Hmm.. Well i would say A or C because water and acid would just soak threw it
Answer:
(a) k = 
(b) τ = 
∝
Explanation:
The moment of parallel pipe rotating about it's axis is given by the formula;
I = ---------------------------------1
(a) The kinetic energy of a parallel pipe is also given as;
k =
--------------------------------2
Putting equation 1 into equation 2, we have;
k = 
k =
(b) The angular momentum is given by the formula;
τ = Iw -----------------------3
Putting equation 1 into equation 3, we have
τ = 
But
τ = dτ/dt = 
------------------4
where
dw/dt = angular acceleration =∝
Equation 4 becomes;
τ = ∝
Answer:
Wet surfaces areaA=+25.3ft^2
Explanation:
Using F= K×A× S^2
Where F= drag force
A= surface area
S= speed
Given : F=996N S=20mph A= 83ft^2
K = F/AS^2=996/(83×20^2)
K= 996/33200 = 0.03
1215= (0.03)× A × 18^2
1215=9.7A
A=1215/9.7=125.3ft^2
Answer:
v = 1/3 m / s = 0.333 m / s
in the direction of the truck
Explanation:
The average speed is defined by the variation of the position between the time spent
v = Δx / Δt
since the position is a vector we must add using vectors, we will assume that the displacement to the right is positive, the total displacement is
Δx = 20 - 15 +20
Δx = 25 m
therefore we calculate
v = 25/75
v = 1/3 m / s = 0.333 m / s
in the direction of the truck
Answer:
a. The electric field lines are linear and perpendicular to the plates inside a parallel-plate capacitor, and always from positive plate to the negative plate. If a positive charge is released near the positive plate, then<em> it will follow a linear path towards the negative plate under the influence of electrostatic force, F = Eq</em>, where q is the charge of the particle. The electric field inside a parallel plate capacitor is constant and equal to
This can be calculated by Gauss' Law.
A positive charge always follow the electric field lines when released. Another approach is that the positive plate repels the positive charge and negative plate attracts the positive charge. Therefore, the positive charge follows a path towards the negative charge.
b. The particle moves from the higher potential to the lower potential. <em>The direction of motion is the same as the direction of the force that moves the particle, so the work done on the particle by that force is positive.</em>
