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
i thinks answers is gap relutance increases linearly with magnetic density
Answer:
f = 3.1 kHz
Explanation:
given,
length of human canal =2.8 cm = 0.028 m
speed of sound = 343 m/s
fundamental frequency = ?
The fundamental frequency of a tube with one open end and one closed end is,
![f = \dfrac{v}{4L}](https://tex.z-dn.net/?f=f%20%3D%20%5Cdfrac%7Bv%7D%7B4L%7D)
![f = \dfrac{343}{4\times 0.028}](https://tex.z-dn.net/?f=f%20%3D%20%5Cdfrac%7B343%7D%7B4%5Ctimes%200.028%7D)
![f = \dfrac{343}{0.112}](https://tex.z-dn.net/?f=f%20%3D%20%5Cdfrac%7B343%7D%7B0.112%7D)
f = 3062.5 Hz
f = 3.1 kHz
hence, the fundamental frequency is equal to f = 3.1 kHz
Answer:
A. 0.289g/mL
Explanation:
Using the equation for density which is d = m/v or density = mass/volume, we input 1.3g/4.5mL and get 0.289g/mL.
Answer:
The maximum amount of work is
Explanation:
From the question we are told that
The temperature of the environment is ![T = 280\ K](https://tex.z-dn.net/?f=T%20%3D%20280%5C%20K)
The volume of container A is ![V_A = 2 m^3](https://tex.z-dn.net/?f=V_A%20%3D%202%20m%5E3)
Initially the number of moles is ![n = 1.2 \ moles](https://tex.z-dn.net/?f=n%20%3D%201.2%20%5C%20moles)
The volume of container B is ![V_B = 3.5 \ m^3](https://tex.z-dn.net/?f=V_B%20%3D%203.5%20%5C%20m%5E3)
At equilibrium of the gas the maximum work that can be done on the turbine is mathematically represented as
Now from the Ideal gas law
![P_A V_A = nRT](https://tex.z-dn.net/?f=P_A%20V_A%20%3D%20%20nRT)
So substituting for
in the equation above
![W = nRT ln [\frac{V_B}{V_A} ]](https://tex.z-dn.net/?f=W%20%3D%20%20nRT%20ln%20%5B%5Cfrac%7BV_B%7D%7BV_A%7D%20%5D)
Where R is the gas constant with a values of ![R = 8.314 \ J/mol](https://tex.z-dn.net/?f=R%20%3D%20%208.314%20%5C%20%20J%2Fmol)
Substituting values we have that