The springs stored energy is transferred to the cube as kinetic energy and then by the slop the KE is converted to height energy.
<span>0.5 . k . x^2 = 0.5 . m . v^2 = m . g . ∆h </span>
<span>0.5 . 50 . (0.1^2) = 0.05 . 9.8 . ∆h </span>
<span>∆h = 0.51 m = 51 cm </span>
<span>This is the height gained </span>
<span>Distance along the slope = ∆h / sin 60 = 0.589 = 59 cm </span>
<span>In the second case, the stored spring energy is converted into height energy AND frictional heat energy. </span>
<span>The height energy is m . g . d sin 60 where d is the distance the cube moves along the slope. </span>
<span>The Frictional energy converted is F . d </span>
<span>F ( the frictional force ) = µ . N </span>
<span>N ( the reaction to the component of the gravity force perpendicular to the surface of the slope ) = m . g . cos60 </span>
<span>Total energy converted </span>
<span>0.5 . k . x^2 = (m . g . dsin60) + (µ . m . g . cos60 . d ) </span>
<span>Solve for d </span>
<span>d = 0.528 = 53 cm</span>
Answer:
The ratio of the resistances of second coil to the first coil is the ratio of square of radius of the first coil to the square of radius of second coil.
And
The ratio of the resistances of fourth coil to the third coil is the ratio of square of radius of the third coil to the square of radius of fourth coil.
Explanation:
The resistance of the coil is directly proportional to the length of the coil and inversely proportional to the area of coil and hence inversely proportional to the square of radius of the coil.
So, the ratio of the resistances of second coil to the first coil is the ratio of square of radius of the first coil to the square of radius of second coil.
And
The ratio of the resistances of fourth coil to the third coil is the ratio of square of radius of the third coil to the square of radius of fourth coil.
I believe it's magnetic stripping. I hope that helps
Multiply these numbers and there’s your answer
Answer:
Explanation:
This problem can be solved easily if we represent velocity in the form of vector.
The velocity of 351 was towards easterly direction so
V₁ = 351 i
The velocity of 351 was towards south west making - 48° with east or + ve x direction.
V₂ = 351 Cos 48 i - 351 sin 48 j
V₂ = 234.86 i - 260.84 j
Change in velocity
= V₂ - V₁ = 234.86 i - 260.84 j - 351 i
= -116.14 i - 260.84 j
acceleration
= change in velocity / time
(-116.14 i - 260.84 j )/ 1
= -116.14 i - 260.84 j
magnitude = 285.53 ms⁻²
Direction
Tan θ = 260.84 / 116.14 = 2.246
θ = 66 degree south of west .
