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sesenic [268]
4 years ago
6

A civil engineer wishes to redesign the curved roadway in the example What is the Maximum Speed of the Car? in such a way that a

car will not have to rely on friction to round the curve without skidding. In other words, a car of mass m moving at the designated speed can negotiate the curve even when the road is covered with ice. Such a road is usually banked, which means that the roadway is tilted toward the inside of the curve. Suppose the designated speed for the road is to be v = 12.8 m/s (28.6 mi/h) and the radius of the curve is r = 37.0 m. At what angle should the curve be banked?
Physics
1 answer:
vlabodo [156]4 years ago
3 0

Answer:

24.3 degrees

Explanation:

A car traveling in circular motion at linear speed v = 12.8 m/s around a circle of radius r = 37 m is subjected to a centripetal acceleration:

a_c = \frac{v^2}{r} = \frac{12.8^2}{37} = 4.43 m/s^2

Let α be the banked angle, as α > 0, the outward centripetal acceleration vector is split into 2 components, 1 parallel and the other perpendicular to the road. The one that is parallel has a magnitude of 4.43cosα and is the one that would make the car slip.

Similarly, gravitational acceleration g is split into 2 component, one parallel and the other perpendicular to the road surface. The one that is parallel has a magnitude of gsinα and is the one that keeps the car from slipping outward.

So gsin\alpha = 4.43cos\alpha

\frac{sin\alpha}{cos\alpha} = \frac{4.43}{g} = \frac{4.43}{9.81} = 0.451

tan\alpha = 0.451

\alpha = tan^{-1}0.451 = 0.424 rad = 0.424*180/\pi \approx 24.3^0

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A ball of radius r rolls on the inside of a track of radius R. If the ball starts from rest at the vertical edge of the track, f
meriva

Answer:

v=\sqrt{\dfrac{10g(R-r)}{7}}

Explanation:

Given that

Radius of track = R

Radius of ball = r

The ball can be treated as solid sphere, so

The moment of inertia of ball

I=\dfrac{2}{5}mr^2

When the ball reach at the lowest position then it will have both angular and linear speed.

Condition for  rolling without slipping       v= ωr

Form energy conservation

mgR=mgr+\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2

 v= ωr

I=\dfrac{2}{5}mr^2

mgR=mgr+\dfrac{1}{2}mv^2+\dfrac{1}{2}\times \dfrac{2}{5}mr^2\omega^2

mg(R-r)=\dfrac{1}{2}mv^2+\dfrac{1}{2}\times \dfrac{2}{5}mv^2

2mg(R-r)=mv^2+\dfrac{2}{5}mv^2

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v=\sqrt{\dfrac{10g(R-r)}{7}}

3 0
3 years ago
(a) Without the wheels, a bicycle frame has a mass of 6.75 kg. Each of the wheels can be roughly modeled as a uniform solid disk
adelina 88 [10]

Answer:

a) Ktotal = 71.85 J

b) Ktotal = 71.85 J

Explanation:

a) The total kinetic energy is that of the total mass of the bicycle plus the rotational kinetic energy of the two wheels. The linear speed of the circumference of the wheels matches the forward speed of the bicycle, so their angular speed is  

ω = v/r

The moment of inertia of one solid disk bicycle wheel is  

I = 0.5*m₂*r²

And the rotational kinetic energy of one wheel is

Kr = 0.5*I*ω² = 0.5*(0.5*m₂*r²)*(v/r)² = 0.25*m₂*v²

The total kinetic energy is then that of the frame and wheels plus the rotational kinetic energy.

Ktotal = 0.5*(m₁ + 2*m₂)*v² + 2*(0.25*m₂*v²)

⇒  Ktotal = 0.5*v²*(m₁ + 3*m₂)

where

m₁ = 6.75 Kg

m₂ = 0.820 kg

v = 3.95 m/s

then

⇒  Ktotal = 0.5*(3.95 m/s)²*(6.75 Kg + 3*0.820 kg)

⇒  Ktotal = 71.85 J

b) We can apply the same equation obtained before

⇒  Ktotal = 0.5*v²*(m₁ + 3*m₂)

where

m₁ = 675 Kg

m₂ = 82.0 kg

v = 0.395 m/s

then

⇒  Ktotal = 0.5*(0.395 m/s)²*(675 Kg + 3*82 kg)

⇒  Ktotal = 71.85 J

3 0
3 years ago
A tiger leaps horizontally from a high rock with a speed of 4.5 m/s. She lands 50 m away
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3 years ago
If the voltage amplitude across an 8.50-nF capacitor is equal to 12.0 V when the current amplitude through it is 3.33 mA, the fr
xxTIMURxx [149]
<h2>Answer:</h2>

5.19 x 10³Hz

<h2>Explanation:</h2>

The capacitive reactance, X_{C}, which is the opposition given to the flow of current through the capacitor is given by;

X_C = \frac{1}{2\pi fC }

Where;

f = frequency of the signal through the capacitor

C = capacitance of the capacitor.

Also, from Ohm's law, the voltage(V) across the capacitor is given by the product of current(I) and the capacitive reactance. i.e;

V = I x X_{C}             [Substitute the value of

=> V = I x \frac{1}{2\pi fC}      [Make f the subject of the formula]

=> f = \frac{I}{2\pi VC}                    ---------------------(i)

From the question;

I = 3.33mA = 0.00333A

C = 8.50nF = 8.50 x 10⁻⁹F

V = 12.0V

Substitute these values into equation (i) as follows;

f = \frac{0.00333}{2 * 3.142 * 12.0 * 8.50 * 10^{-9}}            [Taking \pi = 3.142]

f = 5.19 x 10³Hz

Therefore, the frequency is closest to f = 5.19 x 10³Hz

4 0
3 years ago
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