The work done when a spring is stretched from 0 to 40cm is 4J.
What is work done?
Work done is the magnitude of force multiplied by displacement of an object. It is also the amount of energy transferred to an object when work is done on that.
The work done on the spring to stretch to 40cm is,
F = kx
where F is force, k is force constant.
k = F / x = 10 N / 20 * 10^-2 m = 50 N/m
W = 0.5 * k * (x)^2
where W = work done, k = force constant.
W = 0.5 x 50 x (40 x 10^-2)^2 = 4 J.
Therefore, the work done on the spring when it is stretched to 40cm is 4J.
To learn more about work done click on the given link brainly.com/question/25573309
#SPJ4
To be honest I’m not really sure cause I’m in 5th grade but the red planet is the sun
Answer:=14,160 kJ
Explanation: Let m1 and m2 be the initial and final amounts of mass within the tank, respectively. The steam properties are listed in the table below
Specific Internal SpecificTemp Pressure Volume Energy Enthalpy Quality Phase
C MPa m^3/kg kJ/kg kJ/kg
1 260 4.689 0.02993 2158 2298 0.7 Liquid Vapor Mixture
2 260 4.689 0.0422 2599 2797 1 Saturated Vapor
The mass initially contained in the tank is m1 = V/v1
m1 =0.85 m^3 /0.02993 m^3 /kg
= 28.4 kg
The mass finally contained in the tank is
m2 =V2/v
= 0.85 m^3 /0.0422 m^3 /kg
= 20.14 kg
The heat transfer is then
Qcv = m2u2 − m1u1 − he(m2 − m1)
Qcv = (20.14)(2599) − (28.4)(2158) − (2797)(20.14 − 28.4) = 14,160 kJ
Rw/Ra = MA
18cm/2cm= MA
MA = 9
This means that Fi is 1/9 of the force applied to the axil. The distance travelled by Rw is 9 times more than Ri is that you move 9 times more when turning the wheel using Rw.
Put more simply
Rw/Ra = Fa/Fw
- Rw = Radius of the wheel
- Ra = Radius of the axil
- Fa = Force delivered on the axil
- Fw = Force delivered by the wheel
the answer would be d. volcanic but im not 100% sure
