Ask question

Ask question

All categories

3 years ago

5

A car initially traveling at 27.2 m/s undergoes a constant negative acceleration of magnitude 1.90 m/s2 after its brakes are app

lied. (a) How many revolutions does each tire make before the car comes to a stop, assuming the car does not skid and the tires have radii of 0.325 m

1 answer:

zubka84 [21]3 years ago

6 0

Answer:

Therefore, the revolutions that each tire makes is:

$\Delta \theta=22\: rev$

Explanation:

We can use the following equation:

$\omega_{f}^{2}=\omega_{i}^{2}-2\alpha \Delta \theta$ (1)

The angular acceleration is:

$a_{tan}=\alpha R$

$\alpha=\frac{1.9}{0.325}$

$\alpha=5.85\: rad/s^{2}$

and the initial angular velocity is:

$\omega_{i}=\frac{v}{R}$

$\omega_{i}=\frac{27.2}{0.325}$

$\omega_{i}=83.69\: rad/s$

Now, using equation (1) we can find the revolutions of the tire.

$0=83.69^{2}-2*25.85 \Delta \theta$

$\Delta \theta=135.47\: rad$

Therefore, the revolutions that each tire makes is:

$\Delta \theta=22\: rev$

I hope it helps you!

You might be interested in

Which particle would have the slowest rate of deposition? A.round particle B.very large particle C.particle with sharp ends D.pa

lisabon 2012 [21]

The particle with sharp ends have the slowest rate of deposition

Answer: Option C

<u>Explanation:</u>

As per aerosol physics, deposition is a process where aerosol particles accumulate or settle on solid surfaces. Thereby, it reduces the concentration of particles in the air. Deposition velocity (rate of deposition) defines from F = vc, where v is deposition rate, F denotes flux density and c refers concentration.

Deposition velocity is slowest for particles of intermediate-sized particles because the frictional force offers resistance to the flow. Density is directly proportional to the deposition rate so clearly shows that high-density particles settle faster. Due to friction, round and large-sized particles deposit faster than oval/flattened sediments.

3 0

3 years ago

Read 2 more answers

Two billiard balls having the same mass and speed roll directly toward each other. What is their combined momentum after they me

stealth61 [152]

Their combined momentum after they meet is 0 .

3 0

3 years ago

In your own words, explain how Doppler radar works. Describe the properties of electromagnetic waves and interactions that make

lisov135 [29]

If you stand up in a big room and echo, your voice will echo from the walls. As long as the room is empty. Since the speed of sound is constant, depending on air density, the more humid the air the faster and farther sound travels. The speed of sound is constant, you could measure the time it takes for your voice to echo off the walls. The same thing happens with Doppler radar, but it’s not voice, it has higher frequency signals.<span> </span>

6 0

3 years ago

Read 2 more answers

A copper wire has a square cross section 2.1 mm on a side. The wire is 4.3 m long and carries a current of 3.6 A. The density of

aleksley [76]

Explanation:

Below is an attachment containing the solution.

4 0

3 years ago

Two transverse waves travel along the same taut string. Wave 1 is described by y1(x, t) = A sin(kx - ωt), while wave 2 is descri

Vadim26 [7]

Answer:

6) Wave 1 travels in the positive x-direction, while wave 2 travels in the negative x-direction.

Explanation:

What matters is the part $kx \pm \omega t$ , the other parts of the equation don't affect time and space variations. We know that when the sign is - the wave propagates to the positive direction while when the sign is + the wave propagates to the negative direction, but <em>here is an explanation</em> of this:

For both cases, + and -, after a certain time $\delta t$ ( $\delta t >0$ ), the displacement <em>y</em> of the wave will be determined by the $kx\pm\omega (t+\delta t)$ term. For simplicity, if we imagine we are looking at the origin (x=0), this will be simply $\pm \omega (t+\delta t)$ .

To know which side, right or left of the origin, would go through the origin after a time $\delta t$ (and thus know the direction of propagation) we have to see how we can achieve that same displacement <em>y</em> not by a time variation but by a space variation $\delta x$ (we would be looking where in space is what we would have in the future in time). The term would be then $k(x+\delta x)\pm\omega t$ , which at the origin is $k \delta x \pm \omega t$ . This would mean that, when the original equation has $kx+\omega t$ , we must have that $\delta x>0$ for $k\delta x+\omega t$ to be equal to $kx+\omega\delta t$ , and when the original equation has $kx-\omega t$ , we must have that $\delta x$ for $k\delta x-\omega t$ to be equal to $kx-\omega \delta t$

<em>Note that their values don't matter, although they are a very small variation (we have to be careful since all this is inside a sin function), what matters is if they are positive or negative and as such what is possible or not .</em>

<em />

In conclusion, when $kx+\omega t$ , the part of the wave on the positive side ( $\delta x>0$ ) is the one that will go through the origin, so the wave is going in the negative direction, and viceversa.

4 0

3 years ago