A conjugate acid is formed from the base by accepting a proton from the acid .
A conjugate base is obtained from the Brownstead - Lowry acid when it looses a proton while the conjugate acid is obtained from the Brownstead - Lowry base when it accepts a proton. In the Brownstead - Lowry sense, acid base reaction involves the loss or gain of a proton.
Consider the hypothetical reaction; AH + :B ⇄ BH + :A. The specie BH is the conjugate acid while the specie :B is the Brownstead - Lowry base . The specie :A is the conjugate base while the specie AH is the Brownstead - Lowry acid.
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Answer:
The total work done by the two tugboats on the supertanker is 3.44 *10^9 J
Explanation:
The force by the tugboats acting on the supertanker is constant and the displacement of the supertanker is along a straight line.
The angle between the 2 forces and displacement is ∅ = 15°.
First we have to calculate the work done by the individual force and then we can calculate the total work.
The work done on a particle by a constant force F during a straight line displacement s is given by following formula:
W = F*s
W = F*s*cos∅
With ∅ = the angles between F and s
The magnitude of the force acting on the supertanker is F of tugboat1 = F of tugboat 2 = F = 2.2 * 10^6 N
The total work done can be calculated as followed:
Wtotal = Ftugboat1 s * cos ∅1 + Ftugboat2 s* cos ∅2
Wtotal = 2Fs*cos∅
Wtotal = 2*2.2*10^6 N * 0.81 *10³ m s *cos15°
Wtotal = 3.44*10^9 Nm = <u>3.44 *10^9 J</u>
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The total work done by the two tugboats on the supertanker is 3.44 *10^9 J
Answer:
The mass of the sand that will fall on the disk to decrease the is 0.3375 kg
Explanation:
Moment before = Moment after

where;
I is moment of inertia = Mr² = 0.3 x (0.3)² = 0.027 kg.m²
substitute this in the above equation;
![m = \frac{ 0.027[3(2 \pi) - 2(2 \pi)]} {0.2^2 * 6\pi } = \frac{ 0.027[6 \pi - 4\pi]} {0.2^2 * 4\pi }\\\\m = 0.3375kg](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7B%200.027%5B3%282%20%5Cpi%29%20%20-%202%282%20%5Cpi%29%5D%7D%20%7B0.2%5E2%20%2A%206%5Cpi%20%7D%20%3D%20%5Cfrac%7B%200.027%5B6%20%5Cpi%20%20-%204%5Cpi%5D%7D%20%7B0.2%5E2%20%2A%204%5Cpi%20%7D%5C%5C%5C%5Cm%20%3D%200.3375kg)
Therefore, the mass of the sand that will fall on the disk to decrease the is 0.3375 kg
Grade 1: Stretching or slight tearing of the ligament with mild tenderness, swelling and stiffness. The ankle feels stable and it is usually possible to walk with minimal pain.
Grade 2: A more severe sprain, but incomplete tear with moderate pain, swelling and bruising. Although it feels somewhat stable, the damaged areas are tender to the touch and walking is painful.
Grade 3: This is a complete tear of the affected ligament(s) with severe swelling and bruising. The ankle is unstable and walking is likely not possible because the ankle gives out and there is intense pain.
source - https://www.rushcopley.com/health/physician-articles/varying-degrees-of-ankle-sprains/