Answer:
5.92×10⁷ J
Explanation:
We'll begin by converting 15 tons to Newton. This can be obtained as follow:
1 ton = 9806.65 N
Therefore,
15 ton = 15 ton × 9806.65 N / 1 ton
15 ton = 147099.75 N
Next, we shall convert one-quarter (¼) or 0.25 mile to metre. This can be obtained as follow:
100 mi = 160934 m
0.25 mi = 0.25 mi × 160934 / 100 mi
0.25 mi = 402.335 m
Finally, we shall determine the Workdone. This can be obtained as follow:
Force (F) = 147099.75 N
Distance (d) = 402.335 m
Workdone (Wd) =?
Wd = F × d
Wd = 147099.75 × 402.335
Wd = 5.92×10⁷ J
Thus, the Workdone is 5.92×10⁷ J
Only in metamorphic rocks does this occur.
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
When a candle is burning the candle is releasing thermal and radiant energy
Answer:
(a) The distance-time graph for an object with uniform speed is giving by a straight line sloped graph with a constant positive or negative gradient as shown in the attached diagram
(b) The distance-time graph for an object with non-uniform speed is giving by a curved line sloped graph with varying gradient as shown in the attached diagram
(c) The velocity-time graph for a car with uniform motion is giving by a horizontal line graph at the speed of constant motion with a zero gradient as shown in the attached diagram
(d) The velocity-time graph for a car moving with uniform acceleration is giving by a straight line sloped graph with a constant positive or negative gradient as shown in the attached diagram
(e) The velocity-time graph for a car moving with non-uniform acceleration is giving by a curved line sloped graph with varying gradient as shown in the attached diagram
(f) According to Newton's first law of motion, an object at rest will remain at rest with no motion unless acted by a force, an therefore, will have no motion with time
Explanation:
