Answer:
W= F × d
W= 2kn × 3.6
W= 7.2 J
Work is measured in Joules!
Mass of gold m₁ = 47 g
Initial temperature of gold T₁ = 99 C
Specific heat of gold C₁ = 0.129 J/gC
final temperature T₂ = 38 C
Heat needed by the gold to cool down
Q =m₁ * C₁* ( T₁ - T₂)
Q = (47)(0.129)(99-38)
Q = 369.843 J
This heat will be given by the water
we need to find out mass of water m₂
and initial temperature of water is T₃ = 25 C
Specific heat of water C₂ = 4.184 J/gC
Q = m₂*C₂*(T₂ - T₃)
369.843 = m₂(4.184)(38-25)
m₂ = 6.8 g
The data from different subjects, you compare the data too
Answer: For what objects? And what data?
Explanation:
Answer:
ratio of tangential velocity of satellite b and a will be 0.707
Explanation:
We have given distance of satellite B from satellite A is twice
So 
Tangential speed of the satellite is given by
, G is gravitational constant. M is mass of satellite and r is distance from the earth
We have to find the ratio of tangential velocities of b and a
From the relation we can see that tangential velocity is inversely proportional to square root of distance from earth
So 



So ratio of tangential velocity of satellite b and a will be 0.707