Answer:
The energy absorbed by the atomic electrons in the mercury atom is
J
Explanation:
Given:
Potential
V
According to the conservation law,
Loss in kinetic energy = Gain in potential energy
Here, energy absorbed by the atomic electrons is given by,

Where
( charge of electron )

J
Therefore, the energy absorbed by the atomic electrons in the mercury atom is
J
Answer:
14 m/s
Explanation:
The motion of the book is a free fall motion, so it is an uniformly accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. Therefore we can find the final velocity by using the equation:

where
u = 0 is the initial speed
g = 9.8 m/s^2 is the acceleration
d = 10.0 m is the distance covered by the book
Substituting data, we find

Explanation:
Since the balloon is not accelerating means that the net force on the balloon is zero. This implies that the weight of balloon must be equal to the buoyant force on balloon.
Hence, the buoyant force equals the weight of air displaced by the balloon, also 20,000 N.
Weight of the air displaced = density of air × volume
The density of air at 1 atm pressure and 20º C is 1.2 kg/m³
the volume V = 20,000/(1.2×9.8) = 1700 m³
Answer:
the mass of the lipid content, to the nearest hundredth of a kg, in this solution =0.46 kg
Explanation:
Total heat content of the fat = heat content of water +heat content of the lipids
Let it be Q
the Q= (mcΔT)_lipids + (mcΔT)_water
total mass of fat M= 0.63 Kg
Q= heat supplied = 100 W in 5 minutes
ΔT= 20°C
c_lipid= 1700J/(kgoC)
c_water= 4200J/(kgoC)
then,

solving the above equation we get
m= 0.46 kg
the mass of the lipid content, to the nearest hundredth of a kg, in this solution =0.46 kg
Answer:
0.191 s
Explanation:
The distance from the center of the cube to the upper corner is r = d/√2.
When the cube is rotated an angle θ, the spring is stretched a distance of r sin θ. The new vertical distance from the center to the corner is r cos θ.
Sum of the torques:
∑τ = Iα
Fr cos θ = Iα
(k r sin θ) r cos θ = Iα
kr² sin θ cos θ = Iα
k (d²/2) sin θ cos θ = Iα
For a cube rotating about its center, I = ⅙ md².
k (d²/2) sin θ cos θ = ⅙ md² α
3k sin θ cos θ = mα
3/2 k sin(2θ) = mα
For small values of θ, sin θ ≈ θ.
3/2 k (2θ) = mα
α = (3k/m) θ
d²θ/dt² = (3k/m) θ
For this differential equation, the coefficient is the square of the angular frequency, ω².
ω² = 3k/m
ω = √(3k/m)
The period is:
T = 2π / ω
T = 2π √(m/(3k))
Given m = 2.50 kg and k = 900 N/m:
T = 2π √(2.50 kg / (3 × 900 N/m))
T = 0.191 s
The period is 0.191 seconds.