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adelina 88 [10]

3 years ago

14

Each tire on a car has a radius of 0.330 m and is rotating with an angular speed of 11.7 revolutions/s. Find the linear speed v

of the car, assuming that the tires are not slipping against the ground. v

1 answer:

Eduardwww [97]3 years ago

7 0

Answer:

The linear speed of the car, v, is 24.26 m/s

Explanation:

Given;

radius of the car's tire, r = 0.330 m

angular speed of the car, ω = 11.7 revolutions/s

The angular speed of the car in radian per second:

$\omega = 11.7 \ \frac{rev}{s} \times \frac{2\pi \ rad}{1 \ rev} \\\\\omega = 73.523 \ rad/s$

The linear speed of the car, v, is calculated as;

v = ωr

v = 73.523 rad/s x 0.33 m

v = 24.26 m/s

Therefore, the linear speed of the car, v, is 24.26 m/s

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In a flying ski jump, the skier acquires a speed of 110 km/h by racing down a steep hill and then lifts off into the air from a

matrenka [14]

Answer:

Approximately $\displaystyle\rm \left[ \begin{array}{c}\rm191\; m\\\rm-191\; m\end{array}\right]$ .

Explanation:

Consider this $45^{\circ}$ slope and the trajectory of the skier in a cartesian plane. Since the problem is asking for the displacement vector relative to the point of "lift off", let that particular point be the origin $(0, 0)$ .

Assume that the skier is running in the positive x-direction. The line that represents the slope shall point downwards at $45^{\circ}$ to the x-axis. Since this slope is connected to the ramp, it should also go through the origin. Based on these conditions, this line should be represented as $y = -x$ .

Convert the initial speed of this diver to SI units:

$\displaystyle v = \rm 110\; km\cdot h^{-1} = 110 \times \frac{1}{3.6} = 30.556\; m\cdot s^{-1}$ .

The question assumes that the skier is in a free-fall motion. In other words, the skier travels with a constant horizontal velocity and accelerates downwards at $g$ ( $g \approx \rm -9.81\; m\cdot s^{-2}$ near the surface of the earth.) At $t$ seconds after the skier goes beyond the edge of the ramp, the position of the skier will be:

$x$ -coordinate: $30.556t$ meters (constant velocity;)
$y$ -coordinate: $\displaystyle -\frac{1}{2}g\cdot t^{2} = -\frac{9.81}{2}\cdot t^{2}$ meters (constant acceleration with an initial vertical velocity of zero.)

To eliminate $t$ from this expression, solve the equation between $t$ and $x$ for $t$ . That is: express $t$ as a function of $x$ .

$x = 30.556\;t\implies \displaystyle t = \frac{x}{30.556}$ .

Replace the $t$ in the equation of $y$ with this expression:

$\begin{aligned} y = &-\frac{9.81}{2}\cdot t^{2}\\ &= -\frac{9.81}{2} \cdot \left(\frac{x}{30.556}\right)^{2}\\&= -0.0052535\;x^{2}\end{aligned}$ .

Plot the two functions:

$y = -x$ ,
$\displaystyle y= -0.0052535\;x^{2}$ ,

and look for their intersection. Refer to the diagram attached.

Alternatively, equate the two expressions of $y$ (right-hand side of the equation, the part where $y$ is expressed as a function of $x$ .)

$-0.0052535\;x^{2} = -x$ ,

$\implies x = 190.35$ .

The value of $y$ can be found by evaluating either equation at this particular $x$ -value: $x = 190.35$ .

$y = -190.35$ .

The position vector of a point $(x, y)$ on a cartesian plane is $\displaystyle \left[\begin{array}{l}x \\ y\end{array}\right]$ . The coordinates of this skier is approximately $(190.35, -190.35)$ . The position vector of this skier will be $\displaystyle\rm \left[ \begin{array}{c}\rm191\\\rm-191\end{array}\right]$ . Keep in mind that both numbers in this vectors are in meters.

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3 years ago

If one of two interacting charges is doubled, the force between the charges will _____________.

malfutka [58]

If one of two interacting charges is doubled, the force between the charges will double.

Explanation:

The force between two charges is given by Coulomb's law

$F=\frac{k q1 q2}{r^{2}}$

K=constant= 9 x 10⁹ N m²/C²

q1= charge on first particle

q2= charge on second particle

r= distance between the two charges

Now if the first charge is doubled,

we get $F'=\frac{k (2q1) q2}{r^{2}}$

F'= 2 F

Thus the force gets doubled.

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Answer:

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