In collision that are categorized as elastic, the total kinetic energy of the system is preserved such that,
KE1 = KE2
The kinetic energy of the system before the collision is solved below.
KE1 = (0.5)(25)(20)² + (0.5)(10g)(15)²
KE1 = 6125 g cm²/s²
This value should also be equal to KE2, which can be calculated using the conditions after the collision.
KE2 = 6125 g cm²/s² = (0.5)(10)(22.1)² + (0.5)(25)(x²)
The value of x from the equation is 17.16 cm/s.
Hence, the answer is 17.16 cm/s.
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:


Explanation:
Newton’s First Law of Motion - if an object is at rest, it takes un-
balanced forces to make it move. Conversely, if an object is moving
it takes an unbalanced force to make it change it’s direction or speed.
Newton was the first to see that such apparently diverse phenomena as a satellite moving near the Earth's surface and the planets orbiting the Sun operate by the same principle: Force equals mass multiplied by acceleration, or F=ma.
Mark me as brainlist
Answer: 1.14 N
Explanation :
As any body submerged in a fluid, it receives an upward force equal to the weight of the fluid removed by the body, which can be expressed as follows:
Fb = δair . Vb . g = 1.29 kg/m3 . 4/3 π (0.294)3 m3. 9.8 m/s2
Fb = 1.34 N
In the downward direction, we have 2 external forces acting upon the balloon: gravity and the tension in the line, which sum must be equal to the buoyant force, as the balloon is at rest.
We can get the gravity force as follows:
Fg = (mb +mhe) g
The mass of helium can be calculated as the product of the density of the helium times the volume of the balloon (assumed to be a perfect sphere), as follows:
MHe = δHe . 4/3 π (0.294)3 m3 = 0.019 kg
Fg = (0.012 kg + 0.019 kg) . 9.8 m/s2 = 0.2 N
Equating both sides of Newton´s 2nd Law in the vertical direction:
T + Fg = Fb
T = Fb – Fg = 1.34 N – 0.2 N = 1.14 N