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Colt1911 [192]
3 years ago
14

A car is driving around a banked curve, with the road surface at an angle of 10.0º. If the radius of curvature of the road is 30

.0 m and the coefficient of static friction between the tires of the car and the road is 0.65, what is the maximum speed (in km/hr) the car can go without skidding?
Physics
1 answer:
IRISSAK [1]3 years ago
7 0

Answer:

maximum speed 56 km/h

Explanation:

To apply Newton's second law to this system we create a reference system with the horizontal x-axis and the Vertical y-axis. In this system, normal is the only force that we must decompose

       sin 10 = Nx / N

      cos 10 = Ny / N

      Ny = N cos 10

     Nx = N sin 10

Let's develop Newton's equations on each axis

X axis

We include the force of friction towards the center of the curve because the high-speed car has to get out of the curve

     Nx + fr = m a

     a = v2 / r

     fr = mu N

     N sin10 + mu N = m v² / r

     N (sin10 + mu) = m v² / r

Y Axis  

     Ny -W = 0

     N cos 10 = mg

Let's solve these two equations,

    (mg / cos 10) (sin 10 + mu) = m v² / r

    g (tan 10 + μ / cos 10) = v² / r

    v² = r g (tan 10 + μ / cos 10)

They ask us for the maximum speed

   v² = 30.0 9.8 (tan 10+ 0.65 / cos 10)

   v² = 294 (0.8364)  

   v = √(245.9)

   v = 15.68 m / s

Let's reduce this to km / h

   v = 15.68 m / s (1 km / 1000m) (3600s / 1h)

   v = 56.45 km / h

This is the maximum speed so you don't skid

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A circuit has an impedance of 5 ohms and a voltage of 100 volts. What is the current flow?
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A circuit with an impedance of 5 ohms and a voltage of 100 volts has a current flow of 20 A.

The expression of an electronic component, circuit, or system's resistance to alternating and/or direct electric current is called impedance, indicated by the letter Z. Resistance and reactance are two distinct scalar (one-dimensional) phenomena that combine to form impedance, a vector (two-dimensional) variable.

Z^{} = \frac{V}{I}

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I = \frac{100}{5}

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6 0
2 years ago
(b) A cylinder of cross-sectional area 0.65m2 and
VLD [36.1K]

Answer:

The volume of the cavity is 0.013m^3

Explanation:

To find the volume of the cavity, the major parameter missing is the diameter of the cavity itself. we can obtain this using the following steps:

Step one:

Obtain the volume of the cylinder by dividing the mass of the cylinder by the density.

Volume of the cylinder = 2.1 / 11.053 =0.19m^{3}

Step two:

From the volume of the cylinder, we can get the radius of the cylinder.

radius = \sqrt{\frac{V}{\pi \times h}}  = \sqrt{\frac{0.19}{\pi \times 0.32}} =0.44m

Step three:

From the cross-sectional area, we can obtain the radius of the cavity.

Let the radius of the cavity be = r, while the radius of the cylinder be = R

CSA of cavity =

\pi({R^2}-r^2) = CSA\\0.65 = \pi (0.32^2-r^2)\\r= 0.115m

Step Four:

calculate the volume of the cavity using volume =\pi r^2 \times h

Recall that the cavity has the same height as the original cylinder

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8 0
3 years ago
A car of mass 940.0 kg accelerates away from an intersection on a horizontal road. When the car speed is 42.5 km/hr (11.8 m/s),
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Answer:

0.39 m/s^2

Explanation:

Parameters given:

Mass of car, m = 940 kg

Speed of car, v = 11.8 m/s

Power supplied by engine, P = 4300 W

To get the acceleration, we must define the relationship between Power and velocity.

Power, P, is given in terms of velocity, v, as:

P = F * v

where F = force

This is because Power is given as:

P = \frac{E}{t} \\\\\\P = \frac{F*d}{t} \\\\\\=> P = F * v

(where E = energy. t = time taken, d = distance moved)

Force, F, is given as:

F = m*a

Therefore, Power will be:

P = m * a * v

Acceleration, a, will then be:

a = \frac{P}{m*v}

a = \frac{4300}{940 * 118} \\\\\\a = 0.39 m/s^2

The acceleration of the car at that time is 0.39 m/s^2

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