<h2>Acceleration due to gravity in moon is 1.5 m/s²</h2>
Explanation:
We have equation of motion s = ut + 0.5 at²
Here the ball travels 3 m less distance in fifth second compared to third second.
That is
s₃ = s₅ + 3
Now we have
Distance traveled in third second, s₃ = u x 3 - 0.5 x g x 3² - u x 2 - 0.5 x g x 2²
s₃ = u - 2.5 g
Also
Distance traveled in fifth second, s₅ = u x 5 - 0.5 x g x 5² - u x 4 - 0.5 x g x 4²
s₅ = u - 4.5 g
That is
u - 2.5 g = u - 4.5 g + 3
2 g = 3
g = 1.5 m/s²
Acceleration due to gravity in moon = 1.5 m/s²
Answer:
Archimedes' principle states that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially, is equal to the weight of the fluid that the body displaces. Archimedes' principle is a law of physics fundamental to fluid mechanics. It was formulated by Archimedes of Syracuse.
Answer:
Therefore the correct statement is B.
Explanation:
In the interference and diffraction phenomena, the natural wave of electromagnetic radiation must be taken into account, the wave front that advances towards the slit can be considered as when it reaches it behaves like a series of wave emitters, each slightly out of phase from the previous one, following the Huygens principle that states that each point is compiled as a source of secondary waves.
The sum of all these waves results in the diffraction curve of the slit that has the shape
I = Io sin² θ /θ²
Where the angle is a function of the wavelength and the width of the slit.
From the above, the interference phenomenon can be treated as the sum of two diffraction phenomena displaced a distance equal to the separation of the slits (d)
Therefore the correct statement is B
To solve this problem we will apply the concepts related to Ohm's law and Electric Power. By Ohm's law we know that resistance is equivalent to,

Here,
V = Voltage
I = Current
While the power is equivalent to the product between the current and the voltage, thus solving for the current we have,


Applying Ohm's law


Therefore the equivalent resistance of the light string is 
Answer:
W= -2.5 (p₁*0.0012) joules
Explanation:
Given that p₀= initial pressure, p₁=final pressure, Vi= initial volume=0 and Vf=final volume= 6/5 liters where p₁=p₀ then
In adiabatic compression, work done by mixture during compression is
W=
where f= final volume and i =initial volume, p=pressure
p can be written as p=K/V^γ where K=p₀Vi^γ =p₁Vf^γ
W= 
W= K/1-γ ( 1/Vf^γ-1 - 1/Vi^γ-1)
W=1/1-γ (p₁Vf-p₀Vi)
W= 1/1-1.40 (p₁*6/5 -p₀*0)
W= -2.5 (p₁*6/5*0.001) changing liters to m³
W= -2.5 (p₁*0.0012) joules