How many foot-pounds of work does it take to throw a baseball 90 mph? a baseball weighs 5 oz, or 0.3125 lb?

Physics

1 answer:

NARA [144]3 years ago

5 0

Kinetic energy of the ball is (mv²) / 2, where m is the mass and v is the velocity

So plugging in the mass and the velocity into the kinetic energy expression, you get:

Kinetic energy of the ball = (mv²) / 2

(0.3125/32) times (132)² divided by 2 = 85 ft-lbs

Kinetic energy of the ball = 85 ft-lbs

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A technician builds an RLC series circuit which includes an AC source that operates at a fixed frequency and voltage. At the ope

MariettaO [177]

Answer:

Xc = (0.467 - 0.427j)R

Explanation:

Since the resistance in the circuit is R, the reactance of the inductor is XL and the reactance of the capacitor is XC, then the impedance of the circuit is

Z = √[R² + (XL - XC)²]

Since the inductive reactance XL equals the resistance R, we have that

Z = √[R² + (XL - XC)²]

Z = √[R² + (R - XC)²]

Thus, the current in the circuit is thus I = V/Z = V/√[R² + (R - XC)²]

Now, when the plate separation of the parallel plate capacitor is reduced to one-half its original value, the current doubles. Also, when the plate separation is reduced to half, the capacitance doubles since C ∝ 1/d where C is capacitance and d separation between the plates. Since the capacitance doubles, the new reactance XC' is twice the initial reactance XC. So, XC' = 2XC. Thus the new impedance is thus

Z' = √[R² + (R - XC')²]

Z' = √[R² + (R - 2XC)²]

The new current is I' = V/Z' = V/√[R² + (R - 2XC)²]

Since the current doubles, I' = 2I.

V/√[R² + (R - 2XC)²] = 2V/√[R² + (R - XC)²]

1/√[R² + (R - 2XC)²] = 2/√[R² + (R - XC)²]

√[R² + (R - XC)²] = 2√[R² + (R - 2XC)²]

squaring both sides, we have

[R² + (R - XC)²] = 4[R² + (R - 2XC)²]

expanding the brackets, we have

[R² + R² - 2RXC + XC²] = 4[R² + R² - 4RXC + 4XC²]

[2R² - 2RXC + XC²] = 4[2R² - 4RXC + 4XC²]

2R² - 2RXC + XC² = 8R² - 16RXC + 16XC²

collecting like terms, we have

16RXC - 2RXC + XC² - 16XC² = 8R² - 2R²

14RXC - 15XC² = 6R²

15XC² - 14RXC + 6R² = 0

Using the quadratic formula to find XC, we have

$XC = \frac{-(-14R) +/- \sqrt{(-14R)^{2} - 4 X 15 X 6R^{2} } }{2 X 15}\\= \frac{-(-14R) +/- \sqrt{196R^{2} - 360R^{2} } }{30}\\ \\= \frac{14R +/- \sqrt{- 164R^{2} } }{30}\\ \\= \frac{14R +/- 12.81Ri }{30}\\\\= 0.467R +/- 0.427Ri$