Answer:
Final temperature, 
Explanation:
Given that,
Mass of silver ring, m = 4 g
Initial temperature, 
Heat released, Q = -18 J (as heat is released)
Specific heat capacity of silver, 
To find,
Final temperature
Solution,
The expression for the specific heat is given by :





So, the final temperature of silver is 21.85 degrees Celsius.
<span>The 2nd truck was overloaded with a load of 16833 kg instead of the permissible load of 8000 kg.
The key here is the conservation of momentum.
For the first truck, the momentum is
0(5100 + 4300)
The second truck has a starting momentum of
60(5100 + x)
And finally, after the collision, the momentum of the whole system is
42(5100 + 4300 + 5100 + x)
So let's set the equations for before and after the collision equal to each other.
0(5100 + 4300) + 60(5100 + x) = 42(5100 + 4300 + 5100 + x)
And solve for x, first by adding the constant terms
0(5100 + 4300) + 60(5100 + x) = 42(14500 + x)
Getting rid of the zero term
60(5100 + x) = 42(14500 + x)
Distribute the 60 and the 42.
60*5100 + 60x = 42*14500 + 42x
306000 + 60x = 609000 + 42x
Subtract 42x from both sides
306000 + 18x = 609000
Subtract 306000 from both sides
18x = 303000
And divide both sides by 18
x = 16833.33
So we have the 2nd truck with a load of 16833.33 kg, which is well over it's maximum permissible load of 8000 kg. Let's verify the results by plugging that mass into the before and after collision momentums.
60(5100 + 16833.33) = 60(21933.33) = 1316000
42(5100 + 4300 + 5100 + 16833.33) = 42(31333.33) = 1316000
They match. The 2nd truck was definitely over loaded.</span>
#1
As we know that

now plug in all data into this


now from the formula of strain




#2
As we know that
pressure * area = Force
here we know that


now force is given as

#3
As we know that density of water will vary with the height as given below

here we know that


now density is given as


#4
as we know that pressure changes with depth as per following equation

here we know that

now we will have



here we will have

so it is 20.1 m below the surface
#5
Here net buoyancy force due to water and oil will balance the weight of the block
so here we will have




so it is 3.48 cm below the interface
B. It’s the same roughly at all latitudes