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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 rocket moves upward, starting from rest with an acceleration of 25.4 m/s^2 for 3.39 s. It runs out of fuel at the end of the 3

kolbaska11 [484]

Explanation:

Initial speed of the rocket, u = 0

Acceleration of the rocket, $a=25.4\ m/s^2$

Time taken, t = 3.39 s

Let v is the final velocity of the rocket when it runs out of fuels. Using the equation of kinematics as :

$v=u+at$

$v=25.4\times 3.39=86.10\ m/s$

Let x is the initial position of the rocket. Using third equation of kinematics as :

$v^2=u^2+2ax_o$

$x_o=\dfrac{v^2}{2a}$

$x_o=\dfrac{86.10^2}{2\times 25.4}=145.92\ m$

Let $x_o$ is the position at the maximum height. Again using equation of motion as :

$v^2-u^2=2a(x-x_o)$

Now $a=-g$ and v and u will interchange

$u^2=2g(x-x_o)$

$x=x_o+\dfrac{u^2}{2g}$

$x=145.92+\dfrac{(86.10)^2}{2\times 9.8}$

x = 524.14 meters

Hence, this is the required solution.

5 0

3 years ago

I WILL GIVE BRAINLIEST Which of the following statements is true with regard to transverse and longitudinal waves?

pishuonlain [190]

Answer: Transverse waves have motion perpendicular to velocity, while longitudinal waves have motion parallel to velocity.

Explanation:

Transverse waves are characterized by the fact that the particles of the medium in which they propagate move transversely to the direction of propagation of the wave.

In other words,<u> its displacement is perpendicular to the direction of propagation of the wave</u>, being a good example the circular waves in the water.

On the other hand, Longitudinal waves are characterized by the fact that <u>the oscillation of the particles in the medium is parallel to the direction of propagation of the wave.</u> A good example of this is the sound wave.

8 0

3 years ago

What change happens when matter<br> changes states?

Snowcat [4.5K]

Matter either loses or absorbs energy when it changes from one state to another. For example, when matter changes from a liquid to a solid, it loses energy. The opposite happens when matter changes from a solid to a liquid. For a solid to change to a liquid, matter must absorb energy from its surroundings.

4 0

3 years ago

On Earth, 1 kg = 9.8 N = 2.2 lbs. On the Moon, 1 kg = 1.6 N = 0.37 lbs. Use these relationships to answer the following question

romanna [79]

Answer:

(a) 490 N on earth

(b) 80 N on earth

(c) 45.4545 kg on earth

(d) 270.27 kg on moon

Explanation:

We have given 1 kg = 9.8 N = 2.2 lbs on earth

And 1 kg = 1.6 N = 0.37 lbs on moon

(a) We have given mass of the person m = 50 kg

As it is given that 1 kg = 9.8 N

So 50 kg = 50×9.8 =490 N

(b) Mass of the person on moon = 50 kg

As it is given that on moon 1 kg = 1.6 N

So 50 kg = 50×1.6 = 80 N

(c) We have given that weight of the person on the earth = 100 lbs

As it is given that 1 kg = 2.2 lbs on earth

So 100 lbs = 45.4545 kg

(d) We have given weight of the person on moon = 100 lbs

As it is given that 1 kg = 0.37 lbs

So 100 lbs $\frac{100}{0.37}=270.27kg$

8 0

3 years ago

Consider two children sitting on a merry-go-round, with one closer to the outer edge and one closer to the center. show answer N

dolphi86 [110]

Answer:

They both have the same angular speed.

Explanation:

The mathematical formula for angular speed is:

$w=\frac{2\pi}{T}$

where $w$ is angular speed, $2\pi$ is a constant, and $T$ is the period (the time it takes the marry-go-round to complete a lap).

What we can see from the formula is that, since the $2\pi$ does not change its value, the angular speed depends only on the period T.

In this case for both the children closer to the outher edge and for the children closer to the center, the time to complete a lap is the same, because the time does not depend on where they are sitting in the marry go round. This means that the period for both is the same.

Thus, since the period for both is the same, the angular speed given by

$w=\frac{2\pi}{T}$ will also be the same

4 0

3 years ago