A ski jumper leaves the sky track moving in the horizontal direction with a speed of 25.0 m/s as shown in figure 4.14. The landing incline below him falls off with a slope of 35.0°. Where does he land on the incline? And how long is the jumper in the air and his vertical component of velocity just before he lan

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A: Given Data, Magnitude of charges (q1=q2=q)=1.1×107 C Electrostatic Force(F)=4.2×104 N

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A: A vector can be written as A̅ = Ax î + Ay ĵ     = A Cosθ î + A Sinθ ĵ 

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A: Correct answer is option (b) i.e. Charging by polarization

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A: Given  Stress = 150 M Pa = 150 × 106 Pa  Length = 12 m   E = 200 GPa = 200 × 109 Pa 

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A: acceleration at circular turns =v2R

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A: Given data: Mass m = 50.0 kg Force F = 10.0 N angle with horizontal = 20.0o

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A: Given:ε1=23.5 Vε2=10.5 Vε3= 30.0 VR1= 1.20 kΩR2= 14.0 kΩR3= 6.25 kΩ Let the Current I1 passes upwar...

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A: Current (I) = 2 Sin(2000t + 1.801) Ampere Potential difference = 100 Sin(2000t + 0.873) volts   

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A: Dimension of a quantity and its change has same value. 

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Projectile Motion