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Viktor [21]
1 year ago
7

A 6-ft-long fishing rod AB is securely anchored in the sand of a beach. After a fish takes the bait, the resulting force in the

line is 8.6 lb. Determine the moment about A of the force exerted by the line at B. (Round the final answer to one decimal place.)
Physics
1 answer:
babunello [35]1 year ago
4 0

Answer:

Moment of Force, M = 51.6 lbf-ft

Explanation:

Given,

The fishing rod AB of length, L = 6 ft

The force acting on the rod, F = 8.6 lbf

The moment of force about A is given as the product of force and perpendicular distance. It is given by the formula

                                    M = F x L  

Substituting the given values in the above equation

                                     M = 8.6 lbf  x 6 ft

                                         = 51.6 lbf-ft

The moment about A of the force exerted by the line at B is, M = 51.6 lbf-ft

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Explanation:

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A toy car is given an initial velocity of 5.0 m/s and experiences a constant acceleration of 2.0 m/s656-03-02-00-00_files/i02900
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It would be 17 m/s

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5 0
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In 1991 four English teenagers built an eletric car that could attain a speed of 30.0m/s. Suppose it takes 8.0s for this car to
ivanzaharov [21]

a=Δv/Δt=(30.0-18.0)/8.0=12.0/8.0=1.5 m/s²


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In a series circuit, a generator (1300 Hz, 12.0 V) is connected to a 14.0- resistor, a 4.40-μF capacitor, and a 6.00-mH inductor
klemol [59]

Answer:

(a) 2.8 V

(b) 5.6 V

(c) 9.8 V

Explanation:

Given:

Frequency of the generator (f) = 1300 Hz)

Terminal voltage (V) =12.0 V

Resistance of resistor (R) = 14.0 Ω

Capacitance of capacitor (C) = 4.40 μF = 4.40 × 10⁻⁶ F

Inductance of the inductor (L) = 6.00 mH = 6.00 × 10⁻³ H

In order to find the voltages across each, we first need to find the reactance and impedance.

Reactance of the inductor is given as:

X_L=2\pi f L\\\\X_L=2\times 3.14\times 1300\times 6.00\times 10^{-3}\\\\X_L=49\ \Omega

Reactance of the capacitor is given as:

X_C=\frac{1}{2\pi fC}\\\\X_C=\frac{1}{2\times 3.14\times 1300\times 4.40\times 10^{-6}}\\\\X_C =28\ \Omega

Now, impedance is given as:

Z=\sqrt{X_L^2+X_C^2}\\\\Z=\sqrt{(49)^2+(28)^2}\\\\Z=\sqrt{3185}=56.4\ \Omega

Current across the circuit is given as:

I=\frac{V}{Z}\\\\I=\frac{12}{56.4}=0.2\ A

As resistor, capacitor and inductor are connected in series, the current across each of them is same and equal to total current in the circuit.

(a)

Voltage across the resistor is given as:

V_R=IR\\\\V_R=0.2\times 14=2.8\ V

Therefore, the voltage across resistor is 2.8 V.

(b)

Voltage across the capacitor is given as:

V_C=IX_C\\\\V_C=0.2\times 28=5.6\ V

Therefore, the voltage across the capacitor is 5.6 V.

(c)

Voltage across the inductor is given as:

V_L=IX_L\\\\V_L=0.2\times 49=9.8\ V

Therefore, the voltage across the inductor is 9.8 V.

6 0
2 years ago
Which of the following sketches represents a possible configuration for this problem?
garri49 [273]
Where are the following sketches?
7 0
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