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:
Heat
Friction is what causes heat.
Brainliest always helps.
Let us situate this on the x axis, and let our uniform line of charge be positioned on the interval <span>(−L,0]</span> for some large number L. The voltage V as a function of x on the interval <span>(0,∞)</span> is given by integrating the contributions from each bit of charge. Let the charge density be λ. Thus, for an infinitesimal length element <span>d<span>x′</span></span>, we have <span>λ=<span><span>dq</span><span>d<span>x′</span></span></span></span>.<span>V(x)=<span>1/<span>4π<span>ϵ0</span></span></span><span>∫line</span><span><span>dq/</span>r</span>=<span>λ/<span>4π<span>ϵ0</span></span></span><span>∫<span>−L</span>0</span><span><span>d<span>x/</span></span><span>x−<span>x′</span></span></span>=<span>λ/<span>4π<span>ϵ0</span></span></span><span>(ln|x+L|−ln|x|)</span></span>
Answer:
The percentage of its mechanical energy does the ball lose with each bounce is 23 %
Explanation:
Given data,
The tennis ball is released from the height, h = 4 m
After the third bounce it reaches height, h' = 183 cm
= 1.83 m
The total mechanical energy of the ball is equal to its maximum P.E
E = mgh
= 4 mg
At height h', the P.E becomes
E' = mgh'
= 1.83 mg
The percentage of change in energy the ball retains to its original energy,
ΔE % = 45 %
The ball retains only the 45% of its original energy after 3 bounces.
Therefore, the energy retains in each bounce is
∛ (0.45) = 0.77
The ball retains only the 77% of its original energy.
The energy lost to the floor is,
E = 100 - 77
= 23 %
Hence, the percentage of its mechanical energy does the ball lose with each bounce is 23 %
