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3 years ago

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For a vehicle to negotiate a banked curve in poor weather conditions, where the force of friction f = 0, for a given velocity v

=25 m/s, and the angle at which the curve is banked is 15, find the radius of curvature R.

1 answer:

djverab [1.8K]3 years ago

7 0

Answer:

$R=240m$

Explanation:

From the question we are told that:

Velocity $v=25m/s$

Force of friction f = 0

Angle $\theta=15$

Generally the equation for Radius of curvature is mathematically given by

$R=frac{v^2}{tan\theta *g}$

$R=frac{25^2}{tan 15 *9.81}$

$R=240m$

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A source emits sound with a frequency of 860 Hz. It is moving at 20.0 m/s toward a stationary reflecting wall. If the speed of s

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Answer:

$f_o=860\ Hz$

Explanation:

Given:

original frequency of sound wave, $f=860\ Hz$

speed of the sound source, $v_s=20\ m.s^{-1}$

original speed of sound wave from the source, $s=343\ m.s^{-1}$

<u>According to the Doppler's effect:</u>

$\frac{f_s}{f_o} =\frac{s+v_s}{s-v_o}$

The sound is reflected from the wall and the source is moving towards the wall and observer is also riding the same source.

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3 years ago

Read 2 more answers

In order to determine the coefficients of friction between rubber and various surfaces, a student uses a rubber eraser and an in

Strike441 [17]

Answer:

μs = 0.75

μk = 0.58

Explanation:

From a force diagram:

$m*g*sin \theta - Ff=m*a$ (1)

$N-m*g*cos \theta = 0$ (2)

When it starts slipping, friction force is the maximum and acceleration is 0. Replacing these conditions on (1):

$m*g*sin \theta - \mu*m*g*cos \theta=0$ Solving for μs:

$\mu=tan \theta$

μs = tan 36.7° = 0.75

When it moves at constant speed, friction force is kinetic friction and acceleration is 0. With these conditions the coefficient is:

μk = tan 30.1° = 0.58

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4 years ago

Derive the formula for the moment of inertia of a uniform, flat, rectangular plate of dimensions l and w, about an axis through

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Answer:

A uniform thin rod with an axis through the center

Consider a uniform (density and shape) thin rod of mass M and length L as shown in (Figure). We want a thin rod so that we can assume the cross-sectional area of the rod is small and the rod can be thought of as a string of masses along a one-dimensional straight line. In this example, the axis of rotation is perpendicular to the rod and passes through the midpoint for simplicity. Our task is to calculate the moment of inertia about this axis. We orient the axes so that the z-axis is the axis of rotation and the x-axis passes through the length of the rod, as shown in the figure. This is a convenient choice because we can then integrate along the x-axis.

We define dm to be a small element of mass making up the rod. The moment of inertia integral is an integral over the mass distribution. However, we know how to integrate over space, not over mass. We therefore need to find a way to relate mass to spatial variables. We do this using the linear mass density of the object, which is the mass per unit length. Since the mass density of this object is uniform, we can write

λ = m/l (orm) = λl

If we take the differential of each side of this equation, we find

d m = d ( λ l ) = λ ( d l )

since

λ

is constant. We chose to orient the rod along the x-axis for convenience—this is where that choice becomes very helpful. Note that a piece of the rod dl lies completely along the x-axis and has a length dx; in fact,

d l = d x

in this situation. We can therefore write

d m = λ ( d x )

, giving us an integration variable that we know how to deal with. The distance of each piece of mass dm from the axis is given by the variable x, as shown in the figure. Putting this all together, we obtain

I=∫r2dm=∫x2dm=∫x2λdx.

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I=L/2∫−L/2x2λdx=λx33|L/2−L/2=λ(13)[(L2)3−(−L2)3]=λ(13)L38(2)=ML(13)L38(2)=112ML2.

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3 years ago

A brick lands 10.1 m from the base of a building. If it was given an initial velocity of 8.6 m/s [61º above the horizontal], how

Montano1993 [528]

<h2>Answer: 10.52m</h2><h2 />

First, we have to establish the <u>reference system</u>. Let's assume that the building is on the negative y-axis and that the brick was thrown at the origin (see figure attached).

According to this, the initial velocity $V_{o}$ has two components, because the brick was thrown at an angle $\alpha=61\º$ :

$V_{ox}=V_{o}cos\alpha$ (1)

$V_{ox}=8.6\frac{m}{s}cos(61\º)=4.169\frac{m}{s}$ (2)

$V_{oy}=V_{o}sin\alpha$ (3)

$V_{oy}=8.6\frac{m}{s}sin(61\º)=7.521\frac{m}{s}$ (4)

As this is a projectile motion, we have two principal equations related:

<h2>In the x-axis: </h2>

$X=V_{ox}.t$ (5)

Where:

$X=10.1m$ is the distance where the brick landed

$t$ is the time in seconds

If we already know $X$ and $V_{ox}$ , we have to find the time (we will need it for the following equation):

$t= \frac{X}{ V_{ox}}$ (6)

$t=2.42s$ (7)

<h2>In the y-axis: </h2>

$-y=V_{oy}.t+\frac{1}{2}g.t^{2}$ (8)

Where:

$y$ is the height of the building (<u>in this case it has a negative sign because of the reference system we chose)</u>

$g=-9.8\frac{m}{s^{2}}$ is the acceleration due gravity

Substituting the known values, including the time we found on equation (7) in equation (8), we will find the height of the building:

$-y=(7.521\frac{m}{s})(2.42s)+\frac{1}{2}(-9.8\frac{m}{s^{2}}).(2.42s)^{2}$ (9)

$-y=-10.52m$ (10)

Multiplying by -1 each side of the equation:

$y=10.52m$ >>>>This is the height of the building

3 0

3 years ago