Answer:
The latent heat of fusion of water is 334.88 Joules per gram of water.
Explanation:
Let the latent heat of ice be 'x' J/g
1) Thus heat absorbed by 100 gram of ice to get converted into water equals

2) heat energy required to raise the temperature of water from 0 to 25 degree Celsius equals

Thus total energy needed equals 
3) Heat energy released by the decrease in the temperature of water from 25 to 11 degree Celsius is

Now by conservation of energy we have

Answer:
O bike tires on the road as you ride
Explanation:
is the rolling friction
Answer:
The kilogram (kg) is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015 ×10−34 when expressed in the unit J s, which is equal to kg m2 s−1, where the meter and the second are defined in terms of c and ∆νCs.
The springs stored energy is transferred to the cube as kinetic energy and then by the slop the KE is converted to height energy.
<span>0.5 . k . x^2 = 0.5 . m . v^2 = m . g . ∆h </span>
<span>0.5 . 50 . (0.1^2) = 0.05 . 9.8 . ∆h </span>
<span>∆h = 0.51 m = 51 cm </span>
<span>This is the height gained </span>
<span>Distance along the slope = ∆h / sin 60 = 0.589 = 59 cm </span>
<span>In the second case, the stored spring energy is converted into height energy AND frictional heat energy. </span>
<span>The height energy is m . g . d sin 60 where d is the distance the cube moves along the slope. </span>
<span>The Frictional energy converted is F . d </span>
<span>F ( the frictional force ) = µ . N </span>
<span>N ( the reaction to the component of the gravity force perpendicular to the surface of the slope ) = m . g . cos60 </span>
<span>Total energy converted </span>
<span>0.5 . k . x^2 = (m . g . dsin60) + (µ . m . g . cos60 . d ) </span>
<span>Solve for d </span>
<span>d = 0.528 = 53 cm</span>
Answer:

Explanation:
The acceleration of a circular motion is given by

where
is the angular velocity and
is the radius.
Angular velocity is related to the period, T, by

Substitute into the previous formula.


This acceleration does not depend on the linear or angular displacement. Hence, the amount of rotation does not change it.