1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
horrorfan [7]
2 years ago
8

A racecar experiences a centripetal acceleration of 36.0 m/s2 as it travels at a constant speed of 27.0 m/s along a circular arc

. What is the radius of the circle?
9.3 m

0.18 m

20 m

14 m
Physics
2 answers:
alexdok [17]2 years ago
5 0

Answer:

20.25 m

Explanation:

  • <u>Centripetal acceleration </u>is given by; the square of the velocity, divided by the radius of the circular path.

That is;

         <em><u>ac = v²/r</u></em>

<em>         </em><em><u> Where; ac = acceleration, centripetal, m/s², v is the velocity, m/s and r is the  radius, m</u></em>

Therefore;

r = v²/ac

  = 27²/36

  = 20.25 m

Hence the radius is 20.25 meters

vodka [1.7K]2 years ago
3 0

Answer: Third option.

Explanation: The equation for the centripetal acceleration in a circle is calculated as:

Ac = v^2/r

where v is the velocity of the car and r is the radius of the circle.

We know that Ac = 36 m/s^2 and v = 27 m/s, now we can replace it in the equation for Ac and find the value of r.

36 = 27^2/r

r= 27^2/36 = 20.25m

So the radius of the circle is 20.25m, the answer that is more close to it, is the third one, 20m

You might be interested in
Along the remote Racetrack Playa in Death valley, California, stones sometimes gouge out prominent trails in the desert floor, a
REY [17]

Answer:

   F = 196 N

Explanation:

For this exercise we will use Newton's second law,  we define a reference system with the x axis in the direction of movement of the stones and the y axis vertically

Y axis  

       N- W = 0

       N = mg

X axis

       F -fr = ma

In this case, they ask us for the force to keep moving, so the stones go at constant speed, which implies that the acceleration is zero.

       F- fr = 0

       F = fr

the friction force has the equation

       fr = μ N

       fr = μ mg

we substitute

        F = μ mg

let's calculate

         F = 0.80 9.8 25

         F = 196 N

8 0
3 years ago
A 7600 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.35 m/s2 and feels no appreci
ollegr [7]

Answer:

a) The rocket reaches a maximum height of 737.577 meters.

b) The rocket will come crashing down approximately 17.655 seconds after engine failure.

Explanation:

a) Let suppose that rocket accelerates uniformly in the two stages. First, rocket is accelerates due to engine and second, it is decelerated by gravity.

1st Stage - Engine

Given that initial velocity, acceleration and travelled distance are known, we determine final velocity (v), measured in meters per second, by using this kinematic equation:

v = \sqrt{v_{o}^{2} +2\cdot a\cdot \Delta s} (1)

Where:

a - Acceleration, measured in meters per square second.

\Delta s - Travelled distance, measured in meters.

v_{o} - Initial velocity, measured in meters per second.

If we know that v_{o} = 0\,\frac{m}{s}, a = 2.35\,\frac{m}{s^{2}} and \Delta s = 595\,m, the final velocity of the rocket is:

v = \sqrt{\left(0\,\frac{m}{s} \right)^{2}+2\cdot \left(2.35\,\frac{m}{s^{2}} \right)\cdot (595\,m)}

v\approx 52.882\,\frac{m}{s}

The time associated with this launch (t), measured in seconds, is:

t = \frac{v-v_{o}}{a}

t = \frac{52.882\,\frac{m}{s}-0\,\frac{m}{s}}{2.35\,\frac{m}{s} }

t = 22.503\,s

2nd Stage - Gravity

The rocket reaches its maximum height when final velocity is zero:

v^{2} = v_{o}^{2} + 2\cdot a\cdot (s-s_{o}) (2)

Where:

v_{o} - Initial speed, measured in meters per second.

v - Final speed, measured in meters per second.

a - Gravitational acceleration, measured in meters per square second.

s_{o} - Initial height, measured in meters.

s - Final height, measured in meters.

If we know that v_{o} = 52.882\,\frac{m}{s}, v = 0\,\frac{m}{s}, a = -9.807\,\frac{m}{s^{2}} and s_{o} = 595\,m, then the maximum height reached by the rocket is:

v^{2} -v_{o}^{2} = 2\cdot a\cdot (s-s_{o})

s-s_{o} = \frac{v^{2}-v_{o}^{2}}{2\cdot a}

s = s_{o} + \frac{v^{2}-v_{o}^{2}}{2\cdot a}

s = 595\,m + \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(52.882\,\frac{m}{s} \right)^{2}}{2\cdot \left(-9.807\,\frac{m}{s^{2}} \right)}

s = 737.577\,m

The rocket reaches a maximum height of 737.577 meters.

b) The time needed for the rocket to crash down to the launch pad is determined by the following kinematic equation:

s = s_{o} + v_{o}\cdot t +\frac{1}{2}\cdot a \cdot t^{2} (2)

Where:

s_{o} - Initial height, measured in meters.

s - Final height, measured in meters.

v_{o} - Initial speed, measured in meters per second.

a - Gravitational acceleration, measured in meters per square second.

t - Time, measured in seconds.

If we know that s_{o} = 595\,m, v_{o} = 52.882\,\frac{m}{s}, s = 0\,m and a = -9.807\,\frac{m}{s^{2}}, then the time needed by the rocket is:

0\,m = 595\,m + \left(52.882\,\frac{m}{s} \right)\cdot t + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot t^{2}

-4.904\cdot t^{2}+52.882\cdot t +595 = 0

Then, we solve this polynomial by Quadratic Formula:

t_{1}\approx 17.655\,s, t_{2} \approx -6.872\,s

Only the first root is solution that is physically reasonable. Hence, the rocket will come crashing down approximately 17.655 seconds after engine failure.

7 0
2 years ago
Look at the circuit diagram. Which of these components is part of the circuit?
Roman55 [17]

Answer:

b ac power source

Explanation:

ieififuffucjcjcicogore8e8e9e9e9e9ew0ww0w0w

7 0
3 years ago
Incident rays parallel to the principle axis of a concave mirror will reflect _____. parallel to the principle axis through the
Svetradugi [14.3K]
I’m not sure if its correct but I think it’s focal Ray point


For concave mirrors, some generalizations can be made to simplify ray construction. They are: An incident ray traveling parallel to the principal axis will reflect and pass through the focal point. An incident ray traveling through the focal point will reflect and travel parallel to the principal axis.
8 0
3 years ago
Read 2 more answers
The mass of a string is 7.7 × 10-3 kg, and it is stretched so that the tension in it is 190 N. A transverse wave traveling on th
Scilla [17]

Length of the strings = 2.33 m

3 0
2 years ago
Other questions:
  • What carved the sharp features of this mountain
    9·2 answers
  • A computer monitor accelerates electrons and directs them to the screen in order to create an image. If the accelerating plates
    8·1 answer
  • What is the chemical equation, for li2+h20-li0h, use the drop down menu?
    11·1 answer
  • List two fossil fuels. describe in general how fossil fuels were formed.
    9·1 answer
  • A yo-yo is made from two uniform disks, each with mass m and radius R, connected by a light axle of radius b. A light, thin stri
    12·1 answer
  • I can not find the answer to the one after potassium ?
    11·1 answer
  • (a) What is the best coefficient of performance for a refrigerator that cools an environment at −30.0ºC and has heat transfer to
    14·1 answer
  • A capacitor is made from two hollow, coaxial, iron cylinders, one inside the other. The inner cylinder is negatively charged and
    11·1 answer
  • What is the wavelength of an FM radio
    6·1 answer
  • Easy point...
    12·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!