A. Through the direct contact of particles
F = 130 revs/min = 130/60 revs/s = 13/6 revs/s
t = 31s
wi = 2πf = 2π × 13/6 = 13π/3 rads/s
wf = 0 rads/s = wi + at
a = -wi/t = -13π/3 × 1/31 = -13π/93 rads/s²
wf² - wi² = 2a∅
-169π²/9 rads²/s² = 2 × -13π/93 rads/s² × ∅
∅ = 1209π/18 rads
n = ∅/2π = (1209π/18)/(2π) = 1209/36 ≈ 33.5833 revolutions.
Given :
Mass of water, m = 2 grams.
The temperature of water drops from 31 °C to 29 °C .
The specific heat of water is 4.184 J/(g • °C).
To Find :
Amount of heat lost in this process.
Solution :
We know, heat lost is given by :
Therefore, amount of heat lost in this process is 16.736 J.
Use the Inverse square law, Intensity (I) of a light is inversely proportional to the square of the distance(d).
I=1/(d*d)
Let Intensity for lamp 1 is L1 distance be D1 so on, L2 D2 for Intensity for lamp 2 and its distance.
L1/L2=(D2*D2)/(D1*D1)
L1/15=(200*200)/(400*400)
L1=15*0.25
L1=3.75 <span>candela</span>
Answer:
It is the ratio of the density of a substance to the density of a given reference material.
Explanation:
<em>Specific gravity is also known as relative density.</em>
<u>To find the relative density of substance, you:</u>
- Divide the density of substance measured
- And divide that by the density of the reference
