The answer to this question is "LIES BELOW THE SURFACE" happens or occurs. When one of the main differences between the two which is the Intaglio and the other one is the relief printing processes is that with the Intaglio the ink LIES BELOW the surface of the printing plate.
Answer:

is time required to heat to boiling point form initial temperature.
Explanation:
Given:
initial temperature of water, 
time taken to vapourize half a liter of water, 
desity of water, 
So, the givne mass of water, 
enthalpy of vaporization of water, 
specific heat of water, 
Amount of heat required to raise the temperature of given water mass to 100°C:



Now the amount of heat required to vaporize 0.5 kg of water:

where:
mass of water vaporized due to boiling


Now the power rating of the boiler:



Now the time required to heat to boiling point form initial temperature:


Answer:
E = 2.5 x 10⁻¹⁴ J
Explanation:
given,
diameter = 1.33 x 10⁻¹⁴ m
mass = 6.64 x 10⁻²⁷ kg
wavelength is equal to diameter
de broglie wavelength equal to diameter



v = 7.5 x 10⁶ m/s
Kinetic energy is equal to


E = 2.5 x 10⁻¹⁴ J
The object D is made up of material Lead. The correct option is D.
<h3>What is specific heat?</h3>
The specific heat is the amount of heat required to change the temperature by 1°C. It is denoted by C.
Two 1-kg objects, C and D, increase in temperature by the same amount, but the thermal energy transfer of object C is greater than the thermal energy transfer of object D. The object C has a specific heat of 235 J/kg-K.
Q = m C ΔT
Qc > Qd
The energy transfer is proportional to specific heat.
Specific heat of D must be less. The possible material with specific heat less than the given value is for Lead material.
Thus, the correct option is D.
Learn more about specific heat,
brainly.com/question/11297584
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Answer:
T = 480.2N
Explanation:
In order to find the required force, you take into account that the sum of forces must be equal to zero if the object has a constant speed.
The forces on the boxes are:
(1)
T: tension of the rope
M: mass of the boxes 0= 49kg
g: gravitational acceleration = 9.8m/s^2
The pulley is frictionless, then, you can assume that the tension of the rope T, is equal to the force that the woman makes.
By using the equation (1) you obtain:

The woman needs to pull the rope at 480.2N