The latent heat of fusion refers to the solid to liquid or liquid to solid states.
Answer: Option C
<u>Explanation:
</u>
It is known that the inter conversion process from the states of solid to liquid is referred as fusion. So, for these conversions, the external energy in the heat form should be supplied to solid.
This external energy should be greater than the latent heat of solid in order to successfully break the bonds to form liquid. So the change in the enthalpy of the reaction while conversion from solids to liquids are termed as latent heats of fusion.
Even the inter-conversion from liquid to solid state will undergo change in enthalpy where the heat will be released and that is termed as latent heats of solidification. It is found that latent heat of solidification is equal in magnitude but opposite in direction with the latent heats of fusion.
Answer:
B) 20N.s is the correct answer
Explanation:
The formula for the impulse is given as:
Impulse = change in momentum
Impulse = mass × change in speed
Impulse = m × ΔV
Given:
initial speed = 40m/s
Final speed = -60 m/s (Since the the ball will now move in the opposite direction after hitting the bat, the speed is negative)
mass = 0.20 kg
Thus, we have
Impulse = 0.20 × (40m/s - (-60)m/s)
Impulse = 0.20 × 100 = 20 kg-m/s or 20 N.s
Answer:
<em>The comoving distance and the proper distance scale</em>
<em></em>
Explanation:
The comoving distance scale removes the effects of the expansion of the universe, which leaves us with a distance that does not change in time due to the expansion of space (since space is constantly expanding). The comoving distance and proper distance are defined to be equal at the present time; therefore, the ratio of proper distance to comoving distance now is 1. The scale factor is sometimes not equal to 1. The distance between masses in the universe may change due to other, local factors like the motion of a galaxy within a cluster. Finally, we note that the expansion of the Universe results in the proper distance changing, but the comoving distance is unchanged by an expanding universe.
a.) K 2=K 1 +GmM( r 21− r 11)=2.2×10 7J
b.) K 2 +GmM( r 11− r 21)=6.9×10 7 J
Applying Law of Energy conservation :
K 1+U 1
=K 2+U 2
⇒K 1− r 1GmM
=K 2− r 2 GmM
where M=5.0×10 23kg,r1
=> R=3.0×10 6m and m=10kg
(a) If K 1
=5.0×10 7J and r 2
=4.0×10 6 m, then the above equation leads to
K 2=K 1 +GmM (r 21− r 11)=2.2×10 7J
(b) In this case, we require K 2
=0 and r2
=8.0×10 6m, and solve for K 1:K 1
=K 2 +GmM (r 11− r 21)=6.9×10 7 J
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