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Elena-2011 [213]

3 years ago

11

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 27.5 m/s (about 62 mph ) around the turn, what is the race car's centripetal (radial) acceleration?
What is the force responsible for the centripetal acceleration in this case?
friction, normal, gravity, or weight?

1 answer:

valkas [14]3 years ago

5 0

Answer:

$a_c = 13.26 m/s^2$

Friction Force

Explanation:

As we know that centripetal force is the product of mass and centripetal acceleration

so we know that

$a_c = \frac{v^2}{R}$

so here we have

$v = 27.5 m/s$

$R = 57 m$

so we have

$a_c = \frac{27.5^2}{57}$

$a_c = 13.26 m/s^2$

This acceleration is given by the force which may be towards the center of the circular path

Here in the above case it is possible due to friction force.

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A uniform disk with radius 0.650 m

VashaNatasha [74]

Answer:

$a = 13.758\,\frac{m}{s^{2}}$

Explanation:

First, the instant associated to the angular displacement is:

$(1.10\,\frac{rad}{s} )\cdot t + (6.30\,\frac{rad}{s^{3}} )\cdot t^{2} - 0.628\,rad = 0$

Roots of the second-order polynomial are:

$t_{1} \approx 0.240\,s, t_{2} \approx -0.415\,s$

Only the first root is physically reasonable.

The angular velocity is obtained by deriving the angular displacement function:

$\omega (0.240\,s) = 1.10\,\frac{rad}{s}+ (12.6\,\frac{rad}{s^{2}})\cdot (0.240\,s)$

$\omega (0.240\,s) = 4.124\,\frac{rad}{s}$

The angular acceleration is obtained by deriving the previous function:

$\alpha (0.240\,s) = 12.6\,\frac{rad}{s^{2}}$

The resultant linear acceleration on the rim of the disk is:

$a_{t} = (0.650\,m)\cdot (12.6\,\frac{rad}{s^{2}} )$

$a_{t} = 8.190\,\frac{m}{s^{2}}$

$a_{n} = (0.650\,m)\cdot (4.124\,\frac{rad}{s} )^{2}$

$a_{n} = 11.055\,\frac{m}{s^{2}}$

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

$a = 13.758\,\frac{m}{s^{2}}$

3 0

2 years ago

The whole body metabolic rate of a bottlenose dolphin is 8000kcalsday. The dolphin weighs 190kg. What is the mass specific metab

Komok [63]

Answer: 42.1

Explanation:

Mass specific metabolic rate of a dolphin can be defined as the rate at which the dolphin consume energy per unit mass of body weight.

R = E/M

Where R = mass specific metabolic rate

E = Energy consumption = 8000kcalsday

M = mass = 190kg

R = 8000kcalsday/190kg

R = 42.1

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3 years ago

What is a lunar month

Ira Lisetskai [31]

Answer:

It's the duration between successive new moons. Also called a lunation or synodic month, it has a mean period of 29.53059 days (29 days 12 hours and 44 minutes).

7 0

2 years ago

Read 2 more answers

How long will it take a plane to fly 1256km<br> if it travels 500km/hr?

expeople1 [14]

Answer:

<h3>The answer is 2.51 s</h3>

Explanation:

The time taken can be found by using the formula

$t = \frac{d}{v} \\$

d is the distance

v is the velocity

From the question we have

$t = \frac{1256}{500} \\ = 2.512$

We have the final answer as

<h3>2.51 s</h3>

Hope this helps you

4 0

2 years ago

The parachute on a drag racing car deploys at the end of a run. If the car has a mass of 820 kg and the car is moving 36 m/s, wh

Lelechka [254]

In order to determine the required force to stop the car, proceed as follow:

Calculate the deceleration of the car, by using the following formula:

$v^2=v^2_o-2ax$

where,

v: final speed = 0m/s (the car stops)

vo: initial speed = 36m/s

x: distance traveled = 980m

a: deceleration of the car= ?

Solve the equation above for a, replace the values of the other parameters and simplify:

$\begin{gathered} a=\frac{v^2_o-v^2}{2x} \\ a=\frac{(36\frac{m}{s})^2-(0\frac{m}{s})^2}{2(980m)}=0.66\frac{m}{s^2} \end{gathered}$

Next, consider that the formula for the force is:

$F=ma$

where,

m: mass of the car = 820 kg

a: deceleration of the car = 0.66m/s^2

Replace the previous values and simplify:

$F=(820kg)(0.66\frac{m}{s^2})=542.20N$

Hence, the required force to stop the car is 542.20N

4 0

1 year ago