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3 years ago

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Suppose a NASCAR race car rounds one end of the Martinsville Speedway. This end of the track is a turn with a radius of approxim

ately 57.0 m . If the track is completely flat and the race car is traveling at a constant 30.5 m/s (about 68 mph ) around the turn, what is the race car's centripetal (radial) acceleration

1 answer:

umka21 [38]3 years ago

7 0

Answer:

$a=16.32\ m/s^2$

Explanation:

Given that,

The radius of the track, r = 57 m

The speed of a race car, v = 30.5 m/s

We need to find the centripetal acceleration of the car. Its formula that is use to find it is given by :

$a=\dfrac{v^2}{r}\\\\a=\dfrac{(30.5)^2}{57}\\\\=16.32\ m/s^2$

So, the car's centripetal acceleration is $16.32\ m/s^2$ .

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Which of the following is true concerning hurricanes?

densk [106]

Answer:

B.

Explanation:

B. is correct because every time usually when a hurricane hits it causes flooding causing multiple homes to be ruined. It is a known fact that usually hurricanes start close or in along with tornadoes, the water, meaning once they hit land they have some water to flood that land with.

C. is absolutely false hurricanes have VERY strong wind gusts.

D. is absolutely wrong they do alter landscapes by ripping trees and plants and houses out of the ground making the landscapes look different.

A. is wrong they tend to deposit and remove sediment evenly.

<em><u>~ LadyBrain</u></em>

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3 years ago

Read 2 more answers

The train passes point A with a speed of 30 m/s and begins to decrease its speed at a constant rate of at = - 0.25 m/s^2. Determ

prisoha [69]

Explanation:

At point B, the velocity speed of the train is as follows.

$\nu^{2}_{B} = \nu^{2}_{A} + 2a_{t} (s_{B} - s_{A})$

= $(30)^{2} + 2(-0.25(412 - 0))$

= 26.34 m/s

Now, we will calculate the first derivative of the equation of train.

y = $200 e^{\frac{x}{1000}}$

$\frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}$

Now, second derivative of the train is calculated as follows.

$\frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}$

$\frac{d^{2}y}{dx^{2}} = 0.2 (10^{-3}) e^{\frac{x}{1000}}$

Radius of curvature of the train is as follows.

$\rho = \frac{[1 + (\frac{dy}{dx})^{2}]^{\frac{3}{2}}}{\frac{d^{2}y}{dx^{2}}}$

= $\frac{[1 + 0.2e^{\frac{400}{1000}}^{2}]^{\frac{3}{2}}}{0.2(10^{-3})e^{\frac{400}{1000}}}$

= 3808.96 m

Now, we will calculate the normal component of the train as follows.

$a_{n} = \frac{\nu^{2}_{B}}{\rho}$

= $\frac{(26.34)^{2}}{3808.96}$

= 0.1822 $m/s^{2}$

The magnitude of acceleration of train is calculated as follows.

a = $\sqrt{(a_{t})^{2} + (a_{n})^{2}}$

= $\sqrt{(-0.25)^{2} + (0.1822)^{2}}$

= $0.309 m/s^{2}$

Thus, we can conclude that magnitude of the acceleration of the train when it reaches point B, where sAB = 412 m is $0.309 m/s^{2}$ .

6 0

3 years ago

A biker first accelerates from 0.0 m/s to 6.0 m/s in 6 s, then continues at this speed for 5 s. What is the total distance trave

Svetradugi [14.3K]

Answer:

48m

Explanation:

Given the following data;

Initial velocity = 0m/s

Final velocity = 6m/s

Time, t = 6 secs

Time, T2 = 5 secs

Mathematically, acceleration is given by the equation;

$Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}$

Substituting into the equation;

$a = \frac{6 - 0}{6}$

$a = \frac{6}{6}$

Acceleration, a = 1m/s²

<u>To find the distance covered in the first phase;</u>

<em>Solving for distance, we would use the second equation of motion;</em>

$S = ut + \frac {1}{2}at^{2}$

<em>Substituting the values into the equation;</em>

$S = 0(6) + \frac {1}{2}*1*(6)^{2}$

$S = 0 + \frac {1}{2}*1*36$

$S = 0.5 *36$

Distance, S1 = 18m

<u>For the second phase, time T2 = 5 secs;</u>

<em>Mathematically, speed is given by the equation;</em>

$Speed = \frac{distance}{time}$

<em>Making distance the subject of formula, we have;</em>

$Distance, S = speed * time$

<em>Substituting into the above equation;</em>

$Distance, S = 6 * 5$

Distance, S2 = 30m

Total distance = S1 + S2 = 18m + 30m = 48m

Total distance = 48m

<em>Therefore, the total distance traveled by the biker is 48m.</em>

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3 years ago

Two titanium spheres approach each other head-on with the same speed and collide elastically. After the collision, one of the sp

Thepotemich [5.8K]

I know i did part a correctly. heres what i did: momentum is conserved: m1 * u - m2 * u = m2 * v or (m1 - m2) * u = m2 * v Also, for an elastic head-on collision, we know that the relative velocity of approach = relative velocity of separation (from conservation of energy), or, for this problem, 2u = v Then (m1 - m2) * u = m2 * 2u m1 - m2 = 2 * m2 m1 = 3 * m2 m1 is the sphere that remained at rest (hence its absence from the RHS), so m2 = 0.3kg / 3 m2 = 0.1 kg b) this part confuses me, heres what i did (m1 - m2) * u = m2 * v (.3kg - .1kg)(2.0m/s) = .1kg * v .4 kg = .1 v v = 4 m/s What my teacher did: (.3g - .1g) * 2.0m/s = (.3g + .1g) * v I understand the left hand side but i dont get the right hand side. Why is m1 added to m2 when m1 is at rest which makes its v = zero?? v = +1.00m/s since the answer is positive, what does that mean? Also, if v was -1.00m/s what would that mean? thanks!

<span>Reference https://www.physicsforums.com/threads/elastic-collision-with-conservation-of-momentum-problem.651261...</span>

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3 years ago

A bicyclist was moving at a rate of 3 m/s, and then sped up to 4 m/s. If the cyclist has a mass of 100 kg, how much work was nee

kotegsom [21]

Answer:

i got b but urs is a little different tell me if right

Explanation:

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3 years ago