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vladimir1956 [14]
3 years ago
14

A long jumper can jump a distance of 7.4 m when he takes off at an angle of 45° with respect to the horizontal. Assuming he can

jump with the same initial speed at all angles, how much distance does he lose (in m) by taking off at 429?
Physics
1 answer:
GenaCL600 [577]3 years ago
8 0

Answer:

0.02 m

Explanation:

R₁ = initial distance jumped by jumper = 7.4 m

R₂ = final distance jumped by jumper = ?

θ₁ = initial angle of jump = 45°

θ₂ = final angle of jump = 42.9°

v = speed at which jumper jumps at all time

initial distance jumped is given as

R_{1}=\frac{v^{2}Sin2\theta _{1} }{g}

final distance jumped is given as

R_{2}=\frac{v^{2}Sin2\theta _{2} }{g}

Dividing final distance by initial distance

\frac{R_{2}}{R_{1}}=\frac{Sin2\theta _{1}}{Sin2\theta _{2}}

\frac{R_{2}}{7.4}=\frac{Sin2(42.9)}{Sin2(45))}

R_{2} =7.38

distance lost is given as

d = R_{1} - R_{2}

d = 7.4 - 7.38

d = 0.02 m

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Two capacitors of capacitances 25 µF and 50 µF are connected in series with a 33-V battery. How much energy is stored in the 25-
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6.05×10⁻³ J

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1/Ct = 1/C1+1/C2

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Using

Q = CV.................... Equation 2

Where Q = Charge, V = Voltage.

Given: V = 33 V, C = 16.67 µF = 16.67×10⁻⁶ F

Substitute into equation 2

Q = 33(16.67×10⁻⁶)

Q = 5.5×10⁻⁴ C.

Since both capacitors are connected in series, the same amount of charge flows through them.

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E = 1/2Q²/C.................. Equation 3

Where E = Energy stored in the 25-µF capacitor

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Answer:

a) 3.0×10⁸ m

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