Answer:
547 m
Explanation:
From law of motion
s = ut + ½at²
Where "t" is Time taken to reach Earth
s= distance= 182 m
a= vertical acceleration = 5.82 m / s 2
U= initial velocity in vertical position = 0
182= ½ × 5.82t²
t²=( 2× 182)/ 5.82
= 364/5.82
= 62.54
t= √62.54
t= 7.908s
horizontal distance travelled = speed x time
Horizontal speed= 72.6 m / s
horizontal distance travelled =72.6× 7.908
= 547 m
Hence, the survivor will it hit the waves at 547 m away
Volume = mass/density
Volume = 15000 g/45 g/cm3 ≈ 333.3 cm<span>3</span>
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:
The tensile stress on the wire is 550 MPa.
Explanation:
Given;
Radius of copper wire, R = 3.5 mm
extension of the copper wire, e = 5.0×10⁻³ L
L is the original length of the copper wire,
Young's modulus for copper, Y = 11×10¹⁰Pa.
Young's modulus, Y is given as the ratio of tensile stress to tensile strain, measured in the same unit as Young's modulus.
Therefore, the tensile stress on the wire is 550 MPa.
