The animals need less food people it is winter time.
Answer:
Explanation:
let force exerted by engine be F.Net force =( F-400)N, applying newton law
F-400 = 1.5 x 10³x18 =27000 ,
F = 27400 N.
velocity after 12 s = 0 + 18 x 12 = 216 m/s
Average velocity = (0 + 216 )/2 = 108 m/s
Average power = force x average velocity = 27400 x 108 = 29.6 10⁵ W .⁶
b) At 12 s , velocity = 216 m/s
Instantaneous power = velocity x force = 216 x 27400 = 59.2 x 10⁶ W.
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
The amount of gravitational potential energy acquired by the rock is equal to:
where
m is the mass of the rock
g is the gravitational acceleration
is the increase in height of the rock
Substituting the data of the problem, we find
So, Natalie gave 220.7 J of energy to the rock.
