Answer:
True
Explanation:
Gauss's law:
It is one of the Maxwell's equations which are the foundation of Electrodynamics. According to this law magnetic field has zero divergence and magnetic monopole can't exist. Inside a closed surface, the magnetic flux inward at the south pole will be exactly equal to the outward magnetic flux at the north pole of the magnetic dipole. Thus, the net magnetic flux will be zero.
Answer:
The minimum launch speed is 13366.40 m/s.
Explanation:
Given that,
Mass of planet 
Radius 
Height = 6R
We need to calculate the speed
Using relation between gravitational force and total energy
Put the value into the formula
Hence, The minimum launch speed is 13366.40 m/s.
Answer:
The angular acceleration of the pencil<em> α = 17 rad·s⁻²</em>
Explanation:
Using Newton's second angular law or torque to find angular acceleration, we get the following expressions:
τ = I α (1)
W r = I α (2)
The weight is that the pencil has is,
sin 10 = r / (L/2)
r = L/2(sin(10))
The shape of the pencil can be approximated to be a cylinder that rotates on one end and therefore its moment of inertia will be:
I = 1/3 M L²
Thus,
mg(L / 2)sin(10) = (1/3 m L²)(α)
α(f) = 3/2(g) / Lsin(10)
α = 3/2(9.8) / 0.150sin(10)
<em> α = 17 rad·s⁻²</em>
Therefore, the angular acceleration of the pencil<em> </em>is<em> 17 rad·s⁻²</em>
Answer:
0.699 L of the fluid will overflow
Explanation:
We know that the change in volume ΔV = V₀β(T₂ - T₁) where V₀ = volume of radiator = 21.1 L, β = coefficient of volume expansion of fluid = 400 × 10⁻⁶/°C
and T₁ = initial temperature of radiator = 12.2°C and T₂ = final temperature of radiator = 95.0°C
Substituting these values into the equation, we have
ΔV = V₀β(T₂ - T₁)
= 21.1 L × 400 × 10⁻⁶/°C × (95.0°C - 12.2°C)
= 21.1 L × 400 × 10⁻⁶/°C × 82.8°C = 698832 × 10⁻⁶ L
= 0.698832 L
≅ 0.699 L = 0.7 L to the nearest tenth litre
So, 0.699 L of the fluid will overflow
Answer:
Explanation:
Work in pumping water from the tank is given as
W = ∫ y dF. From a to b
Where dF is the differential weight of the thin layer of liquid in the tank, y is the height of the differential layer
a is the lower limit of the height
b is the upper limit of the height.
We know that, .
F = ρVg
Where F is the weight
ρ is the density of water
V is the volume of water in tank
g is the acceleration due to gravity
Then,
dF = ρg ( Ady)
We know that the density and the acceleration due to gravity is constant, also the base area of the tank is constant, only the height that changes.
Then,
ρg = 62.4 lbs/ft³
Area = L×B = 3 × 9 = 27ft²
dF = ρg ( Ady)
dF = 1684.8dy
The height reduces from 12ft to 0ft
Then,
W = ∫ y dF. From a to b
W = ∫ 1684.8y dy From 0 to 12
W = 1684.8y²/2 from 0 to 12
W = 842.4 [y²] from y = 0 to y = 12
W = 842.4 (12²-0²)
W = 121,305.6 lb-ft
