<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>
Ngan's mass on earth is 85kg.
Ngan has a weight on Mars = 14.5 N
Ngan’s weight on Earth = 833.0 N
Ngan’s mass on Earth = ?
<span>Fg,earth = mg(earth)</span>
<span>M = Fg,earth </span><span>/ g(earth)</span>
<span>M = 833.0 N / 9.8 m/s2</span>
<span>M = 85 kg</span>
Answer:
It remains constant
Explanation:
As we know that buoyant force on an object given as
Fb = ρ Vd g
ρ= Density of fluid
Vd=Volume displace by body
g=10 m/s²
Fb =buoyant force
So from above we can say that buoyant force does not depends on the depth. It only depends on the fluid density and volume displace by body.
So when rock gets deeper and deeper the buoyant force will remain constant.
It remains constant
An antibaryon composed of two antiup quarks
and one antidown quark would have a charge of (2) −1e.
