Answer:
Acceleration of the ship, 
Explanation:
It is given that,
Mass of both ships, 
Distance between two ships, d = 110 m
The gravitational force between two ships is given by :
F = 8.38 N
Let a is the acceleration. Now, using second law of motion as :
So, the acceleration of either ship due to the gravitational attraction of the other is 
. Hence, this is the required solution.
Answer:
2 kg
Explanation:
Remember:
F = m * a re-arrange to
F/a = m substitute in the given values
10 / 5 = 2 kg
<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>
Answer:
c = 1163.34 J/kg.°C
Explanation:
Specific heat capacity:
"Specific heat capacity is the amount of heat energy required to raise the temperature of a substance per unit of mass. The specific heat capacity of a material is a physical property."
Use this equation:
mcΔT = ( mw c + mAl cAl ) ΔT'
Rearranging the equation to find the specific heat (c) you get this:
c = (( mw c + mAl cAl ) ΔT') / (mΔT)
c = (( 0.285 (4186) + (0.15)(900)) (32 -25.1)) / ((0.125) (95 - 32))
c = 1163.34 J/kg.°C
