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 neutral wire is often confused with ground wire, but in reality, they serve two distinct purposes. Neutral wires carry currents back to power source to better control and regulate voltage. Its overall purpose is to serve as a path to return energy.
Answer:
0.61°
Explanation:
Since the box move at constant velocity, it means there is no acceleration then we can say it has a balanced force system.
Pulling force= resistance force
From the formula for pulling force,
F(x)= Fcos(θ)
= 425×cos(35.2)
=347N
The force exerted downward at an angle of 35.2° below the horizontal= Fsin(θ)= 425sin(35.2)
=425×0.567=245N
Resistance force= (325N+ 245N) (α)= 570N(α)
We can now equates the pulling force to resistance force
570 (α)= 347N
(α)= 347/570
= 0.61
Answer:
20cm
Explanation:
A convex lens has a positive focal length and the object placed in front of it produce both virtual and real image <em>(image distance can be negative or positive depending on the nature of the image</em>).
According to the lens equation
where;
f is the focal length of the lens
u is the object distance
v is the image distance
If the magnification is - 0.6
mag = v/u = -0.5
v = -0.5u
since v = 10cm
10 = -0.5u
u = -10/0.5
u =-20 cm
Substitute u = -20cm ( due to negative magnification)and v = 10cm into the lens formula to get the focal length f
Hence the focal length of the convex lens is 20cm
He has 1 1/30 gallons, or 31/30 gallons, you can find this by setting all the fractions to a common denominator and adding them
