pieces of evidence support the conclusion that the total amount of energy in a close system remains the same?</h3>
The car's engine gets hot as it burns gasoline and the car remains full of gasoline if its engine stays off are the two pieces of evidence support the conclusion that the total amount of energy in a close system remains the same because when the gasoline burns, the engine gets hotter and when the car remains full of gasoline if its engine stays off which means no energy leaves the system.
So we can conclude that The two evidence are the car's engine gets hot as it burns gasoline are the car remains full of gasoline if its engine stays off.
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Given:
k = 100 lb/ft, m = 1 lb / (32.2 ft/s) = 0.03106 slugs
Solution:
F = -kx
mx" = -kx
x" + (k/m)x = 0
characteristic equation:
r^2 + k/m = 0
r = i*sqrt(k/m)
x = Asin(sqrt(k/m)t) + Bcos(sqrt(k/m)t)
ω = sqrt(k/m)
2π/T = sqrt(k/m)
T = 2π*sqrt(m/k)
T = 2π*sqrt(0.03106 slugs / 100 lb/ft)
T = 0.1107 s (period)
x(0) = 1/12 ft = 0.08333 ft
x'(0) = 0
1/12 = Asin(0) + Bcos(0)
B = 1/12 = 0.08333 ft
x' = Aω*cos(ωt) - Bω*sin(ωt)
0 = Aω*cos(0) - (1/12)ω*sin(0)
0 = Aω
A = 0
So B would be the amplitude. Therefore, the equation of motion would be x
= 0.08333*cos[(2π/0.1107)t]
Answer:
m = 35 g
Explanation:
The specific heat of a material can be calculated by the following formula:
where,
C = Specific Heat of Wax = 220 J/g
Q = Amount of Heat Supplied by the Heater = 7700 J
m = mass of wax melted = ?
Therefore,
<u>m = 35 g </u>