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4 years ago

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An automobile with a tangential speed of 54.1 km/h follows a circular road that has a radius of 41.6 m. The automobile has a mas

s of 1260 kg. The pavement is wet and oily, so the coefficient of kinetic friction between the car's tires and the pavement is only 0.500.

1 answer:

Viefleur [7K]4 years ago

4 0

1) Available force of friction: 6174 N

2) No

Explanation:

1)

The magnitude of the frictional force between the car's tires and the pavement of the road is given by

$F_f=\mu mg$

where

$\mu$ is the coefficient of friction

m is the mass of the car

g is the acceleration of gravity

For the car in this problem, we have:

$\mu=0.500$ (coefficient of friction)

m = 1260 kg (mass of the car)

$g=9.8 m/s^2$

Therefore, the force of friction is

$F_f=(0.500)(1260)(9.8)=6174 N$

2)

In order to mantain the car in circular motion, the force of friction must be at least equal to the centripetal force.

The centripetal force is given by

$F=m\frac{v^2}{r}$

where

m is the mass of the car

v is the tangential speed

r is the radius of the curve

In this problem, we have

m = 1260 kg

$v=54.1 km/h =15.0 m/s$ is the tangential speed

r = 41.6 m is the radius of the curve

Therefore, the centripetal force is

$F=(1260)\frac{15.0^2}{41.6}=6814 N$

Therefore, the force of friction is not enough to keep the car in the curve, since $F_f$

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Answer:

$d=1.84\ mm$

Explanation:

<u>Capacitance</u>

A two parallel-plate capacitor has a capacitance of

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A = area of the plates = $\pi r^2$

d = separation of the plates

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We need to compute C. We'll use the circuit parameters for that. The reactance of a capacitor is given by

$\displaystyle X_c=\frac{1}{wC}$

where w is the angular frequency

$w=2\pi f=2\pi \cdot 36000=226194.67\ rad/s$

Solving for C

$\displaystyle C=\frac{1}{wX_c}$

The reactance can be found knowing the total impedance of the circuit:

$Z^2=R^2+X_c^2$

Where R is the resistance, $R=15 K\Omega=15000\Omega$ . Solving for Xc

$X_c^2=Z^2-R^2$

The magnitude of the impedance is computed as the ratio of the rms voltage and rms current

$\displaystyle Z=\frac{V}{I}$

The rms current is the peak current Ip divided by $\sqrt{2}$ , thus

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$I_p=0.65\ mA/1000=0.00065\ A$

Now collect formulas

$\displaystyle X_c^2=Z^2-R^2=\left(\frac{\sqrt{2}V}{I_p}\right)^2-R^2$

Or, equivalently

$\displaystyle X_c=\sqrt{\frac{2V^2}{I_p^2}-R^2}$

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3 years ago

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Answer:

(a) The absolute pressure at the bottom of the freshwater lake is 395.3 kPa

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Explanation:

(a)

The absolute pressure is given by the formula

$P = P_{o} + \rho gh$

Where $P$ is the absolute pressure

$P_{o}$ is the atmospheric pressure

$\rho$ is the density

$g$ is the acceleration due to gravity (Take $g = 9.8 m/s^{2}$ )

h is the height

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h = 30.0 m

$\rho$ = 1.00 × 10³ kg/m³ = 1000 kg/m³

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Using the formula

$P = P_{o} + \rho gh$

P = 101300 + (1000×9.8×30.0)

P = 101300 + 294000

P =395300 Pa

P = 395.3 kPa

Hence, the absolute pressure at the bottom of the freshwater lake is 395.3 kPa

(b)

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From

P = F/A

Where P is the pressure

F is the force

and A is the area

Then, F = P × A

Here, The area will be area of the window of the underwater vehicle.

Diameter of the circular window = 34.1 cm = 0.341 m

From Area = πD²/4

Then, A = π×(0.341)²/4 = 0.0913269 m²

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F = 36101.5 N

Hence, the force exerted by the water on the window is 36101.5 N

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3 years ago

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3 years ago

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Citrus2011 [14]

Answer:

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Explanation:

Solution:-

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