The formula to find the kinetic energy is:
Ek= 1/2 × m × v^2
1. Ek= 1/2×15×3^2
= 67.5 J
2.Ek= 1/2×8×4^2
=64 J
3.Ek= 1/2×12×5^2
= 150 J
4.Ek= 1/2×10×6^2
= 180 J
So the fourth dog has the most kinetic energy.
Answer:
i 5.3 cm ii. 72 cm
Explanation:
i
We know upthrust on iron = weight of mercury displaced
To balance, the weight of iron = weight of mercury displaced . So
ρ₁V₁g = ρ₂V₂g
ρ₁V₁ = ρ₂V₂ where ρ₁ = density of iron = 7.2 g/cm³ and V₁ = volume of iron = 10³ cm³ and ρ₂ = density of mercury = 13.6 g/cm³ and V₂ = volume of mercury displaced = ?
V₂ = ρ₁V₁/ρ₂ = 7.2 g/cm³ × 10³ cm³/13.6 g/cm³ = 529.4 cm³
So, the height of iron above the mercury is h = V₂/area of base iron block
= 529.4 cm³/10² cm² = 5.294 cm ≅ 5.3 cm
ρ₁V₁g = ρ₂V₂g
ii
ρ₁V₁ = ρ₃V₃ where ρ₁ = density of iron = 7.2 g/cm³ and V₁ = volume of iron = 10³ cm³ and ρ₃ = density of water = 1 g/cm³ and V₃ = volume of water displaced = ?
V₃ = ρ₁V₁/ρ₃ = 7.2 g/cm³ × 10³ cm³/1 g/cm³ = 7200 cm³
So, the height of column of water is h = V₃/area of base iron block
= 7200 cm³/10² cm² = 72 cm
Answer:
W = 9.93 10² N
Explanation:
To solve this exercise we must use the concept of density
ρ = m / V
the tabulated density of copper is rho = 8966 kg / m³
let's find the volume of the cylindrical tube
V = A L
V = π (R_ext ² - R_int ²) L
let's calculate
V = π (4² - 2²) 10⁻⁴ 3
V = 1.13 10⁻² m³
m = ρ V
m = 8966 1.13 10⁻²
m = 1.01 10² kg
the weight of the tube
W = mg
W = 1.01 10² 9.8
W = 9.93 10² N
The power that the light is able to utilize out of the supply is only 0.089 of the given.
Power utilized = (0.089)(22 W)
= 1.958 W
= 1.958 J/s
The energy required in this item is the product of the power utilized and the time. That is,
Energy = (1.958 J/s)(1 s) = 1.958 J
Thus, the light energy that the bulb is able to produce is approximately 1.958 J.
Answer:
F = 5291.25 N
Explanation:
F = Ma so 1245 times 4.25^2 ,, that equals 5291.25 N
