Answer: 7.38 km
Explanation: The attachment shows the illustration diagram for the question.
The range of the bomb's motion as obtained from the equations of motion,
H = u(y) t + 0.5g(t^2)
U(y) = initial vertical component of velocity = 0 m/s
That means t = √(2H/g)
The horizontal distance covered, R,
R = u(x) t = u(x) √(2H/g)
Where u(x) = the initial horizontal component of the bomb's velocity = 287 m/s, H = vertical height at which the bomb was thrown = 3.24 km = 3240 m, g = acceleration due to gravity = 9.8 m/s2
R = 287 √(2×3240/9.8) = 7380 m = 7.38 km
Answer: Magnetizim
Explanation: Magnetic Atoms collide creating magnetizim
Answer:
220 A
Explanation:
The magnetic force on the floating rod due to the rod held close to the ground is F = BI₁L where B = magnetic field due to rod held close the ground = μ₀I₂/2πd where μ₀ = permeability of free space = 4π × 10⁻⁷ H/m, I₂ = current in rod close to ground and d = distance between both rods = 11 mm = 0.011 m. Also, I₁ = current in floating rod and L = length of rod = 1.1 m.
So, F = BI₁L
F = (μ₀I₂/2πd)I₁L
F = μ₀I₁I₂L/2πd
Given that the current in the rods are the same, I₁ = I₂ = I
So,
F = μ₀I²L/2πd
Now, the magnetic force on the floating rod equals its weight , W = mg where m = mass of rod = 0.10kg and g = acceleration due to gravity = 9.8 m/s²
So, F = W
μ₀I²L/2πd = mg
making I subject of the formula, we have
I² = 2πdmg/μ₀L
I = √(2πdmg/μ₀L)
substituting the values of the variables into the equation, we have
I = √(2π × 0.011 m × 0.1 kg × 9.8 m/s²/[4π × 10⁻⁷ H/m × 1.1 m])
I = √(0.01078 kgm²/s²/[2 × 10⁻⁷ H/m × 1.1 m])
I = √(0.01078 kgm²/s²/[2.2 × 10⁻⁷ H])
I = √(0.0049 × 10⁷kgm²/s²H)
I = √(0.049 × 10⁶kgm²/s²H)
I = 0.22 × 10³ A
I = 220 A
Answer:
m = 56.5 kg
Explanation:
Since the addition of mass on one piston caused a change in pressure head at the other. Diameter of the piston calculated is used as 0.46 m
Δm*g / Area = p * g * Δh ..... Eq1
Answer:
a. 16 s b. -1.866 kJ
Explanation:
a. Since the initial rotational speed ω₀= 3313 rev/min = 3313/60 × 2π rad/s = 346.94 rad/s. Its rotational speed becomes ω₁ = 0.75ω₀ in time t = 4 s.
We find it rotational acceleration using α = (ω₁ - ω₀)/t = (0.75ω₀ - ω₀)/t = ω₀(0.75 - 1)/t = -0.25ω₀/t = (-0.25 × 346.94 rad/s)/4 s = -21.68 rad/s².
Since the turntable stops at ω = 0, the time it takes to stop is gotten from
ω = ω₀ + αt and t = (ω - ω₀)/α = (0 - 346.94 rad/s)/-21.68 rad/s² = (-346.94/-21.68) s = 16 s.
So it takes the turntable 16 s to stop.
b. The workdone by the turntable to stop W equals its rotational kinetic energy change.
So, W = 1/2Iω² - 1/2Iω₀² = 1/2 × 0.031 kgm² × 0² - 1/2 × 0.031 kgm² × (346.94 rad/s)² = 0 - 1865.7 J = -1865.7 J = -1.8657 kJ ≅ -1.866 kJ
