Water evaporates at 100⁰C
So change in temperature = 100-20 = 80⁰C
Amount of water to be evaporated = 1 liter = 1L*1kg/liter = 1 kg
Specific heat of water is 1 calorie/gram ⁰C = 4.186 joule/gram =4186 J/kg
So heat required E = mcΔT = 1 * 4186 *80= 334880 J =334.88 kJ
So amount of heat require to evaporate water = 334.88 kJ
Answer: 4.98 m/s
Explanation:
You solve these kinetic energy, potential energy problems by using the fact P.E.+ K.E. = a constant as long as friction is ignored.
PEi = 0 in this case
KEi = ½mVi² = PEf+KEf = mghf + ½mVf²
½1210*8.31² = 1210*9.8*2.26 + ½1210*Vf²
½1210*Vf² = ½1210*8.31² - 1210*9.8*2.26
Vf² = 8.31² - 2*9.8*2.26 = 4.98² so Vf = 4.98m/s
Answer:
Explanation:
Atmospheric pressure = 7 x 10⁴ Pa
force on a disk-shaped region 2.00 m in radius at the surface of the ocean due to atmosphere = pressure x area
= 7 x 10⁴ x 3.14 x 2 x 2
= 87.92 x 10⁴ N
b )
weight, on this exoplanet, of a 10.0 m deep cylindrical column of methane with radius 2.00 m
Pressure x area
height x density x acceleration of gravity x π r²
= 10 x 415 x 6.2 x 3.14 x 2 x 2
=323168.8 N
c ) Pressure at a depth of 10 m
atmospheric pressure + pressure due to liquid column
= 7 x 10⁴ + 10 x 415 x 6.2 ( hρg)
= 7 x 10⁴ + 10 x 415 x 6.2
(7 + 2.57 )x 10⁴ Pa
9.57 x 10⁴ Pa
Force = Work/distance
Force = 150/10
= 15 Newtons
Force = 15 Newtons
Therefore, 15 newtons of force is applied to the body when 150 joules of work
is done in displacing the body through a distance of 10m in the direction of the force.
Answer:
the work required to turn the crank at the given revolutions is 8,483.4 J
Explanation:
Given;
torque required to turn the crank, T = 4.50 N.m
number of revolutions, = 300 turns
The work required to turn the crank is given as;
W = 2πT
W = 2 x 3.142 x 4.5
W = 28.278 J
1 revolution = 28.278 J
300 revlotions = ?
= 300 x 28.278 J
= 8,483.4 J
Therefore, the work required to turn the crank at the given revolutions is 8,483.4 J
