Answer:
The angle formed of the rope with the surface = 40°
Force applied = 125Newtons
The displacement covered by the box =25metres
W= FDcos theta
[125×40×cos(40°) ] Joules
= [ (3125×0.76604444311)]Joules
= 2393.88888472 joules(ans)
Hope it helps
E=energy=5.09x10^5J = 509KJ
<span>M=mass=2250g=2.25Kg </span>
<span>C=specific heat capacity of water= 4.18KJ/Kg </span>
<span>ΔT= change in temp= ? </span>
<span>E=mcΔT </span>
<span>509=(2.25)x(4.18)xΔT </span>
<span>509=9.405ΔT </span>
<span>ΔT=509/9.405=54.1degrees </span>
<span>Initial temp = 100-54 = 46 degrees </span>
<span>Hope this helps :)</span>
The north vectors add up as so the south vectors. Then subtract the two. For north its 4 + 5 = 9. South is 2 + 5 = 7. Then 9-7 = 2km North (D)
Answer:
Power lines
Explanation:
Generators induce that current that runs the turbines
Turbines also induce currents from water waves that is transmitted.
Transformers change A.C to D.C or vice versa.
0.36 J of work is done in stretching the spring from 15 cm to 18 cm.
To find the correct answer, we need to know about the work done to strech a string.
<h3>What is the work required to strech a string?</h3>
- Mathematically, the work done to strech a string is given as 1/2 ×K×x².
- K is the spring constant.
<h3>What will be the spring constant, if 40N force is required to hold a 10 cm to 15 cm streched spring?</h3>
- The force experienced by a streched spring is given as Kx. x is the length of the spring streched from its natural length.
- Then K = Force / x.
- Here x = 15 - 10 = 5 cm = 0.05 m
- K = 40/0.05 = 800N/m.
<h3>What will be the work required to strech that spring from 15 cm to 18 cm?</h3>
- Work done = 1/2×k×x²
- Here x= 18-15=3cm or 0.03 m
- So, W= 1/2×800×0.03² = 0.36 J.
Thus, we can conclude that the work done is 0.36 J.
Learn more about the spring force here:
brainly.com/question/14970750
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