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Trava [24]
2 years ago
7

a An object is tarown up with a velocity v = 6.02 +7.0j. Calculate the (1) time taken reach the maximum height (ii) the horizont

al range (s = 10m/s2). ​
Physics
2 answers:
AfilCa [17]2 years ago
8 0

Answer:

(i) 0.6s (ii) 8.42m

Explanation:

U² = 6.02² + 7²

U = 9.23

angle of projection

tanø = 6.02/7

ø = 40.7

Time of fligt

t = Usinø/g

t = 9.23 sin 40.7/10

t = 0.6

H range = U²sin2ø/g

H = 9.23²sin 81.4/10

H = 8.42m

Shkiper50 [21]2 years ago
4 0

Answer:

the update is coming down 55th 55 cm long

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Explanation:

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The rope is doing the work against the gravity on the skier to uplift up to the inclined surface. Therefore the work done by the rope is equal to the work done on the skier due to the gravity

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part (b)

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Rate of the work done by the rope is power of the rope.

Power\ =\ \dfrac{\Delta W}{\Delta t}\\\Rightarrow P\ =\ \dfrac{\Delta W}{\dfrac{d}{v}}\\\Rightarrow P\ =\ \dfrac{\Delta W\times v}{d}\\\Rightarrow P\ =\ \dfrac{900\times 1.0}{8.0}\\\Rightarrow P\ =\ 112.5\ Watt

Part (c)

  • Initial speed of the skier = v = 2.0 m/s.

Rate of the work done by the rope is power of the rope.

Power\ =\ \dfrac{\Delta W}{\Delta t}\\\Rightarrow P\ =\ \dfrac{\Delta W}{\dfrac{d}{v}}\\\Rightarrow P\ =\ \dfrac{\Delta W\times v}{d}\\\Rightarrow P\ =\ \dfrac{900\times 2.0}{8.0}\\\Rightarrow P\ =\ 225\ Watt

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14.4 = V(2) - 20 / 10

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