Here when an object is placed on the level floor then in that case there are two forces on the object
1). Weight of object downwards (mg)
2). Normal force due to floor which will counterbalance the weight (N)
so when no force is applied on the box at that time normal force is counter balanced by weight.
Now here it is given that A person tried to lift the box upwards
So now there are two forces on the box
1). Applied force of person
2). Normal force due to ground
So now these two forces will counter balance the weight of the crate
So we can write an equation for force balance like
given that
here
m = 30 kg and
g = acceleration due to gravity = 10 m/s^2
now from above equation
So force applied by the person must be 150 N
Answer:
extrusive I'm pretty sure that's right
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dQ = ρ(r) * A * dr = ρ0(1 - r/R) (4πr²)dr = 4π * ρ0(r² -
r³/R) dr
which when integrated from 0 to r is
total charge = 4π * ρ0 (r³/3 + r^4/(4R))
and when r = R our total charge is
total charge = 4π*ρ0(R³/3 + R³/4) = 4π*ρ0*R³/12 = π*ρ0*R³ / 3
and after substituting ρ0 = 3Q / πR³ we have
total charge = Q ◄
B) E = kQ/d²
since the distribution is symmetric spherically
C) dE = k*dq/r² = k*4π*ρ0(r² - r³/R)dr / r² = k*4π*ρ0(1 -
r/R)dr
so
E(r) = k*4π*ρ0*(r - r²/(2R)) from zero to r is
and after substituting for ρ0 is
E(r) = k*4π*3Q(r - r²/(2R)) / πR³ = 12kQ(r/R³ - r²/(2R^4))
which could be expressed other ways.
D) dE/dr = 0 = 12kQ(1/R³ - r/R^4) means that
r = R for a min/max (and we know it's a max since r = 0 is a
min).
<span>E) E = 12kQ(R/R³ - R²/(2R^4)) = 12kQ / 2R² = 6kQ / R² </span></span>
Answer:
As the pressure inside the pressure cooker increases as it is heated, the temperature also increases.
Explanation:
A pressure cooker contains water and steam in equilibrium at a pressure greater than atmospheric pressure.
According to the ideal gas law, as the pressure increases, the temperature also increases.
As the pressure inside the pressure cooker increases as it is heated, the temperature also increases.
